QUADRATIC FUNCTIONS: Problem Set 1
NOTES:
The motion of projectiles can be represented by a quadratic equation. Here are two examples of this kind of motion:
- A Ball being thrown upwards into the air.
- An aircraft dropping a watermelon.
***
You are going to plot a graph to show the motion of a cannonball being fired into the air at an angle of 45 degrees.
The equation for the motion is:
2
y = ax + bx
Where a = -1/500 and b = 1.
2
In the equation y = (-1/500)x + x, y represents the distance in meters and x is the time in seconds.
The equation describing the motion of a projectile is a quadratic equation.
The graph of a quadratic equation is a parabola.
The value of y rises to a maximum and then falls back to zero.
WORKSPACE:
ANS:
45 , 80 , 105 , 120 , 125 , 120 , 105 , 80 , 45 , 0
SOLVING QUADRATIC EQUATIONS: Factoring
- Solve quadratic equations by factoring and using the zero-product property
NOTES:
In past activities we factored and graphed equations of the form:
2
y = ax + bx + c
This activity focuses on the particular case when y = 0. That is, we will learn to solve equations of the form:
2
0 = ax + bx + c
To do this, we will use factoring.
***
Let's see how to graph the following equations:
2
y = x - 5x - 14
2
x|y y = (-3) - 5(-3) - 14
--+-- 9 + 15 - 14 = 10
-3|10
-2|0
0|-14
***
Before pressing ENTER, describe in your own words the points where the graph crosses the x-axis.
x|y
--+--
-3|4
-2|0
0|-2
1|0
The points where the graph crosses the x-axis are the points where y = 0.
***
Before pressing ENTER, describe in your own words the solutions to equations like 0 = x^2 - 7x + 12.
The solutions to the equation are the x-values of the points where the graph crosses the x-axis.
***
It is not always practical to graph an equation to determine the solutions when y = 0.
In the next part of the lesson, algebraic procedures will be used to find solutions to equations like:
2
ax + bx + c = 0
Without graphing.
First, let's review a few facts that will be needed.
***
Complete the folling products:
3 * _ = 0
_ * 5 = 0
? * ? = 0
Notice that the last product is correct for any pair of numbers as long as one of the numbers is zero.
***
Before pressing ENTER, describe in your own words when:
a * b = 0
Either a or b (or both) must be zero. This is known as the zero-product property.
It also will be true that:
(x - 3)(x + 5) = 0
If either:
x - 3 = 0 OR x + 5 = 0
So x = 3 or x = -5.
***
Before pressing ENTER, describe in your own words when:
(ax + b)(cx + d) = 0
It will be true if either:
ax + b = 0 or cx + d = 0
***
Notice what happens when we factor the following equation:
2
Start with... 3x - x = 0
Then factor... x(3x - 1) = 0
Then either... x = 0 OR 3x - 1 = 0
So either... x = 0 OR x = 1/3
Here are some more examples of solving equations:
2
4x - 9 = 0
(2x + 3)(2x - 3) = 0
x = -3/2 OR 3/2
And:
3 2
3x - 6x - 24x = 0
2
3x(x - 2x - 8) = 0
3x(x - 4)(x + 2) = 0
x = 0, 4 or -2
***
Before pressing ENTER, describe in your own words a way to solve equations of the form:
2
ax + bx + c = 0
Factor the equation, set each factor equal to zero and solve the resulting equations.
WORKSPACE:
2
8x + 5x = 0
x(8x + 5) = 0
0 OR -5/8
***
2
x + 8x + 16 = 0
-4
***
2
x - 2x - 15 = 0
-3, 5
ANS:
0 , -14 , -18 , -18 , -14 , 0 , 4 , 0 , -2 , -2 , 0 , 4 , 10 , 0 , 0 , zero , 0 , 0 ,-9 , 5/2 , 0 , 0 , 0 , 4/3 , -3/2 , 0 -2 , -1/2 3 , 6/5 -4 , 7/3 -9/2 , 0 -2 5 , -3/2 5/4 , 0 -5/8 , -5 5 , -4 , 6 , -3 5 , 1 -1 , 3/2 -9 , ... , 7 3 , -9/5 -1 , 1/3 7 , -4 5 , 1/4 -1/4 , 8 2 , 10/3 -10/3
SOLVING QUADRATIC EQUATIONS: Completing the Square
2
- Solve equations of the form a(x + b) = c.
- Solve quadratic equations by completing the square.
NOTES:
In the last activity we solved quadratic equations by factoring.
2
x + 7x + 12 = 0
(x + 3)(x + 4) = 0
However, not all equations are easily factored. We need some new methods to help us solve quadratics.
The first method we will learn is called COMPLETING THE SQUARE.
***
Sometimes our solutions will involve irrational numbers like:
_ _ _
\/5 2\/3 3 + \/7
Let's see how we get these solutions!
Look at this equation:
2
x = 9
If we take the square root of both sides, we find the two solutions.
__
/ 2 _
\/ x = \/9
|X| = 3
x = 3, -3
Here are some other examples:
2
x = 7
_ _
x = \/7, -\/7
And:
2
4x = 25
2 25
x = --
4
5 -5
x = -, --
2 2
***
Before pressing ENTER, describe in your own words how we solve the following quadratic equation:
2
ax = b
Divide both sides by a and then take the square root of each side:
2
ax = b
2 b
x = -
a
_ _
/b /b
/ - , - / -
x = \/ a \/ a
***
We can use this method to solve the following quadratic equations:
2
If, (x -4) = 36
Then, (x - 4) = 6 OR -6
Here are some more examples:
2
3(x - 6) = 75
2
(x - 6) = 25
(x - 6) = 5 OR -5
Here are some more examples:
2
(x - 2) = 5
_ _
(x - 2) = \/5 OR -\/5
_ _
x - 2 = \/5 OR x - 2 = -\/5
_ _
x = 2 + \/5 OR x = 2 - \/5
***
Before pressing ENTER, use your own words to describe how we solve the following quadratic equation:
2
a(x + b) = c
First we divide by a. Next take the square root of both sides. Then we solve the resulting equations.
2
a(x + b) = c
2 c
(x + b) = -
a
_ _
/c /c
/ - , - / -
(x + b) = \/ a \/ a
_ _
/c /c
/ - OR / -
x = -b + \/ a x = -b - \/ a
The solution steps we have been using don't always work well because some equations like:
2
ax + bx + c = 0
Are not easy to factor.
To solve any equations of this form we will use two facts about PERFECT SQUARE TRINOMIALS.
2 2 2
(x + b) = x + 2bx + b
2 2 2
(x - b) = x - 2bx + b
Notice how we solve the following equation:
2
x + 6x + 7 = 0
Since we can't factor this equation let's change it to a Perfect Square Trinomial which we can factor...
2
x + 6x + 7 = 0
[ 2
/ x + 6x = -7
|
\ 2
[ x + 6x + 9 = -7 + 9
2
(x + 3) = 2
_ _
(x + 3) = \/2 OR -\/2
_ _
x + 3 = \/2 OR x + 3 = -\/2
_ _
x = -3 + \/2 OR x = -3 - \/2
The procedure in brackets is a process called Completing the Square.
Here are some more examples of completing the square:
2
x - 8x + 4 = 0
2
x - 8x = -4
2
x - 8x + 16 = -4 + 16
2
(x - 4) = 12
_ _
x - 4 = 2\/3 OR -2\/3
_
x = 4 +/- 2\/3
And:
2
x + 7x + 2 = 0
2
x + 7x = -2
2 49 -8 49
x + 7x + -- = -- + --
4 4 4
2 41
(x + 7/2) = --
4
WORKSPACE:
2
16x = 9
x = 3/4, -3/4
"This is correct because 16(3/4)(3/4) = 9 and 16(-3/4)(-3/4) = 9.
***
ANS:
4 -4 , 10 -10 , 3\/2 -3\/2 , 2/3 -2/3 , 3/4 -3/4 , 5/7 -5/7 , \/3 -\/3 , 10 , -2 , 11 , 1 , 12 -2 , -6 + \/11 -6 - \/11 , -2 + \/3 -2 - \/3 , 5 -4 , 3 + \/7 3 - \/7 , 5 + \/7 5 - \/7 , 7/3 -5 , 1 + \/3 1 - \/3
The below is continued. Current Date: February 28, 2006
NOTES:
Here are some more examples of completing the square:
2
x - 8x + 4 = 0
2
x - 8x = -4
2
x - 8x + 16 = -4 + 16
2
(x - 4) = 12
_ _
x - 4 = 2\/3 OR -2 \/3
_
x = 4 +/- 2\/3
And:
2
x + 7x + 2 = 0
2
x + 7x = -2
2 49 -8 49
x + 7x + -- = -- + --
4 4 4
2 41
(x + 7/2) = --
4
__ __
\/41 \/41
x + 7/2 = ---- OR - ----
2 2
__
-7 +/- \/41
x = -----------
2
***
Before pressing ENTER, describe in your own words one way to solve polynomials of this form:
2
x + bx + c = 0
Complete the square on the left side:
2 2
2 b b
x + bx + -- = -c + --
4 4
factor,
2
b 2 b
(x + -) = -c + --
2 4
Then take the square root of both sides and solve.
***
Let's see how the transformation for solving an equation by completing the square affect the graph of the equation.
2
x - (3/2)x - 1 = 0
2
x - (3/2)x = 1
2
x - (3/2)x + 9/16 = 1 + 9/16
2
(x - 3/4) = 25/16
x - 3/4 = +/- 5/4
x - 3/4 = 5/4 OR x - 3/4 = - 5/4
x = 2 x = -1/2
Here is another graphic example of solving equations:
2
x + 2x - 1 = 0
2
x + 2x = 1
2
x + 2x + 1 = 1 + 1
2
(x + 1) = 2
_
x + 1 = +/- \/2
_ _
x + 1 = \/2 OR x + 1 = -\/2
_ _
x = -1 + \/2 x = -1 - \/2
~ ~
x = .41 x = -2.41
***
Before pressing ENTER, describe in your own words how transformations of equations affect the solutions to the equations.
As we have seen, the solutions are not affected, but always lie on the same lines.
WORKSPACE:
2
x - 4x + 1 = 0
2
x - 4x = -1
2
x - 4x + 4 = -1 + 4
2
(x - 2) = 3
_ _
x - 2 = \/3 OR \/3
_ _
x = 2 + \/3, 2 - \/3
*Why do they pick that certain number for completing the square? I'm not sure... I'll have to figure that out*
***
2
x + 2x - 1 = 0
2
x + 2x = 1
2
x + 2x + 1 = 1 + 1
2
(x + 1) = 2
_ _
x + 1 = \/2 OR -\/2
_
x = -1 + \/2
***
2
x + 3x - 8 = 0
2
x + 3x = -8
2
x + 3x + 9/4 = 41/4
2
(x + 3/2) = 41/4
__ __
x + 3/2 = \/41/2 OR -\/41/2
__ __
x = (-3 + \/41)/2, (-3 - \/41)/2
ANS:
2 + \/3 2 - \/3 , -1 + \/2 -1 - \/2 , (-3 + \/41)/2, (-3 - \/41)/2 , 9 -9 , 3/7 -3/7 , 2\/2 -2\/2 , 11 3 , -2 -8 , 9/2 -3/2 , -3 + \/2 -3 - \/2 , 4 + \/7 4 - \/7 , (-3 + \/33)/2 (-3 - \/33)/2
SOLVING QUADRATIC EQUATIONS: The Quadratic Formula
2
- Solve equations of the form ax + bx + c = 0 using the quadratic formula.
NOTES:
In the last activity we learned the method of completing the square which can use to solve all quadratic equations of the form:
2
ax + bx + c = 0
In this activity we will learn a "shortcut" for solving these same quadratic equations. This shortcut is the Quadratic Formula.
***
Here is a quadratic equation:
2
3x + 2x + 1 = 0
Written in the form:
2
ax + bx + c = 0
In The below equation,
a = 3
b = 2
c = 1
***
2
ax + bx + c = 0
Determine the coefficients a, b, and c in the following quadratic equation:
2
4x + 6x + 2 = 0 a = 4, b = 6, c = 2
***
Using algebraic transformations, and completing the square, we can solve the following quadratic equation for x:
2 /
ax + bx + c = 0 a = 0
The result is:
_______
/ 2
-b +/- \/ b - 4ac /
x = ----------------- a = 0
2a
This is a formula for finding the solution of all quadratic equations. It is called the quadratic formula. We will derive this formula in Algebra II.
***
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
The way we read the quadratic formula is:
"x equals the opposite of b plus or minus the square root of b squared minus 4ac, all divided by 2a."
***
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
Let's see how the quadratic formula works.
2
x - 3x + 2 = 0
a = 1
b = -3
c = 2
_____________
/ 2
-(-3) +/- \/(-3) - 4(1)(2)
x = --------------------------
2(1)
_
3 +/- \/1
= ---------
2
3 +/- 1
= -------
2
So,
3 + 1 3 - 1
x = ----- OR ----- = 2 OR 1
2 2
***
Let's solve the same equation by completing the square.
2
x - 3x + 2 = 0
2
x - 3x = -2
2
x - 3x + (9/4) = -2 + (9/4)
2
[x - (3/2)] = 1/4
x - (3/2) = +/- 1/2
x - (3/2) = 1/2 OR x - (3/2) = -1/2
x = 4/2 = 2 x = 2/2 = 1
Notice that we get the same answers.
*Huh? They look different to me...*
***
Let's do an equation together...
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
2
2x + 6x + 4 = 0
a = 2
b = 6
c = 4
____________
/ 2
-(6) +/- \/(6) - 4(2)(4)
x = ------------------------
2(2)
_
-6 +/- \/4
x = ----------
4
-6 + 2 -6 - 2
x = ------ OR ------
4 4
x = -1 OR -2
***
Before pressing ENTER, use your own words to describe how we use the quadratic formula to solve equations like:
2
x - 5x + 9 = 0
Substitute 1, -5, and 9 for a, b, c in the formula, then simplify.
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
***
WORKSPACE:
a = 1
b = 6
c = -16
_______
/ 2
-6 +/- \/ 6 - 4(-16)
x = --------------------
2
___
-6 +/- \/100
------------
2
-6 +/- 10
---------
2
2 OR -8
***
a = 1
b = 7
c = -3
______
/ 2
-7 +/- \/ 7 + 12
x = ----------------
2
__
-7 +/- \/61
-----------
2
***
a = 3
b = 35
c = 50
________
/ 2
-35 +/- \/ 35 - 600
x = -------------------
6
35 +/- 25
---------
6
***
a = 5
b = 1
c = -18
_______
/ 2
-1 +/- \/ 1 + 360
x = -----------------
10
-1 +/- 19
---------
10
9/5 OR -2
***
a = 1
b = 11
c = 24
_______
/ 2
11 +/- \/ 11 - 96
x = -----------------
2
11 +/- 5
--------
2
3 OR 8
***
a = 6
b = 17
c = 7
________
/ 2
-17 +/- \/ 17 - 168
x = -------------------
12
-17 +/- 11
----------
12
-1/2 OR -7/3
***
a = 5
b = 29
c = -6
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
ANS:
4 6 2 , 3 -2 5 , 2 1/2 3 , 1 -4 -7 , 1 2 -4 , 2 6 4 , 6 , 6 , 2 , 4 , 2 , 4 , 2 , 2 , -1 , -2 , 1 -6 , 2 -8 , -1/2 1/3 , -3 + \/5 -3 - \/5 , 5/2 -7/3 , (-7 + \/61)/2 (-7 - \/61)/2 , 5/3 10 , 9/5 -2 , 3 8 , -1/2 -7/3
The below is continued. Current Date: March 6, 2006
NOTES:
This is the format:
2
ax + bx + c = 0
_______
/ 2
-b +/- \/ b - 4ac
x = -----------------
2a
WORKSPACE:
a = 5
b = 29
c = -6
________
/ 2
-29 +/- \/ 29 + 120
x = -------------------
10
-29 +/- 31
----------
10
***
a = 2
b = 19
c = 35
________
/ 2
-19 +/- \/ 19 - 280
x = -------------------
4
-19 +/- 9
---------
4
***
a = 1
b = -7
c = 12
_______
/ 2
7 +/- \/ -7 - 48
x = ----------------
2
7 +/- 1
-------
2
***
a = 28
b = -1
c = -15
_________
/ 2
1 +/- \/ -1 + 1680
x = -------------------
56
1 +/- 41
--------
56
***
a = 9
b = -22
c = 8
_________
/ 2
22 +/- \/ -22 - 288
x = -------------------
18
22 +/- 14
---------
18
***
a = 8
b = 25
c = 3
________
/ 2
-25 +/- \/ 25 - 96
x = ------------------
16
-25 +/- 23
----------
16
***
ANS:
1/5 -6 , -5/2 -7 , 4 3 , -5/7 3/4 , 2 4/9 , -1/8 -3
QUADRATIC FUNCTIONS: Problem Set 2
NOTES:
In Problem Set 1 we graphed points belonging to the quadratic equations that described the motion of a cannonball fired into the air.
2
y = (-1/500)x + x
In this problem imagine that the cannonball is being fired on Jupiter where the gravity is about 2.5 times that of the Earth.
As you might expect, gravity affects the flight of the cannonball.
The equations for the motion of the cannonball on Jupiter is:
2
y = (-1/200)x + x
Plot some points for this equation and later we'll compare the graphs for the effects of gravity.
WORKSPACE:
ANS:
0 , 22 , 38 , 47 , 50 , 47 , 38 , 22 , 0
QUADRATIC FUNCTIONS: The Distance Formula
- Excursion
NOTES:
Suppose one day you find an old map that looks like this:
--+----------+
R | |
| a b |
I| c |
| |
V | d |
| e |
E| |
|f |
R | g |
-----R---A----
-------O---D--
Written on the map is the following description:
"In a field 10 miles by 10 miles square,
Are found 7 trees, A-B-C-D-E-F-G
If you please.
If you're smart,
Find two trees 5 miles apart.
Halfway between these two
Is a treasure buried for you."
Between which two trees will you find the buried treasure?
At the end of this lesson we will return to this problem and you will be able to find the buried treasure!
***
In a previous lesson you learned about the PYTHAGOREAN THEOREM and some of its applications.
One of the most important applications of the Pythagorean Theorem is finding the DISTANCE between any two points in a plane.
The distance between any two points is always a positive number.
For example, to find the distance between these two points on the number line, you must compute the absolute value of the difference in their coordinates.
The distance between two points will always be the same.
This is true whether you go from point 1 to point 2 or from point 2 to point 1.
***
If two points are on the same horizontal line, the distance between them is the absolute value of the difference between their x-coordinates.
D = |10 - 2| = 8
If two points are on the same vertical line, the distance between them is the absolute value of the difference between their y-coordinates.
D = |5 - 2| = 3
***
But what if two points are NOT on the same vertical or horizontal line?
Now we have to find another way to determine the distance between the two points.
Let's add a VERTICAL line to the graph.
Now add a HORIZONTAL line to the graph.
You have made a RIGHT triangle.
The process of finding the lengths of the vertical and horizontal sides of a right triangle is similar to the process of finding the slope of a line, a skill that you learned earlier.
To calculate the slope of a line, you found the change in y-coordinates and divided it by the change in x-coordinates.
You can use the same information here to find the length of the third side of the triangle.
Tx = x - x Ty = y - y
2 1 2 1
a = |Tx| b = |Tx|
a - |x - x| b = |y - y|
2 1 2 1
Let side a equal 3 and side b equal 6, then solve for c:
2 2 2
c = a + b
2 2 2
c = 3 + 6
2
c = 9 + 36
2
c = 45
__ _
c = \/45 OR 3\/5
[This is the distance from (2,5) to (8,2)]
***
Let's review the procedure for finding the distance between any two points.
We will call these points point 1 (x^1,y^1) and point 2 (x^2,y^2) (Do not square, these are just labels!).
But remember that it doesn't matter which point is chosen as point 1 or point 2.
- First, make a right triangle.
- Measure length of the vertical and horizontal sides.
Now you can use the Pythagorean Theorem to find the length of the hypotenuse of this triangle.
Let a = |x - x| and b = |y - y|
2 1 2 1
Then solve for c:
2 2 2
c = |x - x| + |y - y|
2 1 2 1
Since any number squared is positive, the absolute value signs are not needed.
2 2 2
c = (x - x) + (y - y)
2 1 2 1
Finally, this is the result when we solve for c.
_________________
/ 2 2
c = \/(x - x) + (y - y)
2 1 2 1
This is the length of the third side of the triangle (the hypotenuse).
It is also the distance from (x^1,y^1) to (x^2,y^2).
_________________
/ 2 2
c = \/(x - x) + (y - y)
2 1 2 1
If we let D represent the distance between any two points (x^1,y^1) and (x^2,y^2) on a coordinate plane then we have the following equation:
_________________
/ 2 2
D = \/(x - x) + (y - y)
2 1 2 1
This equation is known as the Distance Formula.
The distance is the same no matter which point you start with.
***
To find the distance from the point (4,-2) to (-5,3), let:
x = 4 x = -5
1 2
y = -2 y = 3
1 2
Use these values in the Distance Formula.
_________________
/ 2 2
D = \/(x - x) + (y - y)
2 1 2 1
____________________
/ 2 2
D = \/(-5 - 4) + (3- (-2))
__________
/ 2 2
D = \/(-9) + (5)
_______
D = \/81 + 25
___
D = \/106
***
At the beginning of the lesson you were examining an old map.
To locate the buried treasure, you must find two trees that are 5 miles apart.
Tree A. (4,8)
Tree B. (8,8)
Tree C. (2,7)
Tree D. (5,5)
Tree E. (9,4)
Tree F. (1,2)
Tree G. (7,1)
Now you can use the Distance Formula to find the distance between any two trees.
Which two trees are 5 miles apart? Tree D and F.
X marks the spot halfway between the two trees where treasure is buried. HAPPY DIGGING!!!
*Editor's Note: Hehe, :-) ... O_o ... ;-)*
WORKSPACE:
ANS:
11 , 8 , 12 , 7 , 10 , 7 , 3 , 6 , \/40 , #3 , #1 , 5 , 13 , 10 , 4,8 , 8,8 , D F
QUADRATIC EQUATIONS AND THE DISTANCE FORMULA II: Test
This lesson will evaluate your skills in the solutions to quadratic equations and using the distance formula.
NOTES:
WORKSPACE:
ANS:
a , a , NOT:a,d , b , d , c , d , a , c , b
DIVINDING POLYNOMIALS: Long Division
- Divide polynomials using long division.
NOTES:
In our unit on polynomials we learned how to factor some polynomails by "guess and test."
In this topic we are going to investigate another method for determining factors of polynomials. That method is the method of long division.
We say that 13 is a factor of 32513 if 13 divides into 32,513 evenly.
32513/13 = 2501
Since the remainder is zero, 13 is a factor of 32,513.
We can use the method of long division to determine if 3K + 1 is a factor of 3k^2 + 4k + 1. Notice how it is similar to dividing 341 by 31.
1k +1
______________
3k + 1 )3k^2 + 4k + 1
-(3k^2 + k)
--------
3k + 1
-(3k + 1)
------
0
11
____
31 )341
-(31)
--
31
-(31)
--
0
***
Before pressing ENTER, describe in your own words one way to tell if "polynomial a" is a factor of "polynomial b."
Divide "polynomial b" by "polynomial a" using long division. If the remainder is zero, then "a" is a factor of "b."
***
WORKSPACE:
ANS:
k , 2k^2 + 5k , 4 , 8k + 20 , 3k^2 , 3k^3 + 12k^2 , -2k , -2k^2 - 8k , -5 , -5k - 20 , 2x , 10x^2 + 8x , -3 , -15x - 12 , x^2 , x^3 - 4x^2 , 2x^2 - 7x , 2x , 2x^2 - 8x , x - 4 , +1 , x - 4 , 0 , 3x^2 , 3x^3 + 6x^2 , -5x^2 - 3x , -5x , -5x^2 - 10x , 7x + 8 , +7 , 7x + 14 , -6 , 3x^2 , 6x^3 - 12x^2 , 12x^2 - 38x , +6x , 12x^2 - 24x , -14x + 28 , -7 , -14x + 28 , 0 , 3x^2 , 15x^3 - 6x^2 , -35x^2 - x , -7x , -35x^2 + 14x , -15x + 6 , -3 , -15x + 6 , 0 , 2x^2 , 6x^3 + 12x^2 , x^2 - 10x , +(1/3)x , x^2 + 2x , -12x - 4 , -4 , -12x - 24 , 20 , x^2 + 3x + 4 , 0 , 3x^3 - 2x - 6 , 7 , 5x^2 + 2x - 9 , 26 , 2x^2 + (1/2)x - 8 , 0 , 7x^3 + 5x^2 + 6x + 1 , 5
The below is continued (some parts above are from this date, though). Current Date: March 14, 2006.
DIVIDING POLYNOMIALS: Exponents
- Simplify expressions of the form a^m / a^n.
- Review the meaning of zero and negative exponents.
NOTES:
In this topic we will learn how to simplify expressions like these:
7 3 3
a a a
-- -- --
3 7 3
a a a
Let's see how we simplify the quotient of expressions with the same base.
***
Write a general rule for the quotient of two expressions with the same base, where the exponent in the numerator is greater than the expeonent in the denominator.
m - n
***
Write a general rule for the quotient of two expressions with the same base, where the exponent in the denominator is greater than the exponent in the numerator (REVERSED).
n - m
***
Here is another rule for simplifying expressions like a^m / a^n where n > m.
We subtract the exponent in the denominator from the exponent in the numerator.
5
x 5-10 -5
--- = x = x
10
x
Negative exponents can be used to simplify expressions like x^5 / x^10.
***
8
x 8-14 -6
--- = x = x
14
x
Negative exponents can be used to simplify expressions like x^8 / x^14.
We know that:
5
x 1 1
--- = ----- = --
10 10-5 5
x x x
And:
5
x 5-10 -5
--- = x = x
10
x
So:
-5 1
x = --
5
x
***
Notice how we simplify the following expressions.
5 6
x y 2 5
---- = x y
3
x y
3 5 3
a b -1 3 b
---- = a b = --
4 2 a
a b
***
Before pressing ENTER, describe in your own words how we simplify expressions like:
7
a b
----
4 2
a b
Subtract the exponents of corresponding variables and then simplify.
WORKSPACE:
ANS:
a*a*a , a^2 , k*k*k , k^5 , a^4 , b^6 , x , y^7 , m-n , a*a*a*a*a*a*a , 4 , k*k*k*k*k*k , 2 , 1 a^3 , 1 b^5 , 1 x , 1 y^8 , n-m , x^-1 , x^-4 , x^-5 , x^-3 , m-n , x^-6 , 3^-4 , d^-2 , r^-7 , x^-8 , -n , n t^2 , z^5 y , 1 n^2k^2 , g^4h , n m^2 , 1 , 2 , 4 , 3 , 1 , 3 , 9 , 1 , 1 , r^3y , x^-5 , 1 , n^3m^-6 , x^-1 , y^-3 , p^2k^-4 , x^9 , 1
RATIONAL EXPRESSIONS: Problem Set 1
NOTES:
Problem 1:
A 1979 television commericial for Timex watches contained the statement that the watches were 99.97% accurate. In a day, how much time might be lost or gained by a watch with that accuracy? Round to the nearest second.
ANS: 26 seconds
***
In 1983 an economist stated that all of the world's population could easily fit in the state of Texas, with plenty of room for homes and yards. The world's population is 4.7 billion. The area of texas is 267,300 square miles. How many square feet would there be for each person?
3
ANS: 1.585 * 10 sq ft/person
(OR: 1,585 sq ft/person)
***
A 30-foot by 60-foot ballroom is to be turned into a discotheque by placing an 18-foot by 42-foot dance floor in the middle and carpeting the remainder. What percent of the floor is dance area? Round the nearest percent.
ANS: 42%
WORKSPACE:
ANS:
26 , 1.585 * 10^3 , 42
POLYNOMIALS AND EXPONENTIAL OPERATIONS II: Test
This lesson will evaluate your skills in dividing polynomials and simplifying exponential expressions.
NOTES:
WORKSPACE:
ANS:
*Editor's Note: Made some mistakes, taken out*
OPERATIONS ON RATIONAL EXPRESSIONS: Simplifying Rational Expressions
- Simplify rational expressions by canceling common factors
- Analyze graphs of rational functions
NOTES:
In this activity we will simplify rational expressions like:
2
x + 8x + 7
----------
2
x - 2x - 3
By finding common factors and canceling. We also will investigate how this affects the graph of the expression.
***
Let's talk about equivalence.
Sometimes things that are equivalent under one condition are not necessarily equivalent under another condition.
Let's look at an example...
Two halves are not always equivalent to one whole.
Suppose we are given these two halves of a coat.
*Picture of left and right side of coat torn in half*
We can say that the sum of the halves is equivalent to one whole coat.
However, suppose we are given these two halves.
*Picture of two left sides ^_^ *
We cannot say that the sum of these halves is equivalent to one whole coat.
So, under some conditions we can say that two halves are equivalent to one whole, and other conditions they are not.
***
A similar case arises in simplifying rational expressions.
2x 2(x) 2
------ = -------- = -----
2 x(x - 2) x - 2
x - 2x
According to the example above,
2x
------ should be equivalent to,
2
x - 2x
2
-----
x - 2
We will see that for some values of x the expressions are equivalent, but this is not true for ALL values of x.
*Shows picture of a graph impossible to reproduce*
These graphs are equivalent EXCEPT at x = 0. For the graph on the left there is a hole at x = 0, and for the one on the right there is not.
If we want the graph on the right to be equivalent to the one on the left we RESTRICT the expression.
2 /
----- x = 0
x - 2
By excluding the x-value where the hole occurs, our graph on the left and our graph on the right become equivalent.
***
Let's look at some more examples:
x + 1 x + 1 1
----- = -------------- = -----
2 (x + 1)(x - 1) x - 1
x - 1
Here are the graphs of these expressions.
*Sigh, can't show pictures...*
The hole in the left graph indicates that in order for these two expressions to be equivalent:
/
x = -1
***
Let's simplify this expression:
2
x + 5x + 6 (x + 2)(x + 3) x + 3
----------- = -------------- = -----
2 (x + 2)(x - 5) x - 5
x - 3x - 10
How must we restrict
x + 3
-----
x - 5
so that it is equivalent too
2
x + 5x + 6 /
----------- ? x = -2
2
x - 3x - 10
***
Before pressing ENTER, descirbe in your own words one way to make the simplified form of an expression equivalent to the original expression.
Look at the graphs of both expressions. If there are "holes" in the original graph, restrict the simplified expression at the x-value of the holes.
***
We can find the "holes" of a graph WITHOUT graphing!
*Yay!*
x - 3 x - 3
----------- = --------------
2 (x - 3)(x + 7)
x + 4x - 21
This expression has a hole in its graph at x = 3.
2
x - 2x - 8 (x + 2)(x - 4)
---------- = --------------
2 (x + 1)(x - 4)
x - 3x - 4
This expression has a hole in its graph at x = 4.
2
x - 4 (x - 2)(x + 2)
------ = --------------
3 x(x - 2)(x + 2)
x - 4x
This expression has holes in its graph at x = 2 and x = -2.
***
At what x-values are there "holes" in the graphs of the following expressions?
x + 2
---------
2
x + x - 2
There is a hole at x = ... -2.
*Next time factor it, sigh.*
***
Before pressing ENTER, describe in your own words one way to determine the values for the holes in the graph of an expression.
The holes occur at the x-values which make both the numerator and the denominator zero. This occurs because there are common factors in the numerator and the denominator.
***
In addition to restricting a simplified expression at the x-values where there are holes, we must also restrict it at those x-values that make it undefined.
x + 4 x + 4 1
---------- = -------------- = -----
2 (x + 4)(x - 3) x - 3
x + x - 12
/
x = -4
Notice: If x = 3, the expression is undefined since the denominator would be zero.
2x - 8 2(x - 4) 2
------ = -------------- = -----
2 (x + 4)(x - 4) x + 4
x - 16
/
x = 4, -4
***
Before pressing ENTER, describe in your own words the restricted values of a simplified rational expression.
The restricted values of the simplified expression are the values of x which make the denominator of the original expression zero.
WORKSPACE:
ANS:
-1 , x + 3 x - 5 , -2 , -2 , 1 , -3 , -2 , 1 , -3 , 4 , -5 , 3 x - 2 , -2 2 , x + 3 , 5 , x - 1 x + 2 , -6 -2 , 2x - 2 x + 8 , 3 -8 , x + 2 x - 1 , 4 1 , x + 5 x + 2 , -2 5 , 1 2x^3 , -3 -1 , x + 3 x + 1 , -3 -1 , x , 0 3 , x - 1 x + 9 , -1 -9 , x + 2 x - 2 , 2 4 , x + 4 x - 4 , 4 -5 , x + 1 x - 5 , 5 6 , x + 7 x^3 , 0 -2 , x - 8 x - 7 , 7 -7
The below is continued (some parts above are from this date, though). Current Date: March 15, 2006.
OPERATIONS ON RATIONAL EXPRESSIONS: Multiplication and Division
- Multiply and divide rational expressions
NOTES:
In this activity we will use what we've learned about simplifying rational expressions to help us multiply and divide rational expressions.
Remember how we multiply rational numbers.
(We multiply numerator by numerator and denominator by denominator.)
3 8 24 2
- * - = -- = -
4 9 36 3
We could simplfy this process by canceling common factors first.
3 8 2 * 3 * 4 2
- * - = --------- = -
4 9 3 * 3 * 4 3
We follow the same process for rational expressions.
x + 4 x - 3 (x + 4)(x - 3) x + 4
----- * ----- = -------------- = -----
x - 3 x + 6 (x - 3)(x + 6) x + 6
Here we multiply and cancel common factors.
- We factor first on this one.
2 2
x + 4x - 5 x + 5x - 14
---------- * -----------
2 x + 5
x + 6x - 7
Equals...
(x + 5)(x - 1) (x + 7)(x - 2)
-------------- * --------------
(x - 1)(x + 7) (x + 5)
- Then cancel common factors.
- Finally we simplify.
= x - 2
***
Before pressing ENTER, describe in your own words one way to multiply rational expressions.
Factor the numerator and denominator of each expression and "cancel" like factors before multiplying.
***
Recall how we divide rational numbers.
3 9 3 4 12 1
- / - = - * - = -- = -
8 4 8 9 72 6
/\ \\
|| \\
We first Then we simplify.
invert and
multiply.
We could simplify this process by canceling common factors first.
/ /
3 9 3 4 3 * 4 1
- / - = - * - = ------------- = -
8 4 8 9 2 * 4 * 3 * 3 6
/ / /
We follow the same process for the rational expressions.
x + 3 2x + 6 x + 3 3x + 12
----- / ------- = ----- * -------
x + 4 3x + 12 x + 4 2x + 6
\\
\\
We invert
and multiply.
/////// ///////
(x + 3) * 3(x + 4) 3
= ------------------ = - We cancel
(x + 4) * 2(x + 3) 2 to simplify.
/////// ///////
We follow the same process for the rational expressions.
2
x - 2 x + 3x - 10
---------- / -----------
2 2
x + 2x - 3 x + 8x + 15
We invert
and multiply.
\\ 2
x - 2 x + 8x + 15
= ---------- * -----------
2 2
x + 2x - 3 x + 3x - 10
/////////////////////
(x - 2)(x + 5)(x + 3) 1
= ---------------------------- = -----
(x + 3)(x - 1)(x + 5)(x - 2) x - 1
/////// //////////////
- We cancel to simplify.
***
Before pressing ENTER, describe in your own words one way to divide rational expressions.
Change the quotient to a product by inverting the divisor, factor each expression completely, cancel common factors and simplify.
***
WORKSPACE:
3(x + 1) -2
-------- * -------
(x + 4) (x + 5)
*Note to self: Simplify ALL THE WAY*
***
ANS:
6 , -6x - 6 x^2 + 9x + 20 , 5 x - 4 , x + 4 x + 1 , 3x^2 - 9x - 30 2x - 2 , 1 3 , x + 2 x + 3 , x - 2 x + 4 , 3x - 15 2x - 8 , x + 3 x - 1 , x - 5 x + 4 , 9x - 18 2x - 6 , 1 1 , 1 1 , x^2 - 6x + 9 x^2 + 7x + 10 , x^2 - 7x + 6 x^2 - 4 , -x + 2 x - 1 , x^2 - 9x + 20 -2x^2 - 16x - 30 , x^2 - 2x + 1 x^2 + 4x + 3 , x + 1 x + 3 ,
The below is continued (some parts above are from this date, though). Current Date: March 16, 2006.
OPERATIONS ON RATIONAL EXPRESSIONS: Addition and Subtraction
- Add and subtract rational expressions
NOTES:
In this activity we will extend what we learned about multiplying and dividing rational expressions to help us add and subtract rational expressions.
***
Remember how we add and subtract common denominators.
3 5 8
- + - = - = 4
5 2 2
We simply add the numerators and simplify.
14 2 12 4
-- - - = -- = -
9 9 9 3
Here we subtract the numerators.
***
We follow the same process for rational expressions with common denominators.
Addition:
2x + 9 4x - 7 (2x + 9) + (4x - 7)
------ + ------ = -------------------
2x - 4 2x - 4 2x - 4
6x + 2 3x + 1
= ------ = ------
2x - 4 x - 2
Subtraction:
5x + 3 4x - 5 (5x + 3) - (4x - 5)
------ - ------ = -------------------
3x + 1 3x + 1 3x + 1
x + 8
= ------
3x + 1
***
Before pressing ENTER, describe in your own words how we add and subtract rational expressions with common denominators.
We add or subtract the numerators and leave the denominator the same. We then simplify if possible.
***
Remember how we add and subtract rational numbers with different denominators.
5 2 5 7 2 4 35 8 43
- + - = - * - + - * - = -- + -- = --
4 7 4 7 7 4 28 28 28
- We find a common denominator.
- We add the numerators.
Another:
4 5 4 5
- - - = ----- - -----
9 6 3 * 3 2 * 3
- First we find a common denominator.
4 2 5 3
= - * - - - * -
9 2 6 3
- Next we subtract the numerators.
8 15 -7
-- - -- = --
18 18 18
We follow the same process for rational expressions with different denominators.
x - 6 x + 5
----- + -----
x + 4 x - 4
- Find the common denominator.
x - 6 (x - 4) x + 5 (x + 4)
----- * ------- + ----- * -------
x + 4 (x - 4) x - 4 (x + 4)
2 2
x - 10x + 24 x + 9x + 20
= -------------- + --------------
(x + 4)(x - 4) (x - 4)(x + 4)
- Add the numerators and simplify.
2
2x - x + 44
= -----------
2
x - 16
Another:
2
x + 3x + 2 1
---------- - -----
2 x + 3
x + 4x + 3
- We find a common denominator.
2
x + 3x + 2 1 (x + 1)
-------------- - ----- * -------
(x + 3)(x + 3) x + 3 (x + 1)
2
x + 3x + 2 - (x + 1)
--------------------
(x + 3)(x + 1)
- Then we subtract and simplify.
(x + 1)(x + 1) x + 1
-------------- = -----
(x + 3)(x + 3) x + 3
***
Before pressing ENTER, describe in your own words how we add and subtract rational expressions with different denominators.
We factor each denominator to determine the factors of the least common denominator. Then we change each expression so it has the same denominator and then simplify.
***
Sometime's it is best to reduce each expression BEFORE finding a common denominator.
2
x - 9 8x
----- + ------
x + 3 2
x + 2x
(x + 3)(x - 3) 8x
= -------------- + --------
(x + 3) x(x + 2)
x - 3 8
----- + -----
1 x + 2
(x - 3)(x + 2) 8
= -------------- + -----
(x + 2) x + 2
2 2
(x - x - 6) + 8 x - x + 2
= --------------- = ---------
x + 2 x + 2
Another:
2
x - 3x - 10 x + 6
----------- - -----------
x - 5 2
x + 3x - 18
(x + 2)(x - 5) (x + 6)
= -------------- - --------------
(x - 5) (x + 6)(x - 3)
1 (x + 2)(x - 3) 1
= x + 2 - ----- = -------------- - -----
x - 3 (x - 3) x - 3
2 2
(x - x - 6) - 1 x - x - 7
= --------------- = ---------
x - 3 x - 3
***
Before pressing ENTER, describe in your own words one way to simplify a sum or difference of rational expressions.
Reduce each expression to its simplest form before finding a common denominator.
WORKSPACE:
x x - 1
----- - -----
x + 2 x + 3
x(x + 3) (x - 1)(x + 2)
-------------- - --------------
(x + 2)(x + 3) (x + 3)(x + 2)
2 2
x + 3x x + x - 2
---------- - ----------
2 2
x + 5x + 6 x + 5x + 6
***
4x 2
----- + -----------
x + 2 2
x + 7x + 10
4x 2
----- + --------------
x + 2 (x + 2)(x + 5)
4x 2
----- + --------------
x + 2 (x + 2)(x + 5)
2
4x + 20x + 2
------------
2
x + 7x + 10
***
2
3x - 4 x + 3x + 2
------ - ----------
x - 1 2
x - 1
2
3x - 4 x + 3x + 2
------ - ----------
x - 1 (x + 1)(x - 1)
2x - 6
------
x - 1
***
2
2x - 7x - 4 x + 3
----------- + -----
x - 4 x - 7
2
2x - 7x - 4 x + 3
----------- + -----
x - 4 (x - 7)(x - 4)
(2x + 1)(x - 4) (x + 3)
--------------- + -------
(x - 4) (x - 7)
(2x + 1)(x - 4) (x + 3)
--------------- + -------
(x - 4) (x - 7)
2
2x - 12x - 4
------------
x - 7
***
2 2
3x + x - 2 x + 5x - 24
---------- - -----------
2 x - 3
x + 9x + 8
(3x - 2)(x + 1) (x + 8)(x - 3)
--------------- - --------------
(x + 1)(x + 8) (x - 3)
2
-x - 13x - 66
-------------
x + 8
***
2 2
x + 6x + 13 x + x - 12
----------- - ----------
x + 5 x + 5
5x + 25 5(x + 5)
------- = -------- = 5
x + 5 (x + 5)
***
2
x - 4x + 3 3x + 12
---------- + -------
2 2
x - 3x + 2 x - 16
2
x - 4x + 6
----------
2
x - 6x + 8
***
ANS:
-x + 1 2x + 1 , x + 4 , 12x - 11 x + 4 , 6x + 2 2x - 3 , 10x - 5 3x - 1 , 2x + 2 x^2 + 5x + 6 , 2x^2 - 6x + 3 x^2 + 2x - 15 , -2x - 1 x^2 - 5x - 6 , 4x^2 + 20x + 2 x^2 + 7x + 10 , 2x - 6 x - 1 , 2x^2 - 12x - 4 x - 7 , -x^2 - 13x - 66 x + 8 , x^2 + 16x + 48 x^2 + 3x - 54 , 9x + 54 x^2 - 8x - 20 , x^2 - 3x - 27 x - 7 , 5 , 2 x^2 + 8x + 12 , x^2 - 4x + 6 x^2 - 6x + 8 , 3x^2 - 10x - 23 3x + 6 , 2x - 3 x + 8 , -x - 5 12x + 36 , 2x^2 - 4 x - 2 , x^2 - 2x - 12 x + 2 , -4x + 10 x^2 - 2x - 8 , -6x - 22 x^2 + 12x + 35
The below is continued (some parts above are from this date, though). Current Date: March 20-21, 2006.
RATIONAL EXPRESSIONS: Problem Set 2
NOTES:
Problem 3:
The 1970 census from the western states showed a population size of 34,838,243. The 1980 census for the same region showed 43,172,490.
What is the percent increase? Round your answer to the nearest percent. ANS: 24%
43,172,490 - 34,838,243 = 8,334,247
The increase is a fraction of the original population. We'll convert the fraction to a decimal and the decimal to a percent.
8,334,247
---------- = .2392 = 24%
34,838,243
WORKSPACE:
ANS:
Biology , 1/3 , 24
RATIONAL EXPRESSIONS: Number Bases
- Excursion
NOTES:
Different Bases:
Our numbering system is known as the base, ten, or decimal system. It's called that because all numbers can be represented by the combination of ten different symbols:
0 1 2 3 4 5 6 7 8 9
The group of symbols 362 represents the same thing to everyone using the base ten system. The symbols 362 stand for:
3 - 100's
6 - 10's
2 - 1's
Why don't the columns stand for fives, or twenties instead of ones, tens, and hundreds?
The reason is that the ones column represents 10^0, the tens column represents 10^1, the hundreds column 10^2, etc.
Base 5 uses only these symbols:
0 1 2 3 4
What would the group of symbols 341 (base 5) represent in our normal base ten system?
Let's first identify what each column stands for.
0
5 = 1
1
5 = 1
2
5 = 25
3
5 = 125
GRAPH:
3
(5) 125's 0 = 0
2
(5) 25's 3 = 75
1
(5) 5's 4 = 20
0
(5) 1's 1 = 1
----------------------
TOTAL: 96 (in base 10)
Now try it for 1432 (base 5):
125's 1 = 125
25's 4 = 100
5's 3 = 15
1's 2 = 2
-----------------
TOTAL: 242 (in base 10)
On to base 8...
***
We have seen that the symbols 1432 can mean different things in different bases.
1432 (base 10) = 1432
1432 (base 5) = 96
1432 (base 8) = 794
Now we'll look at a very practical numbering scheme using base two.
Base Two:
Pick a number from 1 to 15. Now answer the questions below using this table:
A | B | C | D
-----+-----+-----+-----
8 | 4 | 2 | 1
9 | 5 | 3 | 3
10 | 6 | 6 | 5
11 | 7 | 7 | 7
12 | 12 | 10 | 9
13 | 13 | 11 | 11
14 | 14 | 14 | 13
15 | 15 | 15 | 15
The trick is that the table is cleverly constructed so that your yes/no responses are changed into 1's or 0's (base 2 symbols).
If you want to learn a magic trick to impress your friends copy the table down on a card or a piece of paper.
Give the table to a friend and ask them to think of a number from 1 to 15. Then tell them that you will ask them a series of questions about their number.
To guess their number just ask them if their number is in each of the columns. If their answer is yes, add the power of two at the top of the column. If the answer is no, add nothing. You will quickly come up with their answer, almost magically!
It takes only four guesses to identify any number from 1 to 15. If you had a computer to help add the powers of 2 you could guess any number up to a million with just twenty guesses!
Now, if you can specify (guess) such large numbers so easily you could do the same thing with letters (if you gave each letter a number).
You could do the whole alphabet with 6 guesses, or a whole keyboard of letters and punctuation makrs with 7 guesses.
You could even identify every person in the United States with just 30 guesses the same way!
Another nice thing about base 2 (or binary) numbers is that it's easy to build electronic devices that use them.
If you say that when a switch is off that means 0 and when a switch is on that means 1, you can represent huge amounts of numbers or people or words with just 20 or 30 switches.
These electronic switches don't have to be as large as ordinary mechanical switches, like a light switch.
In fact, in modern computers switches have been made so small that this computer's microchip can hold hundreds of thousands of them. Each switch is either on (1) or off (0).
That's why computers can hold so much information, because they use base 2.
WORKSPACE:
ANS:
1 , 125 , 100 , 15 , 2 , 0 1 2 3 4 5 6 7 , 1 , 8 , 64 , 512 , 512 , 256 , 24 , 2 , NO , YES , YES , NO
The below is continued (some parts above are from this date, though). Current Date: March 22, 2006.
RATIONAL OPERATIONS II: Test
This lesson will evaluate your skills in performing operations with polynomials and simplifying expressions in the form of fractions.
NOTES:
WORKSPACE:
125 + 75 + 15 + 4
ANS:
a , a , c , b , d , c , c , d , 9 , 219
Now Entering: PRE-CALC. ALGEBRA II
LINEAR SYSTEMS AND MATRICES: Linear Functions and Graphs Pre-Test
This test has 10 questions, each worth 10 points. You have one chance to answer each question, so be sure the answer you choose is the answer you want graded.
NOTES:
WORKSPACE:
ANS:
(2,1) , NOT:one , (1,-2) , NOT:1 , NOT:#4 , NOT:#4 , #3 , #2 , NOT:#3 , NOT:#1
LINEAR FUNCTIONS AND GRAPHS: Forms of Equations
- Determine the slope and the intercepts of a line from its equations or graph
- Transform a linear equation in one form to an equivalent equation in standard form
NOTES:
In Algebra I you studies equations and graphs of linear functions.
In this activity you will review graphs of linear functions and write their equations in several different forms.
***
Use your own words to describe the graph and the equation of a direct variation.
The graph is a LINE through the origin.
The slope m is equal to y/x for any point P(x,y) on the line, so: y = mx.
***
The graph of a direct variation will be shifted if a constant is added to the equation.
***
The slope-intercept form of a linear equation is y = mx + b where m is the slope and b is the y-intercept.
The graph of the equation crosses the y-axis at (0,b) and the slope is:
y - b
m = -----
x
Where P(x,y) is another point on the line.
***
You used the y-intercept to determine the slope-intercept form of a linear equation.
You use both intercepts to determine the two-intercept form of the equation.
You can find the two-intercept form of the equation for this line by locating the intercepts.
Since the x-intercept is -3 and the y-intercept is 2, the equation is:
x y
-- + - = 1
-3 2
***
Sometimes it is difficult to determine the intercepts from the graph of a line.
To find the equation of the line, you can use any two points on the line and the two-point form of the equation.
y - y y - y
1 1 2
------ = ------
x - x x - x
1 1 2
To find the equation for this line, use the points (-1,8) and (3,-3) and the two-point form.
y - y y - y
1 1 2
------ = ------
x - x x - x
1 1 2
y - 8 8 + 3
----- = ------
x + 1 -1 - 3
or
y - 8 -11
----- = ---
x + 1 4
***
There are many forms for writing linear equations.
y = mx ----------------- direct variation
y = mx + b -------------- slope-intercept
x y
- + - = 1 ----------------- two-intercept
a b
y - y y - y
1 1 2
------ = ------ --------------- two-point
x - x x - x
1 1 2
Each of these forms can also be written as an equivalent equation in standard form, ax + by = c.
***
You can transform the equation y = (4/3)x - 4 from slope-intercept form to standard form.
y = (4/3)x - 4
3y = 4x - 12 (multiply by 3)
-4x + 3y = -12 (subtract 4x)
This equation is in standard form.
***
Subtracting x from both sides of the equation completes the transformation from two-point form to standard form.
y - 3 1
----- = -
x + 2 3
x + 2
y - 3 = -----
3
3y - 9 = x + 2
3y = x + 11
-x + 3y = 11
***
By performing the same operations on both sides of the equation, you can transform any form of a linear equation into an equivalent equation in standard form.
WORKSPACE:
ANS:
15 , origin , 4 , -2/3 , line , mx , 0 , -3 , b , -1 , 2 , a , 5 , -3 , multiplication , multiplication , addition , 6 , 7x , 0 , -10 , -2 , -3 , 4 and 3 , 7y , -6
LINEAR FUNCTIONS AND GRAPHS: Algebraic and Geometric Transformations
- Build the graph of a linear equation by adding the graphs of its terms
- Transform a linear equation algebraically and geometrically
NOTES:
In the last activity you studies linear equations and their graphs.
Algebraically you can build a linear equation by adding terms.
y = x + 3
Geometrically, you can build the graph of this linear function by adding the graphs of its terms.
Watch as the graphs y and y = 3 are added
1 2
together.
At x = 1, 3 + 1 = 4.
At x = 2, 3 + 2 = 5.
At x = 3, 3 + 3 = 6.
At x = -1, 3+(-1) = 2.
At x = -2, 3+(-2) = 1.
At x = -3, 3+(-3) = 0.
The equation of the sum of y = x and y
1 2
= 3 is y = x + 3.
Here are the graphs of y = x and y = -4.
1 2
At x = -1, the value for y is -1 and the
1
value for y is -4.
2
y + y = -1+(-4)
1 2
= -5
***
You have added the graphs of a constant term and a linear term to determine the graph of a linear equation of the form y = mx + b.
Conversely, you can determine the two terms that were added to form the graph of a linear function.
***
Algebraically, a linear equation like y = 7x + 3 is the sum of a linear term and a constant term.
Geometrically, you can graph this function by adding the graphs of:
y = 7x and y = 3
***
Now you will transform and graph linear equations. As you perform the algebraic transformations, the graph shows the geometric transformations.
***
Here are the graphs of y = x - 4 and y = -x + 6.
Substituting for y, you get x - 4 = -x + 6.
Notice that the lines intersect at x = 5, the solution of x - 4 = -x + 6.
***
Performing algebraic operations on both sides of an equation transforms it into equivlanet equations.
Although the equations change in form, the graphs show that the solutions are preserved.
WORKSPACE:
ANS:
-4 , -2 , 4 , 2 , 1 , #1 , #4 , 3 , 6 , 10 , 5 , -7 , 5/4 , 28/5 , ... , #2 , #3 , #1 , 3 , -1
LINEAR FUNCTIONS AND GRAPHS: Solving Systems of Linear Equations
- Solve systems of linear equations by using the substitution method
- Solve systems of linear equations by using the linear combination method
NOTES:
In the last activity, you performed algebraic transformations on equations and followed the corresponding geometric transformations on the graphs.
This mapping back and forth between algebra and geometry can often be used to study and solve mathematical problems.
***
Sometimes you can solve systems of equations by reading the solution from the graph.
y = x - 4
y = -x + 6
The point (5,1) lies on both lines and satisfies each of the equations.
1 = 5 - 4
1 = -5 + 6
***
When the solution is difficult to determine from the graph, you can use algebraic methods to find the ordered pair that satisfies both equations.
The substitution method can be used to solve this system.
5y = 4x + 20
y = 2x + 3
You can substitute 2x + 3 for y in the first equation.
5(2x + 3) = 4x + 20
10x + 15 = 4x + 20
6x = 5
x = 5/6
y = 2(5/6) + 3
= 10/6 + 18/6
= 28/6 = 14/3
The solution is the ordered pair (5/6,14/3).
***
Use your own words to describe how the substitution method solves a system of equations like these.
x - 2y = 10
y = 3x
Substitute 3x for y in the first equation and transform it into a single equation for x. Then solve for y.
x - 6x = 10
-5x = 10
x = -2
y = 3x
y = 3(-2)
y = -6
Point (-2,-6) satisfies both equations.
***
Sometimes neither equations in a system has x or y alone on one side of the equation.
3x + 6y = 4
y - 3x = -4
When this happens, one of the equations must be transformed before the substitution method can be used.
3x + 6y = 4
y - 3x = -4
You can isolate y in the second equation.
y = 3x - 4
***
5x - 2y = 4
3x + 2y = 1
Although you solved this system of equations using the substitution method, there is another way to eliminate one of the variables.
To form a single equation in x, you can add the equations.
5x - 2y = 4
3x + 2y = 1
-----------
8x + 0y = 5
8x = 5
x = 5/8
3(5/8) + 2y = 1
15/8 + 2y = 1
2y = 8/8 - 15/8
2y = -7/8
y = -7/16
The ordered pair (5/8,-7/16) is the same solution obtained by using the substitution method.
***
This linear combination method of solving systems is convenient to use when both equations are in standard form.
Sometimes one or both equations must be multiplied by a constant before they can be added or subtracted to eliminated one of the varaiables.
***
2x + y = -2
x - 3y = 2
You can multiply the first equation by 3.
6x + 3y = -2
x - 3y = 2
------------
7x = -4
x = -4/7
-4/7 - 3y = 2
-3y = 14/7 + 4/7
-3y = 18/7
y = -6/7
***
Systems of linear equations can be solved geometrically by graphing the lines and determining the point of intersection.
Systems of equations can also be solved algebraically by using the substitution method or the linear combination method.
***
Use your own words to describe when you might use each algebraic method.
If one of the equations can be easily expressed in terms of a single varaiable, the substitution method works well.
If both equations are in standard form, with coefficients other than 1, the linear combination method may be easier to use.
WORKSPACE:
ANS:
#4 , #3 , 4/3 , 8/3 , 4/5 , -12/5 , 4/3 , 0 , -6/7 , 16/7 , 5/8 , -7/16 , 0 , -5/2 , 10/3 , 1/12 , 7/2 , -8 , standard , 9/4 , -15/8 , -1/6 , 19/6 , 9/4 , 1/2 , -1/3 , -4 , 9/2
The below is continued (some parts above are from this date, though). Current Date: March 23, 2006.
LINEAR FUNCTIONS AND GRAPHS: Problem Set 1
NOTES:
WORKSPACE:
ANS:
135 , 8 , 38
LINEAR SYSTEMS AND MATRICES: Linear Functions and Graphs Post-Test
NOTES:
WORKSPACE:
The point (2,-2) could NEVER be the solution to a system of equations which includes the line...
#4 y = (1/2)x - 1
***
A system of two linear equations, whose graphs intersect in a single point, has HOW MANY solution?
#2
***
Choose the equations that is NOT equivalent to the other equations.
#2 5y + 30 = 3x
***
The ordered pair which satisfies this system of linear equations is...
#2 (3,2)
***
Use the substition method to solve this system of linear equations.
#4 (-3,-5)
***
The standard form for the linear equation
-x y
-- + - = 1
3 2
is...
#4
***
The standard form for the linear equation
y + 2 3
----- = -
x - 1 4
is...
#2
***
Use the linear combination method to solve this system of equations.
#4 (3,0)
***
Use the linear combination method to solve this system.
#3
***
To eliminate the y-term, multiply the first equation by...
#3 2
ANS:
See above.
SOLVING SYSTEMS WITH MATRICES: Row Operations
- Represent systems of linear equations using matrices
- Solve a linear system by transforming the matrix of the system
In the last lesson you saw that linear equations can always be transformed to the standard form of the equation. You also solved systems of two linear equations.
In this lesson you will use matrices to solve systems of two linear equations in standard form.
***
The pair of linear equations:
3x + 2y = 5
2x + 7y = 1
Is represented by this matrix:
[3 2 5]
[2 7 1]
NOTES:
Linear equations such as these can be represented by a matrix because the equations are in standard form.
ax + by = c
dx + ey = f
The matrix which corresponds to this system is written:
[a b c]
[d e f]
Watch what happens to the corresponding matrix as linear combinations are used to solve this pair of equations.
2x + 3y = 5 [2 3 5]
4x + 3y = 9 [4 3 9]
-2x - 3y = -5 [-2 -3 -5]
4x + 3y = 9 [ 4 3 9]
2x + 0 = 4 [2 0 4]
4x + 3y = 9 [4 3 9]
-4x + 0 = -8 [-4 0 -8]
4x + 3y = 9 [ 4 3 9]
The second row is replaced by adding the two equations.
-4x + 0 = -8 [-4 0 -8]
0 + 3y = 1 [0 3 1]
The first equation is multiplied by -1/4 and the second equation by 1/3.
x + 0 = 2 [1 0 2 ]
0 + y = 1/3 [0 1 1/3]
x = 2, y = 1/3 x = 2, y = 1/3
The point (2,1/3) satisfies the system.
***
When transforming a system of equations or its representative matrix to solve the system, you can:'
1. Multiply either equation (or row) by a constant.
2. Add the equations (or rows) to eliminate varaiables (produce zeros).
To solve a system of equations such as:
x + 3y = 7
2x + 9y = 20
You can perform ROW OPERATIONS on the matrix.
[1 3 7]
[2 9 20]
The first row is multiplied by -3.
[1 3 7] x + 3y = 7
[2 9 20] 2x + 9y = 20
[-3 -9 -21]
[ 2 9 20]
| | | | | | Indicate the operation
performed. ANS: Addition
[-1 0 -1]
[ 2 9 20]
| | | | | | Indicate the operation
performed. ANS: Multiplied
[-2 0 -2]
[ 2 9 20]
| | | | | | Indicate the operation
performed. ANS: Addition
[-2 0 -2]
[ 0 9 18]
***
Sometimes two or more operations can be applied in one step to speed up the matrix method of solving a system.
To solve the system:
3x - 2y = -6
-2x + y = 5
Determine the matrix:
[ 3 -2 -6]
[-2 1 5]
Multiply 2nd row by 2 and add rows.
[ 3 -2 -6]
[-1 0 4]
Multiply 2nd row by 3 and add rows.
[ 0 -2 6]
[-1 0 4]
Multiply 1st row by -1/2, 2nd row by -1.
[0 1 -3]
[1 0 -4]
The solution point is (-4,-3).
***
To solve a system of linear equations using a matrix of the form:
[a b c]
[d e f]
You can perform row operations on the matrix until it is transformed to a solution matrix of the form:
[0 1 p] OR [1 0 q]
[1 0 q] [0 1 p]
Where x = q and y = p.
If a matrix is transformed to an equivalent solution matrix of the form:
[0 1 p] OR [1 0 q]
[1 0 q] [0 1 p]
The values for x and y can be checked in the original equations to see if they are correct.
The matrix...
[0 1 4/3]
[1 0 -1/3]
...is the solution matrix for the system:
3x + 6x = 7
-2x + 4y = 6
because 3(-1/3) + 6(4/3) = -3/3 + 24/3
= 7
and
-2(-1/3) + 4(4/3) = 2/3 + 16/3
= 6
***
To solve a system of linear equations using matrix methods, first determine the matrix for the system, then find a solution matrix of the form:
[0 1 q] OR [1 0 p]
[1 0 p] [0 1 q]
By performing ROW operations on the matrix.
WORKSPACE:
ANS:
#4 , #2 , #4 , #3 , #2 , #2 , standard , addition , multiplication , addition , multiplication , 1,2 , -6 , 2 , -2 , #3 , #3 , #2 , row , #3 , #2 , #4 , #1 , #4
SOLVING SYSTEMS WITH MATRICES: Cramer's Rules and Determinants
- Solve a system of linear equations using Cramer's Rule
- Find the determinant of a 2 by 2 matrix
NOTES:
In the last activity you saw that a matrix can be used to represent a system of linear equations.
ax + by = c --\ [a b c]
dx + ey = f --/ [d e f]
Furthermore, row operations can be performed to transform that matrix.
[a b c] --\ [1 0 p]
[d e f] --/ [0 1 q]
The solution x = p and y = q can be read from the matrix.
Now, you can use matrix transformations to find a general rule that gives the solution to any pair of linear equations in standard form.
ax + by = c
dx + ey = f
Start with the matrix for this system and perform row operations to find the solution for the system.
[a b c]
[d e f]
Multiply the top row by e.
[a b c] ax + by = c
[d e f] dx + ey = f
[ae be ce]
[ d e f]
[ ae be ce]
[-bd -be -bf]
[ae-bd 0 ce-bf]
[ -bd -be -bf ]
[ 1 0 ce-bf]
-----] ---\ ce - bf
ae-bd] ---/ x = -------
[-bd -be -bf ] ae - bd
A similar process can be used to obtain a solution matrix for y.
[a b c] --\ ax + by = c
[d e f] --/ dx + ey = f
[-ad -bd -cd ]
[ 0 1 af-cd]
[ -----]
[ ae-bd]
The solution for y read from this matrix is:
af - cd
y = -------
ae - bd
The general formulas for x and y can be used to solve this system.
2x + 3y = 2
4x - 9y = -1
ce - bf af - cd
x = ------- y = -------
ae - bd ae - bd
2(-9) - 3(-1) -18 + 3
x = ------------- = -------- = 1/2
2(-9) - 3(4) -18 - 12
2(-1) - 2(4) -2 - 8
y = ------------ = -------- = 1/3
2(-9) - 3(4) -18 - 12
The solution to this system is the ordered pair (1/2,1/3).
***
Use the general formulas for x and y to solve this system of equations.
4x - 3y = 11
5x - 6y = 9
ax + by = c
dx + ey = f
ce - bf af - cd
x = ------- y = -------
ae - bd ae - bd
11(-6) + 3(9)
x = -------------
4(-6) + 3(5)
...
-39 13
x = --- = --
-9 3
4(9) - 11(5)
y = ------------
4(-6) + 3(5)
...
-19 19
y = --- = --
-9 9
***
The formulas for finding the solution of any system of two linear equations are known as CRAMER'S RULE.
ce - bf af - cd
x = ------- y = -------
ae - bd ae - bd
Because Cramer's Rule is difficult to remember, the concept of "determinant" may be of help.
The determinant of the matrix:
[5 3] is denoted by |5 3|
[7 2] |7 2|
And the value of the determinant is 5(2) - 7(3) = 10 - 21 = -11.
***
Now you can apply derminants to Cramer's Rule and to the solution of a system of linear equations.
Once again, consider the linear equations:
ax + by = c
dx + ey = f
And the corresponding matrix:
[a b c]
[d e f]
***
IMPORTANT, KEEP THIS:
Cramer's Rule:
ce - bf
x = -------
ae - bd
af - cd
y = -------
ae - bd
[a b c]
[d e f]
Find the determinant of the x and y coefficient matrix.
[a b] D = ae - bd
[d e]
Notice that this determinant is the denominator for both values of Cramer's Rule.
NEXT:
Find the determinant of the matrix formed by replacing the x coefficients with the constant terms.
[c b] D = ce - bf
[f e]
Notice that this determinant is the "x-numerator" of Cramer's Rule.
NEXT:
Find the determinant of the matrix formed by replacing the y coefficients with the constant terms.
[a c] D = af - cd
[d f]
Notice that this determinant is the "y-numerator" of Cramer's Rule.
THINK:
This is the determinant form of Cramer's Rule.
|c b| |a c|
|f e| |d f|
x = ----- y = -----
|a b| |a b|
|d e| |d e|
[a b c]
[d e f]
...[animation of taking the c and f away]...
[a b]
[d e]
The denominator of both equations is the determinant of the coefficient matrix.
...[another animation, switched some stuff around]...
[c b]
[f e]
The numerator of the x equation is the determinant of the matrix formed by replacing the x coefficients.
...[another animation, switched more around]...
[a c]
[d f]
The numerator of the y equation is the determinant of the matrix formed by replacing the y coefficients.
***
To solve the system:
3x + 4y = 8
2x - 2y = 7
Using the determinant form of Cramer's Rule, we follow these steps:
Evaluate the DENOMINATOR DETERMINANT.
|3 4| = 2(-2) - 4(2) = -14
|2 -2|
Then the X-NUMERATOR DETERMINANT.
|8 4| = 8(-2) - 4(7) = -44
|7 -2|
Then the Y-NUMERATOR DETERMINANT.
|3 8| = 3(7) - 8(2) = 5
|2 7|
We simplify to find the solution.
-44 5
x = --- = 22/7 y = --- = -5/14
-14 -14
***
The general formulas used to solve a system of two linear equations are known as Cramer's Rule.
ax + by = c
dx + ey = f
ce - bf af - cd
x = ------- y = -------
ae - bd ae - bd
Since the rule can difficult to remember, the determinant form of the rule is more easily applied.
|c b| |a c|
|f e| |d f|
x = ----- y = -----
|a b| |a b|
|d e| |d e|
WORKSPACE:
2x - 3y = 3
4x + 6y = -2
x = 1/2
| 3 -3|
|-2 6|
x = -------
| 2 -3|
| 4 6|
...
x = 12/24 = 1/2
|2 3|
|4 -2|
y = ------
|2 -3|
|4 6|
...
y = -16/24 = -2/3
ANS:
13/3 , 19/9 , -2 , 11 , ae - bd , ce - bf , af - cd , 5 , 20 , 10 , 1/2 , -2/3 , #3 , #4 , #1 , #2 , #1
The below is continued (some parts above are from this date, though). Current Date: March 28, 2006.
QUADRATIC FUNCTIONS: Plotting Points
- Evaluate quadratic functions
- Plot points using x and f(x) values
NOTES:
In Unit 1 you graphed linear functions of the form y = mx + b. All the points which satisfy a linear function lie on a straight line.
If only two points of the function have been plotted, you can predict the location of other points.
These points lie on the graph of a linear function.
For x = 3, you can guess the value of y.
y = 3
All three points satisfy the function y = 2x - 3.
***
Next, you will be given points on the graphs of other functions which may not be linear. Try to guess the location of other points on the graphs of these "mystery" functions.
***
Use your own words to describe the results of predicting the location of other points on the graph of a function when only a few of the points have been plotted.
Unless the function is known to be linear, errors may result when predicting the location of other points.
To avoid errors, the rule or function must be known.
***
When a function is given, it can be evaluated to plot points on its graph.
You can plot some points of f(x) = x^2 - 6x + 11 by completing the table.
x | f(x)
--+-----
1 | 6
2 | 3
3 | 2
4 | 3
5 | 6
Complete this table to plot points of:
2
f(x) = x - 2x - 2
x | f(x)
--+-----
-1| 1
0 | -2
1 | -3
2 | -2
3 | 1
Complete this table to plot points of:
2
f(x) = -2x + 4x - 3
x | f(x)
--+-----
-1| -9
0 | -3
1 | -1
2 | -3
3 | -9
***
Describe how to plot points of the graph when the rule or function is given.
To plot points of a given function, make a table of x and f(x) values that satisfy that function.
***
You have been evaluating functions of the form:
2
y = ax + bx + c
These equations are called QUADRATIC EQUATIONS.
These are examples of QUADRATIC FUNCTIONS:
2 2
f(x) = 2x - 5x - 12 f(x) = x + 4
2 2
f(x) = -7x f(x) = 3x - x
2 2
f(x) = 16 - x f(x) = 6x - 2 + 3x
Notice the x-squared terms.
These are NOT quadratic functions:
f(x) = 2x + 5 f(x) = 5x
3
f(x) = -3 f(x) = x - x + 2
***
Before pressing ENTER, use your own words to describe a quadratic function.
It is a second degree function that can be written in the form:
2 /
f(x) = ax + bx + c, a = 0
It must have an x-squared term but no term raised to a higher power.
WORKSPACE:
ANS:
9 , 1 , 3 , 2 , -1 , 2 , 3 , 6 , 1 , -2 , -3 , -2 , 1 , -9 , -3 , -1 , -3 , -9 , NO , YES , NO , NO , YES , 1 , 6 , .25 , YES , NO , #2 , #2 , #3 , #4 , #4
The below is continued (some parts above are from this date, though). Current Date: March 29, 2006.
GRAPHING QUADRATIC FUNCTIONS: Patterns in Graphs
- Find the maximum or minimum point of a quadratic function
- Identify the line of symmetry
- Determine where the function is increasing or decreasing
NOTES:
In the last activity, you plotted points of a quadratic function by evaluating the function for several values of x.
To sketch the entire graph, you could plot many points using x and f(x).
However, to save time, you can look for critical points and patterns when evaluating the function.
***
You may have noticed that the x-values in the tables clearly indicated a maximum or minimum point for the functions.
Unless certain x-values are chosen, the maximum or minimum point may not be evident.
***
Given the funtion:
2
f(x) = x + 4x + 6
One value of x whose f(x)-value is in the shaded area is x = -1.
f(-1) = 3
The point (-1,3) is the minimum point of this function.
***
Use your own words to describe how you can identify the maximum or minimum point of a quadratic function.
If the appropiate values for x have been chosen, you can look at the f(x)-values of the table or graph.
***
Once the maximum or minimum point of a quadratic function has been located, you can look for patterns to find other points that satisfy the function.
***
Describe how the axis of symmetry can be used to predict other points that lie on the graph of a quadratic function.
Any point on one side of the axis of symmetry has a mirror image on the other side of the axis.
***
You have seen how maximum or minimum points and the axis of symmetry produce patterns in the graphs of quadratic functions.
Now, you will learn about another pattern that allows you to complete the sketch of a graph.
These points satisfy:
2
f(x) = x
x | f(x)
--+-----
0 | 0
1 | 1
2 | 4
3 | 9
5 | 25
For x > 0, as x increases f(x) also increases. When x > 0, f(x) = x^2 is an INCREASING FUNCTION.
These points also satisfy:
2
f(x) = x
x | f(x)
--+-----
-5| 25
-3| 9
-2| 4
-1| 1
0 | 0
For x < 0, as x increases f(x) decreases. When x < 0, f(x) = x^2 is a DECREASING FUNCTION.
***
Describe how you can determine where a quadratic function is increasing or decreasing.
If f(x) increases as x increases, then the function is increasing. If f(x) decreases as x increases, then the function is decreasing.
***
Using the maximum or minimum point, the line of symmetry, and the interval where the function increases or decreases, you are now ready to sketch the graph of a quadratic function.
The graph of any quadratic function is a PARABOLA.
***
Given a quadratic function that increases for all values of x > 4 and has a minimum point, only parabola B satisfied these conditions.
[B is a graph of a line opening towards the north]
***
Before pressing ENTER, describe how the graph of a quadratic function is sketched.
To sketch the graph, plot a few points to determine the maximum or minimum point and the line of symmetry. Decide where the function is increasing and decreasing to connect the points of the parabola.
WORKSPACE:
ANS:
#1 Min. , #1 Min. , #2 Max. , #1 Min. , #2 Max. , #4 , #1 , 5 , 4 , 6 , 0 , 3 , 8 , 3 , -4 , 2 , #1 Inc. , #2 Dec. , #2 Dec. , #1 Inc. , #1 Inc. , #2 Dec. , 4,2 , 4 , 4 , #2 , #3 , #3 , #3 (4,1) , #4 , 4 , #2 , #4
THE CRITICAL POINT FORM: Exploring Graphs of:
2
y = a(x - h) + k
- Learn how values for a, h, and k in the equations...
2
y = a(x - h) + k
...determine the shape and location of the graph
NOTES:
In the last activity you plotted points to graph quadratic function in STANDARD FORM, f(x) = ax^2 + bx + c.
In this activity you will graph quadratic functions in the CRITICAL POINT FORM:
2
f(x) = a(x - h) + k
Noting the effects of a, h, and k on the graph.
***
In the next 2 explorations, you will choose values for h and k to explore their graphs.
These are examples of some of the values you might try.
VALUE EXAMPLE
-----------------------------------------
positive integer 4
negative integer -6
fraction 2/3
decimal numeral 5.7
WORKSPACE:
ANS:
THE CRITICAL POINT FORM: Analyzing y = a(x - h)^2 + k
- Graph quadratic equations in critical point form
- Determine the critical point form of a quadratic equation from its graph
NOTES:
In the last activity you learned values for "a", "h", and "k" effect the graph of:
2
y = a(x - h) + k
Now you will use these values to graph parabolas.
Given the equation:
2
y = 2(x + 1) + 3
/ | \
a = 2 h = -1 k = 3
The point (-1,3) is the vertex of the graph.
***
You can determine the value of "a" and identify the vertex (h,k) from the equation y = a(x - h)^2 + k.
After plotting the vertex, the value of "a" can be used to plot other points and sketch the graph.
***
The vertex of:
2
y = 2(x - 3) + 4
Is (3,4) and a = 2. Since "a" is POSITIVE, the graph opens UPWARD. With a = 2, points one over and 2 up on either side of the vertex also like on the graph.
2
y = 2(x - 3) + 4
These three points will show the direction and location of the parabola.
***
Use the value for a to find the points whose x-values are -1 and -3.
2
y = -(x + 2) + 5
Vertex: (-2,5)
a = -1
(-1,4),(-3,4)
AND:
What is the vertex of?
2
y = 3(x - 3) - 4
Vertex: (3,-4)
a = 3
(2,-1),(4,-1)
***
Given a quadratic equation, you used the vertex and the value of "a" to plot some points and sketch the graph.
Even if the quadratic equation is not known, you can sketch the graph if you know the vertex and the value of "a".
***
To sketch a graph of an equation of the form:
2
y = a(x - h) + k
Plot the vertex (h,k) and find the value of "a".
Then use the value of "a" to determine the direction of the parabola and to plot other points.
Given a quadratic equation in critical point form, you have learned to graph the parabola.
Now, given the graph of a quadratic function, you can determine the values of "a", "h", and "k" to write the equation for the parabola.
***
The vertex (h,k) of this graph is (3,-2).
Points one over and 1 up also like on the graph so a = 1.
Substituting these values, the critical point form is:
2
y = (x - 3) - 2
***
Describe how you can determine the equation of a quadratic function from its graph, then press ENTER.
Locate the vertex (h,k), determine the value of a, and substitute these values in the equation:
2
y = a(x - h) + k
WORKSPACE:
ANS:
-1 , 2,4 , 1 , -3,-2 , 3 , 0,1 , -2 , 1,0 , 1/2 , -5,-9 , -2,5 , -1 , 4,4 , 3,-4 , 3 , -1,-1 , 3 , 3 , 1 , 1 , -1 , -1 , 0 , 0 , h,k , -4,6 , -2 , -2,4,6 , #1 , #4 , #4 , #2 , 3,1 , -4,3 , 0,-2 , 1,0 , 2 , -1 , 1 , #1 , #3 , #2
THE CRITICAL POINT FORM: Transforming Quadratic Equations
- Convert quadratic equations from standard form to critical point form
- Build quadratic functions graphically
- Illustrate algebraic transformations with graphs
NOTES:
In previous activities, you graphed quadratic functions using the standard form f(x) = ax^2 + bx + c and the critical point form f(x) = a(x - h)^2 + k.
In this activity, you will study transformations of these functions algebraically and graphically.
***
To convert equations like...
2
y = 2x + 20x + 47
...from standard form to critical point form, you can use the method of COMPLETING THE SQUARE.
First, factor out the coefficient of the x^2 term from the first two terms.
2
y = 2x + 20x + 47
2
y = 2(x + 10x) + 47
Next divide 10 by 2 and square the result.
10/2 = 5 5^2 = 25
2
y = 2(x + 10x + 25) + 47 - 50
| / /
2 * 25 = 50
2
y = 2(x + 5) - 3
The equation has been transformed from standard form to critical point form by completing the square.
***
Convert this equations from standard form to critical point form.
2
y = 3x - 6x - 1
2
y = 3(x - 2x) - 1
2
y = 3(x - 2x + 1) - 1 - 3
2
y = 3(x - 1) - 4
Another:
2
y = -4x + 24x - 35
2
y = -4(x - 6x) - 35
2
y = -4(x - 6x + 9) - 35 + 36
2
y = -4(x - 3) + 1
***
Before pressing ENTER, use your own words to describe how to transform a quadratic equation from standard form to critical point form.
To transform:
2
y = ax + bx + c
Into:
2
y = a(x - h) + k
Complete the square and simplify.
***
Notice, in standard form a quadratic equation is the sum of three terms:
2
ax + bx + c
|| || |
||2 || |
ax bx c
...[animation]...
You can build quadratic expressions algebraically by adding the terms.
2
3x - 4x + 8
You can also build the graph of a quadratic function by adding the graphs of its terms.
This is the graph of a quadratic term.
2
y = x
1
This is a graph of a constant term.
y = 3
2
And...
2
y = x
1
y = 3
2
For x = -2, the sum of the two y-values is 7.
For x = 0, the sum of the two y-values is 3.
For x = 1, the sum of the two y-values is 4.
If the two y-values are added for each x, the resulting graph is the graph of:
2
y = x + 3
***
Desribe how the graph of a quadratic function can be built by adding the graphs of its terms.
Sketch the graph of each term. Add the y-values for any given x. These sums determine the graph of the sum of the terms.
***
In Unit 1, you transformed linear equations and graphed each step of the transformation.
Now you will transform and graph quadratic equations. As you perform the algebraic transformations, the graph shows the geometric transformations.
***
Performing algebraic operations on both sides of an equation transforms it into equivalent equations.
Although the equations change in form, the graphs show that the solutions are preserved.
WORKSPACE:
ANS:
2 , 1,3 , 1,4 , 6 , 9,36 , 3,1 , 2 , 5 , 0 , -3 , -4 , -3 , 0 , -2 , -2 , 0 , 4 , 7 , 6 , 2 , 2 , 3 , 2 , 2 , 4 , #3 , #3
SOLVING QUADRATIC EQUATIONS: Roots and Factors
- Determine the roots and factors of a quadratic function from its graph
- Solve a quadratic equation by factoring
- Build the graph of a quadratic function by multiplying the graphs of its linear factors
NOTES:
In previous activities you graphed quadratic functions. Now you will use those graphs to determine the roots or solutions of the functions.
***
The graph of:
2
y = x - 6x + 5
Intersects the x-axis twice.
The function has two real ROOTS.
You can tell the number of real roots by determining the number of x-intercepts.
Sometimes it is easy to read the roots from the graph.
***
The real ROOTS or SOLUTIONS of an equation are the points where the graph intersects the x-axis.
The values of x when y = 0 are also called the ZEROS of a quadratic function.
The roots or zeros of a quadratic function can be used to determine the factored form of its equation.
***
You can often factor a quadratic equation by reading the roots from its graph.
By reversing the process, you can find the roots of an equation by factoring.
2
-3x + 18x - 27 = 0
2
-3(x - 6x + 9) = 0
-3(x - 3)(x - 3) = 0
The double factor (x - 3) tells you the single root is 3.
Compare the x-intercept with the solution obtained by factoring.
***
Describe in your own words how to solve a quadratic equation by factoring.
Let y = 0 and factor the equation. Solving each factor for x results in the solutions for that equation.
***
You've seen the relationship between roots and factors. If the roots are 2 and -1, the factors are (x - 2) and (x + 1).
You can build a quadratic function algebraically by multiplying these factors.
2
(x - 2)(x + 1) = x - x - 2
You can also build a quadratic function graphically by multiplying the graphs of its linear factors.
***
In Unit 1 you studied first degree equations of the form y = mx + b.
The graph of a linear equation is always a line.
In this unit you studied second degree equations of the form y = ax^2 + bx + c.
The graph of a quadratic function is always a parabola.
A parabola can be built from the graphs of its linear factors.
y = (x + 3)(x - 4)
Here are the graphs of:
y = x + 3 AND y = x - 4
1 2
Notice what happens when y and y are
1 2
multplied for given values of x.
At x = -4, y = -1 and y = -8, their
1 2
product is 8.
For any x-values less than -3, y and y
1 2
are both negative. Since (-)*(-) = (+), the product of the two y-values lies above the x-axis.
WORKSPACE:
ANS:
#1 , #2 , #3 , #2 , #3 , 1,5 , -9,3 , -3,4 , -2,3 , 0,4 , 2 , x , -3,-6 , +2,-8 , -1,-4 , -3,-3 , +4,+3 , 5,-3 , -4,1 , 1/3,-1/2 , 1/3 , 6,5/2 , 0 , 0 , #2 , #1 , factors , -4,2 , x+4,x-2 , -3/2,2 , 2x+3,x-2 , -1/2 , 2x+1,2x+1 , #1 , #3 , #4 , #2
SOLVING QUADRATIC EQUATIONS: Completing the Square and the Quadratic Formula
- Solve quadratic equations by completing the square
- Solve quadratic equations by using the quadratic formula
NOTES:
In the last activity, you solved quadratic equations by reading the x-intercepts from the graph or by factoring the equation.
Sometimes the roots are not easily read from the graph or the equation can not be easily factored by inspection. This activity looks at two other methods used to solve quadratic equations.
***
COMPLETING THE SQUARE can be used to find the roots of an equation.
2
x - 6x - 2 = 0
2
x - 6x = 2
2
x - 6x + 9 = 2 + 9
2
(x - 3) = 11
__
x - 3 = +/- \/11
__
x = 3 +/- \/11
__ __
x = 3 + \/11 OR x = 3 - \/11
***
When the roots can not be read from the graph or when the equation is not easily factored by inspection, you can solve a quadratic equation by completing the square.
Although the method of completing the square can be used to solve any quadratic equation, it can be difficult to use when fractions are involved.
2
5x - 6x - 2 = 0
2
5x - 6x = 2
2 6x 2
(x - --) = -
5 5
2 6x 9 2 9
(x - -- + -- ) = - + --
5 25 5 25
3 2 19
(x - -) = --
5 25
__
3 /19
x - - = +/- / --
5 \/ 25
__
3 /19
x = - +/- / --
5 \/ 25
__
3 /19
x = - +/- / --
5 \/ 5
__
3 +/- \/19
x = ----------
5
***
To derive a general formula for solving quadratic equations, you can solve ax^2 + bx + c = 0 for x.
2
ax + bx + c = 0
2
ax + bx = -c
2 b -c
x + -x = --
a a
2 2
2 b b b c
x + -x + --2 = --2 - -
a 4a 4a a
2
b 2 b - 4ac
(x + --) = ----2--
2a 4a
_2_____
b /b - 4ac
x + -- = +/- /----2---
2a \/ 4a
_2_____
-b \/b - 4ac
x = -- +/- ---------
2a 2a
_2_____
-b +/- \/b - 4ac
x = ---------------- This is the
2a QUADRATIC FORMULA.
You solved:
2
ax + bx + c = 0
For x.
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
You can solve ANY quadratic equation for x by using the quadratic formula.
_2_____
-b + \/b - 4ac
x = --------------
2a
OR:
_2_____
-b - \/b - 4ac
x = --------------
2a
***
You can use the quadratic formula to solve:
2
x - 8x - 5 = 0
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
_2______________
(8 +/- \/(64 - 4(1)(-5)))
x = -------------------------
2(1)
__ __
(8 +/- \/84) (8 +/- 2\/21)
x = ------------ OR -------------
2 2
__
x = 4 +/- \/21
So:
__ __
x = 4 + \/21 OR x = 4 - \/21
TEMPLATES:
2
ax + bx + c = 0
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
***
Before pressing ENTER, use your own words to describe how to solve a quadratic equation.
You can find the roots of a quadratic equation by looking at the graph or by factoring, if the equation factors easily. Otherwise, complete the square or use the quadratic formula.
WORKSPACE:
2
ax + bx + c = 0
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
ANS:
6 , 4 , 2 , 2 , 6 , 2 , 3 , 2 , square , 17 , 3 , 2,-1/2 , -1/2,2/3 , 2,-7 , 10 , 3/2,1 , 3 , -3/4,1 , 0,3/8 , 2 , 13 , 7 , 0,5/3 ,
The below is continued (some parts above are from this date, though, and others are not because I forgot to place one of these "placeholders"). Current Date: April 4, 2006.
SOLVING QUADRATIC EQUATIONS: Problem Set 1
NOTES:
WORKSPACE:
ANS:
14 , 49 , 2
CIRCLES AND PARABOLAS: The Pythaforean Theorem and the Distance Formula
- Derive the distance formula from the Pythagorean Theorem
- Use the Pythaforean Theorem or the distance formula to find the distance between two points
NOTES:
Geometric figures such as circles, parabolas, ellipses, and hyperbolas are sets of points defined in terms of distance in a coordinate plane.
In this activity, you will use the Pythagorean Theorem and the distance formula to find distances between points.
***
Given the lengths of the sides of a triangle, you can use the Pythagorean Theorem to determine if it is a right triangle.
2 2 2
a + b = c
2 __ 2
3 + (2\/10)
9 + (4 * 10)
9 + 40 = 49
2 2
c = 7 = 49
2 2 2
Since a + b = c, this is a right triangle.
***
Given the lengths of three sides of a triangle, you used a^2 + b^2 = c^2 to see if it was a right triangle.
If the lengths of two sides of a right triangle are known, you can use the Pythagorean Theorem to find the length of the third side.
2 2 2
c = a + b
2 2
3 + 7
9 + 49 = 58
__
c = +/- \/58
Since lengths are positive, the length of the hypotenuse is \/58.
***
Before pressing ENTER, describe in your own words how the lengths of the sides of a right triangle are related.
Given the lengths a and b of the legs and the length c of the hypotenuse.
2 2 2
a + b = c
This is called the PYTHAGOREAN THEOREM.
***
In a right triangle you found the lengths of the hypotenuse given the lengths of the two legs.
In a similar manner the Pythagorean Theorem can be used to find the distance between two points in the coordinate plane.
The horizontal distance is 4 since:
|5 - 1| = 4
OR
|1 - 5| = 4
The vertical distance is 3 since:
|6 - 3| = 3
OR
|3 - 6| = 3
Applying the Pythagorean Theorem...
2 2 2
AB = 4 + 3
16 + 9 = 25
__
AB = \/25 = 5
The distance from A to B is 5.
***
Describe how the Pythagorean Theorem is used to find the distance between two points in the coordinate plane.
Think of the segment between the two points as the hypotenuse of a right triangle. Find the horizontal and vertical components of the distance and apply the formula.
2 2 2
a + b = c
***
2 2 2
You used a + b = c to find the distance between two points.
The Pythagorean Theorem can now be used to derive a general formula for finding distance between any two points P and Q in the coordinate plane.
The horizontal distance is:
PR = |x - x|
1 2
(OR |x - x|)
2 1
The vertical distance is:
QR = |y - y|
1 2
(OR |y - y|)
2 1
2 2 2
(PQ) = (PR) + (QR)
____2______2
PQ = \/(PR) + (QR)
_______2_________2
PQ = \/|x - x| + |y - y|
1 2 1 2
_______2_________2
PQ = \/(x - x) + (y - y)
1 2 1 2
This is the DISTANCE FORMULA.
- The absolute value signs are not required because the quantity is squared
_______2_________2
d = \/(x - x) + (y - y)
1 2 1 2
________2__________2
d = \/(-3 - 1) + (-2 - 6)
____2______2
d = \/(-4) + (-8)
_______
d = \/16 + 64
__
d = \/80
Simplified, the distance AB is:
_
d = 4\/5
WORKSPACE:
ANS:
NO , YES , YES , 10 , 3 , 5 , 10 , 5 , 4 , c^2 , 5 , 17 , 3 , 37 , hypotenuse , 10 , 13 , 8 , 6 , 5 , 2 , 3 , 10 , 2 , 5
The below is continued (some parts above are from this date, though). Current Date: April 5, 2006.
- Use the distance formula to find the radius of a circle
- Determine the equation of a circle in the coordinate plane
CIRCLES AND PARABOLAS: Circles
NOTES:
Intersections of a plane with a cone are called CONIC SECTIONS. If the plane intersects the cone perpendicular to its axis, a CIRCLE is formed.
The DISTANCE FORMULA can be used to find the radius of a circle with center at the origin and passing through the point (-2,\/21).
_______2_________2
r = \/(x - x) + (y - y)
1 2 1 2
________2____________2
\/(-2 - 0) + (\/21 - 0)
____2________2
\/(-2) + (\/21)
______
\/4 + 21
__
\/25 = 5
The radius of this circle is 5.
***
The distance formula can be used to derive the standard equation for any circle with center at the origin and containing the point (x,y).
_______2_________2
r = \/(x - 0) + (y - 0)
2 2 2
r = (x - 0) + (y - 0)
2 2 2 2 2 2
r = x + y OR x + y = r
This is the standard form of the equation of a circle with center at the origin and radius r.
***
When evaluating x^2 + y^2 = 25 for some values of x, you will find two values for y.
2 2 2
y = r - x
_2___2
y = +/- \/r - x
Therefore, circles are quadratic RELATIONS, not functions.
***
The distance formula can be used to find the radius of a circle with center at (4,-3) and containing the point (6,-3 + 3\/5).
_______2____________________2
r = \/(4 - 6) + (-3 - (-3 + 3\/5))
____2________2
\/(-2) + (3\/5)
______
\/4 + 45
__
\/49 = 7
The radius of this circle is 7.
***
The distance formula can be used to derive the standard equation for a circle with a center (h,k) and containing a point (x,y).
_______2_________2
r = \/(x - h) + (y - k)
2 2 2
r = (x - h) + (y - k)
OR
2 2 2
(x - h) + (y - k) = r
This is the standard equation for any circle in the coordinate plane.
***
The equation for this circle with center (-5,2) and radius 4 is (x + 5)^2 + (y - 2)^2 = 16.
WORKSPACE:
ANS:
3,2 , 1.3 , 5/3 , 16 , 10 , .64 , .25 , #1 , #3 , 4 , 13 , 4 , 12 , -2,-2 , .25 , +2,-4 , center , radius , 3 , 2 , 0,0 , NOT KNOWN , 5 , +1,-4 , 2 , +2,+2 , 3 , -3,3
CIRCLES AND PARABOLAS: Parabolas
- See how the focus and directrix of a parabola determine the graph and equation of the parabola
In this activity you will study some geometric and algebraic properties of the parabola.
NOTES:
Point P is equidistant from the point (0,1) and the line y = -1.
2 = 2
d = d
1 2
The set of all points which satisfy the rule d = d form a PARABOLA.
1 2
A parabola can be described as a set of points which are equidistant from a given point and a given line.
All such points lie on the parabola.
The point (0,1) is called the FOCUS. The line y = -1 is called the DIRECTRIX.
***
You saw points on a parabola when the focus was (0,1) and the directrix was the line y = -1.
Now, using the x-axis as directrix, you will choose several values for the focus to see how the parabola changes.
***
The vertex of a parabola always lies equidistant from the directrix and the focus.
***
Describe what happens to the parabola as the distance between focus and directrix is increased.
The parabola opens wider as the distance is increased.
***
You have explored geometrically the set of points called a parabola. If a point P is on the parabola, the distance from P to the focus and the directrix is the same.
Now you will use the distance formula to derive an algebraic equation for a parabola.
***
Given focus (2,0) and directrix x = -2.
d = d
1 2
OR
_______2_________2
\/(x + 2) + (y - y)
_______2_________2
\/(x - 2) + (y - 0)
_______2 _______2___2
\/(x + 2) = \/(x - 2) + y
2 2 2
(x + 2) = (x - 2) + y
2 2 2
x + 4x + 4 = x - 4x + 4 + y
2
8x = y
2
x = (1/8)y
This is the equation for this parabola.
***
Describe in your own words how to derive the equation for a parabola whose vertex is at the origin and whose directrix is parallel to one of the axes.
Choose a point P(x,y) which lies on the parabola. Set the distances from P to the focus and directrix equal, and solve for x or y.
***
You have used the distance formula to derive the equation for parabolas with center at the origin.
The distance formula can also be used to derive the equation of any parabola in the plane.
***
Given directrix y = 5 and focus (3,1),
d = d OR
1 2
_______2_________2
\/(x - 3) + (y - 1)
_______2_________2
\/(x - x) + (y - 5)
2 2 2
(x - 3) + (y - 1) = (y - 5)
2 2 2
(x - 3) + y - 2y + 1 = y - 10y + 25
2
(x - 3) = -8y + 24
2
-8y = (x - 3) - 24
2
y = (-1/8)(x - 3) + 3
***
You can determine the equation for any parabola by setting d!1 = d!2. If d!1 is the distance from a point on the parabola to the directrix, then d!2 is the distance from that point to the focus.
WORKSPACE:
ANS:
focus , 0 , 2 , -1/4 , 1/12 , 1/12 , -1/12 , focus , #4 , -1 , -1,0 , 1/4 , -1/8
The below is continued (some parts above are from this date, though). Current Date: April 6, 2006.
CIRCLES AND PARABOLAS: Problem Set 2
NOTES:
WORKSPACE:
ANS:
6 , 16 , 4
ELLIPSES AND HYPERBOLAS: Ellipses
- Determine the graph of an ellipse from its equation
- Determine the equation of an ellipse from its graph
NOTES:
The sum of the distances from P to each of the points (-5,0) and (5,0) is 14.
9 + 5 = 14
d + d = 14
1 2
An ellipse can be described as a set of points P such that the sum of the distances from P to each of two given points is a constant.
The points (-5,0) and (5,0) are the FOCI of this ellipse.
You saw points on the ellipse when the given distance was 14 and the foci were (-5,0) and (5,0).
Now, keeping the same foci, you will choose several values for d to see how the ellipse changes.
***
Before pressing ENTER, describe how the shape of an ellipse changes as the value for d changes.
As d approaches the distance between the foci, the ELLIPSE FLATTENS.
As d is increased the ELLIPSE BECOMES MORE CIRCULAR.
***
You just chose different values for the distance d when the foci remained constant.
Now choose values for the foci c when the distance remains the same.
***
As the distance between the foci approaches 0, the ellipse becomes more like a circle.
***
You have explored geometrically the set of points called an ellipse. If a point P is on the ellipse, the sum of distances from P to each of the foci is constant.
Now you will use the distance formula to derive an algebraic equation for an ellipse.
Given the constant sum 10 and the foci (-3,0) and (3,0):
d + d = d
1 2
OR
_______2_________2
\/(x + 3) + (y - 0)
_______2_________2
+ \/(x - 3) + (y - 0) = 10
_______2___2 _______2___2
\/(x + 3) + y + \/(x - 3) + y = 10
_______2___2 _______2___2
\/(x + 3) + y = 10 - \/(x - 3) + y
_______2___2 _______2___22
(\/(x + 3) + y) = (10 - \/(x - 3) + y)
2 2
(x + 3) + y
______2____2 2 2
100 - 20\/(x - 3) + y + (x - 3) + y
Expanding and simplifying:
_______2___2
12x = 100 - 20\/(x - 3) + y
_______2___2 _______2___2
\/(x + 3) + y + \/(x - 3) + y = 10
_______2___2 _______2___2
\/(x + 3) + y = 10 - \/(x - 3) + y
2 2
(x + 3) + y
_______2___2 2 2
= 100 - 20\/(x - 3) + y + (x - 3) + y
_______2___2
12x = 100 - 20 \/(x - 3) + y
_______2___2
3x = 25 - 5\/(x - 3) + y
_______2___2
3x - 25 = -5\/(x - 3) + y
2 _______2___22
(3x - 25) = (-5\/(x - 3) + y)
Expanding and simplifying:
2 2
400 = 16x + 25y
2 2
x y
-- + -- = 1
25 16
You used to distance formula and d!1 + d!2 = 10 to derive the standard form of the equation for this ellipse.
To find the x-intercepts, let y = 0.
2
x 0
-- + -- = 1
25 16
2
x
-- = 1
2
x = 25
x = +/- 5
To find the y-intercepts, let x = 0.
2
0 y
-- + -- = 1
25 16
2
y
-- = 1
16
2
y = 16
y = +/- 4
***
The graph of:
2 2
x y
-- + -- = 1
9 36
Has foci on y-axis, x-intercepts at +/- 3, and y-intercepts at +/- 6.
***
You have used the equation of an ellipse to find the intercepts and determine the graph for the ellipse.
Conversely, you can determine the equation of an ellipse from its graph.
This ellipse has intercepts at (+/-3,0) and (0,+/-6).
2
3 = 9
2
6 = 36
The equation for this ellipse is:
2 2
x y
-- + -- = 1
9 36
***
Describe how to determine the equation for an ellipse with center at the origin.
Read the intercepts from the graph of the ellipse and substitute these values for a and b in the equation:
2 2
x y
-- + -- = 1
2 2
a b
WORKSPACE:
ANS:
circle , 3 , 2 , 2 , 5 , a , b , #2 , #1 , #3 , #2 , #3 , #2 , #1 , #3 , 4
ELLIPSES AND HYPERBOLAS: Hyperbolas
- Determine the graph of a hyperbola from its equation
- Determine the equation of a hyperbola from its graph and asymptotes
In this activity you will study some geometric and algebraic properties of the hyperbola.
NOTES:
The difference of the distance from P to each of the points (-5,0) and (5,0) is 6.
9 - 3 = 6
d - d = 6
1 2
The set of all points which satisfy the rule |d!1 - d!2| = 6 form a hyperbola.
A hyperbola can be described as a set of points P such that the difference of the distances form P to each of two given points is a constant.
The points (-5,0) and (5,0) are the FOCI of this hyperbola.
***
You saw points on a hyperbola when the constant difference was 6 and the foci were (-5,0) and (5,0).
Now, keeping the same foci, you will choose several values for d to see how the hyperbola changes.
***
Before pressing ENTER, describe how the shape of a hyperbola changes as the value for d changes.
When d is close to 0, the branches look like this.
[picture of vertical lines on a graph that are slightly curved and very close to the center]
As d approaches the distances between the foci, the brances look more like this.
[picture of vertical lines that are EXTREMELY curved outwards, x-axis is close to the foci]
***
You just chose different values for the distance d when the foci remained constant.
Now you will choose values for the foci c when the distance remains constant.
***
Describe how the branches of a hyperbola change as the distance between the foci is increased.
For a constant difference d, the branches of a hyperbola are less rounded when the foci are far apart.
***
You have explored geometrically the set of points called a hyperbola. If a point P is on the hyperbola, the difference of the distances from P to each of the foci is constant.
Now you will use the distances formula to derive an algebraic equation for a hyperbola.
***
Given the constant 4 and the foci (-3,0) and (3,0):
d - d = d
1 2
OR:
_______2_________2 _______2_________2
\/(x + 3) + (y - 0) - \/(x - 3) + (y - 0)
equals 4.
_______2___2 _______2___2
\/(x + 3) + y - \/(x - 3) + y = 4
_______2___2 _______2___2
\/(x + 3) + y = 4 + \/(x - 3) + y
_______2___2 2 _______2___2 2
(\/(x + 3) + y ) = (4 + \/(x - 3) + y )
2 2
x + 6x + 9 + y = ...
______2____2 2 2
16 + 8\/(x - 3) + y + x - 6x + 9 + y
_______2___2
12x - 16 = 8\/(x - 3) + y
_2____________2
3x - 4 = 2\/x - 6x + 9 + y
2 _2____________2 2
(3x - 4) = (2\/x - 6x + 9 + y )
Expanding and simplifying:
2 2
5x - 4y = 20
This equation can also be written as:
2 2
x y
-- - -- = 1
4 5
***
You used the distance formula and d!1 - d!2 = 4 to derive the standard form of the equation for this hyperbola.
2 2
x y
-- - -- = 1
4 5
To find the x-intercetps, let y = 0.
2
x 0
-- - - = 1
4 5
2
x
-- = 1
4
2
x = 4
x = +/- 2
2 2
x y
-- - -- = 1
4 5
To find the y-intercetps, let x = 0.
2
0 y
- - -- = 1
y 5
2
y
- -- = 1
5
2
y = -5
Since this equation has no real solutions, there are no y-intercepts.
***
If the equation for a hyperbola is of the form:
2 2
x y
-- - -- = 1
2 2
a b
...the foci and vertices of the hyperbola lie on the x-axis.
If the equation is of the form:
2 2
y x
-- - -- = 1
2 2
b a
..the foci and vertices lie on the y-axis. The y-intercepts are +/- b.
***
Given the equation:
2 2
x y
-- - -- 1
16 9
...you know the vertices are (-4,0) and (4,0).
To sketch the entire graph, you need to know something about the shape of the branches of the hyperbola.
***
The equation:
2 2
x y
-- - -- = 1
16 9
2
...can be solved for y.
2 2
9x - 16y = 144
2 2
16y = 9x - 144
2
2 9x - 144
y = --------
16
Consider the values of:
2
9x - 144
--------
16
...as x gets very large.
***
For the equation:
2
2 9x - 144
y = --------
16
...as x gets larger (5, 10, 15, ...), 9x^2 grows very fast (225, 900, 2025, ...) and the term -144 becomes insignificant.
2 2
9x - 144 9x
-------- approximates ---
16 16
Therefore:
2
2 9x
y ~ ---
16
...for large values of x.
2
2 9x
y ~ ---
16
3x
y ~ +/- --
4
The two lines, y = (3/4)x and y = -(3/4)x serve as guide lines for sketching the hyperbola.
2 2
x y
-- - -- = 1
16 9
As |x| gets larger the branches get closer to y = +/- 3/4x.
These two lines, y = (3/4)x and y = -(3/4)x, are called the asymptotes of the hyperbola.
***
The asymptotes for:
2 2
x y
-- - -- = 1
25 4
...are:
2
y = +/- -x
5
With x-intercepts at +/- 5, the hyperbola can be accurately sketched.
***
Find the asymptotes for:
2 2
x y
-- - -- = 1
4 9
3
y = +/- -x
2
***
Use your own words to describe the asymptotes of a hyperbola with center at the origin and foci on either axis.
2 2 2 2
x y y x
-- - -- = 1 OR -- - -- = 1
2 2 2 2
a b b a
The asymptotes are the lines y = (b/a)x and y = -(b/a)x. The branches of the hyperbola come very close to the asymptotes but never touch them.
***
You have learned to identify the vertices and the asymptotes of a hyperbola from its equation in standard form.
Now you will determine the equation of a hyperbola from its graph and its asymptotes.
***
This hyperbola has x-intercepts at +/- 3. The equation will be of the form:
2 2
x y
-- - -- = 1
2 2
a b
Where a = +/- 3.
This hyperbola has x-intercepts at +/- 3.
2 2
x y
-- - -- = 1
9 ??
When x = 3, the y-values of the asymptotes are +/- 4.
y = (4/3)x
AND: y = -(4/3)x
The equation is:
2 2
x y
-- - -- = 1
9 16
***
Describe how to determine the equation of a hyperbola from its graph.
If the hyperbola is of the form:
2 2
x y
-- - -- = 1
2 2
a b
...read the x-intercepts to find a. Then determine the slopes (+/- b/a) of the asymptotes to find b.
WORKSPACE:
ANS:
12x - 16 , 20 , 3 , 2 , x , 2 , 2 , #2 , #4 , #3 , #1 , #3 , #3 , 16
The below is continued (some parts above are from this date, though). Current Date: April 25, 2006.
...
The below is continued. Current Date: May 2, 2006.
ELLIPSES AND HYPERBOLAS: Transformations and Systems of Quadratic Equations
- Identify the standard form of the equation of a conic section from its graph
- Transorm quadratic equations from standard form to the general form:
2 2
ax + by + cxy + dx + ey + f = 0
- Solve systems that include quadratic equations
NOTES:
In previous activities you determined equations for each of the CONIC SECTIONS by locating center points, focal points, x and y intercepts, vertices, or other critical points on the graph.
Now, let's review the standard forms by matching equations to graphs of the conic sections.
***
HYPERBOLAS cenetered at the origin have equations of the form:
2 2 2 2
x y y x
-- - -- = 1 OR -- - -- = 1
2 2 2 2
a b b a
Since the vertices and the foci lie on the x-axis, the standard form of the equation for this hyperbola is:
2 2
x y
-- - -- = 1
2 2
a b
***
Now let's transform equations in standard form to the GENERAL FORM for any quadratic relation:
2 2
ax + by + cxy + dx + ey + f = 0
2 2
(x - 2) + (y - 2) = 16
2 2
x + y - 4x - 4y - 8 = 0
These two equations are equivalent.
***
To find the general equation of a circle with radius 4 and center at the point (-1,3), we first determine the equation in STANDARD FORM.
2 2
(x + 1) + (y - 3) = 16
We then SQUARE the x and y terms and SIMPLIFY to an equation of the form:
2 2
ax + by + cxy + dx + ey + f = 0
2 2
(x + 1) + (y - 3) = 16
2 2
x + y + 2x - 6y - 6 = 0
In the general form of the equation for a circle, a is always equal to b.
***
Transform this equation to the form:
2 2
ax + by + cxy + dx + ey + f = 0
2 2
x y
-- + -- = 1
25 9
2 2
9x + 25y - 225 = 0
***
Transform this equation to the form:
2 2
ax + by + cxy + dx + ey + f = 0
2 2
x y
-- - -- = 1
4 9
2 2
9x - 4y - 36 = 0
***
Transform this equation to the form:
2 2
ax + by + cxy + dx + ey + f = 0
2
x = -2(y + 3) + 2
2
2y + x + 12y + 16 = 0
***
Transform this equation to the form:
2 2
ax + by + cxy + dx + ey + f = 0
2 2
(x - 2) (y + 3)
-------- - -------- = 1
9 9
2 2
x - y - 4x - 6y - 14 = 0
***
Any standard form equation for a conic section can be transformed to the general form for quadratic relations:
2 2
ax + by + cxy + dx + ey + f = 0
Although the general form for a conic an likewise be converted to the standard form, the values for a and b in the general form are sufficient to determine the type of conic described.
***
To determine the type of conic section described by this equation, we note that a and b have opposite signs. This condition limits our possibilities to a HYPERBOLA.
2 2
-8x + 5y - 80 = 0
We can check out conclusions by converting the equation to standard form:
2 2
2 2 -x y
-8x + 5y - 80 = --- + -- - 1
10 16
OR
2 2
y x This is the standard
-- - -- = 1 form for a hyperbola.
16 10
***
Before pressing ENTER, describe in your own words how the values for a and b in the general quadratic equation...
2 2
ax + by + cxy + dx + ey + f = 0
...Indicate the type of conic section that it represents.
Circle a = b (but not 0)
/
Ellipse a = b (but same sign)
Hyperbola a and b opposite signs
Parabola a = 0 or b = 0 (but not both)
***
Now let's consider systems of equations which include quadratic equations.
To solve this system which includes a line and an ellipse, we can read the two points of intersection from the graph.
The solutions are (5,0) and (0,-3).
***
The SUBSTITION METHOD can be used to solve this system of equations.
We can solve the linear equation for x and substition in the equation for the ellipse:
2 2
(7 - 2y) + 4y = 25
2 2
49 - 28y + 4y + 4y = 25
2
8y - 28y + 24 = 0
2
2y - 7y + 6 = 0
(2y - 3)(y - 2) = 0
y = 3/2 AND y = 2
So the solutions are (4,3/2) and (3,2).
***
To solve this system which includes two quadratics, we use the LINEAR COMBINATION METHOD. We can add the equations and factor the result.
2 2
x + y = 25
2
-x + y = -5
-----------
2
y + y = 20
2
y + y - 20 = 0
(y + 5)(y - 4) = 0
So... y = -5 and y = 4
If y = 4, x = +/- 3
If y = -5, x = 0
The solutions are:
(3,4), (-3,4), and (0,-5)
***
Before pressing ENTER, describe in your own words how to solve a system of equations which includes quadratics.
If a graph is available, the intersection points can be located and the results tested in the equations. If no graph is available (or if the intersection points are difficult to determine) the substition or linear combination method can be used to solve the system.
WORKSPACE:
ANS:
#2 , #4 , #3 , #2 , #3 , 255 , 9 , 16 , 14 , #1 , #4 , #2 , -4 , 4 , 4 , -1,3 , 5,4 , #4 , 11 , 2,0 , 9/5 , 6
The below is continued. Current Date: May 4, 2006.
ELLIPSES AND HYPERBOLAS: Problem Set 3
NOTES:
The equation d = t^2 - 12t describes the dive of a submarine, where t is the time in minutes after beginning the dive and d is the depth in fathoms. How many minutes after beginning the dive does the submarine resurface? ANS: 12 (minutes)
2
d = t - 12t = 12
***
If d = t^2 - 12t describes the dive, what is the greatest depth reached by the submarine? ANS: 36
2 2 2
d = t - 12t = t - 12t = t - 12t + 36 - 36
***
The sun is one focus of an elliptical orbit of a planet. If the maximum distance from the planet to the sun is 6.8*10^7 miles and the minimum distances is 6.7*10^7 miles, find the distances from the sun to the other focus. ANS: 1 (millions miles)
WORKSPACE:
ANS:
12 , 36 , 1
QUADRATIC FUNCTIONS: Ellipses and Hyperbolas Test
NOTES:
The ellipse whose equation is...
2 2
x y
-- + -- = 1
36 20
...has foci on the.
ANS: #1 - x-axis
***
Given the equation...
2 2
x y
-- + -- = 1
16 25
...of an ellipse, find the x-intercepts of the graph.
ANS: #1 - +/- 4
***
Where does the graph of...
2 2
x y
-- - -- = 1
9 16
...intersect the x-axis?
ANS: #2 - +/- 3
***
Complete the equation for this hyperbola:
2 2
y x
-- - -- = 1
?? 9
ANS: #4 - 16
***
How many solution points are indicated by the graph of this system?
ANS: #4 - 4
***
Convert this equation from standard form to general form.
2
y = (x - 2) + 2
ANS: #2 - x^2 - 4x - y + 6 = 0
***
The conic section described by the equation...
2 2
x y
-- + -- = 1
4 4
...is a.
ANS: #1 - Circle
***
The conic section described by the equation...
2
-2x + 8x - y + 27 = 0
...is a.
ANS: #2 - Parabola
***
The conic section described by the equation...
2
x = (y - 2)
...is a.
NOT: #3 - Hyperbola , #1 - Ellipse
***
Choose the equation that describes this graph.
ANS: #4 - y^2/16 - x^2/4 = 1
WORKSPACE:
ANS:
IMAGINARY AND COMPLEX NUMBERS: Imaginary Numbers
- Define imaginary numbers
- Use imaginary numbers to solve quadratic equations
In Unit 2, you found real solutions to quadratic equations.
In this activity, you will solve some equations that have NO real roots.
NOTES:
The REAL SOLUTIONS on:
2
x - 18 = 0
2
x = 18
Are:
_ _
3\/2 OR -3\/2
The graph intersects the x-axis at:
_
x = +/- 3\/2
***
Solve this quadratic equation:
2
x - 27 = 0
2
x = 27
_
x = +/- 3\/3
The graph of f(x) = x^2 - 27 intersects the x-axis at:
_
x = +/- 3\/3
The real roots of a function occur where the graph crosses the x-axis.
***
You just saw graphs for f(x) = x^2 - 18 and f(x) = x^2 - 27.
Now consider:
2
f(x) = x + 9
How many x-intercepts does the graph of this function have? ANS: 0
The graph of this quadratic function does not intersect the x-axis.
Now let's try to solve the equation:
2
x + 9 = 0
2
x = -9
To solve this equation, we need a number whose square -9.
Try 3. 3^2 = 9.
Try -3. (-3)^2 = 9.
So, x can't equal 3 or -3.
***
2
x = -9
Until the eighteenth century, mathematicians agreed that there were no solutions to equations like these because the square of a non-zero real number is always positive.
To solve x^2 = -9, you need a number called an imaginary number. \/-9 is an IMAGINARY NUMBER.
__ 2
(\/-9) = -9
__ 2
(-\/-9) = -9
__ __
So, x = \/-9 or x = -\/-9
__ __
Therefore, either \/-9 or -\/-9 satisfies the equation x^2 + 9 = 0.
***
__ ___ __ ___
\/-2, \/-15, \/-3, and \/-16 are other examples of imaginary numbers.
***
Rene Descartes in the 17th century, distinguished between real and imaginary numbers. Later, Leonhard Euler used the letter "i" to represent \/-1.
__
i = \/-1
2 __ 2
i = (\/-1) = -1
You can simplify by using i.
__ _ __
\/-4 = \/4 = \/-1 = 2i
___ __ __
\/-16 = \/16 * \/-1 = 4i
___ __ __ _
\/-27 = \/27 * \/-1 = 3\/3i
____ ___ __ _
\/-300 = \/300 * \/-1 = 10\/3i
__ _ __ _
-\/-6 = -\/6 * \/-1 = -\/6i
__ _ __ _
-\/-5 = -\/5 * \/-1 = -\/5i
***
Now that you are familiar with imaginary numbers, you can solve equations like x^2 + 9 = 0.
Notice the IMAGINARY ROOTS.
2
x + 16 = 0
2
x = -16
___
x = +/- \/-16
x = 4i OR x = -4i
Recall that 4i and -4i are both solutions because (4i)^2 = -16 and (-4i)^2 = -16.
***
Evaluations like...
2
x - 9 = 0
2
x - 25 = 0
...have real roots.
Equations like...
2
x + 9 = 0
2
x + 25 = 0
...have imaginary roots.
***
Now that you can solve equations with imaginary roots, let's look at the graphs of equations with imaginary roots and real roots.
***
In your own words, describe how the graph of an equation and the nature of its roots are related.
When the roots are imaginary, there are no x-intercepts.
When there are two real roots, there are two x-intercepts.
When there is one real root, there is one x-intercept.
WORKSPACE:
ANS:
3,3 , x , 0 , 9 , 4i , 6i , 3i , 5i , 6i , 5i,-5i , \/3i,-\/3i , #3 , #2 , #1 , no , two , 6i , 5i , 4i,-4i , 2i , 5i,-5i , NO , YES , 0 , 2 , 1
IMAGINARY AND COMPLEX NUMBERS: Complex Numbers
- Define complex numbers
- Use the quadratic formula to solve equations that have complex roots
- Use the discriminant to determine the number and nature of roots
NOTES:
2
0 = x + 9
2
x = -9
__
x = +/- \/-9
x = +/- 3i
The function has two imaginary roots.
***
Some equations with no real roots, like x^2 + 25 = 0, are easy to solve using imaginary numbers.
Other equations with no real roots cannot be solved using only imaginary numbers.
***
To find the roots of:
2
x + 2x + 2 = 0
First complete the square.
2
(x + 2x + 1) - 1 + 2 = 0
2
(x + 1) - 1 + 2 = 0
2
(x + 1) + 1 = 0
2
(x + 1) = -1
Take the square root of both sides.
_______2 __
\/(x + 1) = +/- \/-1
x + 1 = +/- i
x = ?
***
2
f(x) = x + 2x + 2
x + 1 = +/- i
x = ?
The real numbers together with the pure imaginary numbers form a set. To be able to add two numbers inthis set, such as -1 and i, a new number, namely -1+i, must be invented.
-1+i is neither a real number nor a pure imaginary number; it is a COMPLEX NUMBER. Any number of the form:
a + bi
Is called a complex number.
***
2
f(x) = x + 2x + 2
Now you can complete the problem.
_______2 __
\/(x + 1) = \/-1
x + 1 = +/- i
x = -1 +/- i
***
Any number in the form a+bi is called a complex number.
"a" represents the REAL part of the number;
"bi" represents the IMAGINARY part.
***
If b = 0 in a+bi, the complex number is a real number.
The comple number -6+0i is the same as the real number -6.
If a = 0 in a+bi, the complex number is a pure imaginary number.
The complex number 0+2i is the same as the imaginary number 2i.
***
Complex Numbers
a + bi
REAL NUMBERS
a + 0i
OTHER COMPLEX NUMBERS
a + bi
/
a = 0
/
b = 0
PURE IMAGINARY NUMBERS
0 + bi
***
Now that you are more familiar with complex numbers, you will find the complex roots of some quadratic equations.
***
Earlier, you found complex roots of a function by completing the square.
2
f(x) = x + 2x + 2
2
0 = (x + 2x + 1) - 1 + 2
2
0 = (x + 2x + 1) + 1
2
-1 = (x + 1)
x = -1 + i OR x = -1 - i
***
You can also use the QUADRATIC FORMULA to find complex roots.
For any equation:
2
ax + bx + c = 0
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
Notice how the quadratic formula:
_2_____
-b +/- \/b - 4ac
----------------
2a
Is used to find the complex roots of the equation.
2
f(x) = x + 2x + 2
_2___________
-2 +/- \/2 - 4 * 1 * 2
x = ----------------------
2 * 1
_____
-2 +/- \/4 - 8
x = --------------
2
__
-2 +/- \/-4
x = -----------
2
-2 +/- 2i
x = ---------
2
x = -1 +/- i
***
In the quadratic formula, let's see what the DISCRIMINANT, the number under the radical, indicates about the roots.
_2_____
-b +/- \/b - 4ac
x = ----------------
2a
Consider the equation:
2
x - 7x + 12 = 0
____2____________
7 +/- \/(-7) - 4 * 1 * 12
x = -------------------------
2 * 1
The discriminant is:
2
(-7) - 4 * 1 * 12
49 - 48 = 1
The equation has two real roots.
7 +/- 1
x = -------
2
x = 3 OR x = 4
***
Before continuing, use your own words to describe the discriminant and what it tells you about the roots of an equation.
WORKSPACE:
ANS:
5i , 1,1 , i , i , 4 , 2i , -6 , 0i , 3 , 5i , 0 , 2i , -6 , 2i , 2i, 7i , i , 0 , 5 , 33 , 33 , 8 , 2 , -19 , 19i , two , one , imaginary , #2 , #3 , #1 , #1 , #3 , #2 , 63 , 3i , 6 , -4i , 0 + 4i , 12 + 0i , #2
The below is continued (some parts above are from this date, though). Current Date: May 23, 2006.
IMAGINARY AND COMPLEX NUMBERS: Operations
- Add, subtract, and multiply imaginary numbers
- Add, subtract, and multiply complex numbers
NOTES:
In the previous activities, you solved equations that had imaginary roots.
In this activity, you will learn to simplify expressions that contain imaginary and complex numbers.
***
Suppose you wanted to SIMPLIFY this expression:
4 3 2
2i - i + 4(i + 2) + 6i + 3
You would need to be able to ADD, SUBTRACT, and MULTIPLY imaginary and complex numbers.
You can add, subtract, and multiply IMAGINARY NUMBERS like this:
7i + 8i = 15i
3i - 8i = -5i
4(2i) = 8i
***
In your own words, describe how adding and subtracting imaginary numbers compares to adding and subtracting variables.
To add or subtract imaginary numbers, treat i as if it were a variable, then simplify.
2x + 3x = 5x
2i + 3i = 5i
2x - 3x = -x
2i - 3i = -i
***
You multiplied a real and an imaginary number. 3(8i) = 24i
Two imaginary numbers can be multiplied like this:
2
2i * 3i = 6i = 6(-1) = -6
__
Remember, i = \/-1
2
i = -1
***
You can add COMPLEX NUMBERS.
(5 + 2i) + (3 + i) = (5 + 3) + (2i + i)
(5 + 3) + (2 + 1)i
8 + 3i
You also can SUBTRACT complex numbers.
(5 + 2i) - (3 + i) = (5 + 2i) + (-3 - i)
(2 + i)
***
Notice how complex numbers are multiplied.
(5 + 2i)(3 + i)
5(3) + 5(i) + 2i(3) + 2i(i)
2
15 + 5i + 6i + 2i
15 + 11i + 2(-1)
13 + 11i
***
Before continuing, use your own words to describe how to perform operations with complex numbers.
Add, subtract, and multiply complex numbers as if they were binomials, then simplify.
WORKSPACE:
ANS:
14i , 17i , 12i , 24i , i , -24 , 7 , 8 , -i , -1 , 2i , -27i , imaginary , real , #3 , #2 , 3i , -1 , #3 , 37 , 13 , 13 , 17 + 23i , 1 + i , 9i , 10i , -2i , -i , -16 , -36 , 7 + 5i , 14 + 23i , 5 - 4i , 10
The below is continued. Current Date: May 24, 2006.
IMAGINARY AND COMPLEX NUMBERS: Graphs
- Graph imaginary numbers
- Graph complex numbers
NOTES:
Mathematicians use many kinds of numbers to reprensent the points in a coordinate system.
These points may be represneted by the positive integers like this:
|--1--2--3--4-->
Using this number line, you can show many mathematical concepts.
***
What if the cat walked three miles in the opposite direction?
To have a unique number assigned to every point on the number line requires some new numbers. The new numbers are NEGATIVE NUMBERS.
***
There is a ONE-TO-ONE CORRESPONDENCE between a point on the line and the number assigned to that point. On this line, there is only one point called "2".
***
What if the cat walks up?
To preserve a one-to-one correspondence between points and numbers, you cannot use positive or negative numbers.
New numbers are needed to represent the new points: IMAGINARY NUMBERS.
***
Imaginary numbers, like real numbers, may be represented in a coordinate system.
Points on the HORIZONTAL axis are REAL numbers.
Points on the VERTICAL axis are IMAGINARY numbers.
***
Since complex numbers (3 + 4i) consist of a real and an imaginary part, you can use them to represent any point in the plane formed by the real and imaginary axes.
WORKSPACE:
ANS:
3 , 4 , 3/4 , 2 , 5i , (NAVIGATE) , (NAVIGATE) , (NAVIGATE) , 0 + 2i , 0 - 3i , -2 + 0i , 3 - 4i , -4 + i , Real , Imaginary , 4 + 4i , 0 + 3i , -6 - 9i , -8 + 0i , 7 + 7i
EVALUATE AND GRAPH POLYNOMIALS: Evaluating Polynomials
- Evaluate functions at real and complex numbers, and at literal expressions
NOTES:
In Units 1 and 2 you graphed linear and quadratic functions by evaluating the functions and plotting points (x,f(x)).
f(x) = x + 3
f(-2) = (-2) + 3
f(-2) = 1
So, the point (-2,1) lies on the graph.
***
In this activity you will evaluate POLYNOMIALS of higher degree at values for x that include irrational and complex numbers, and literal expressions.
***
You evaluated functions at rational, irrational, and imaginary values of x.
You can also evaluate functions at the literal expression.
***
Describe in your own words how to evaluate a polynomial function.
Substitute the value of the variable in the expression and simplify.
WORKSPACE:
3
x - 2x + 4
ANS:
#2 , 4 , 0 , -6 , 2 , -4 , 11 - 3\/3 , 11 + 3\/3 , #1 , #2 , #3 , #2 , #4 , a^2 - 4 , a^2 - 2a - 3 , \/3 + 4 , a , 2 , #1 , 12
The below is continued. Current Date: May 24, 2006.
EVALUATE AND GRAPH POLYNOMIALS: Graphing Polynomials
- Plot points to determine the nature of graphs of polynomials
- Determine the number of possible x-intercepts of polynomial functions
NOTES:
In Unit 2, you saw that graphs of quadratic equations followed patterns.
The patterns helped you sketch their graphs.
For example, this parabola increases as x increases when x > 0.
[picture of curved line going up with it's only root at (0,0)]
In this activity, you will find patterns for graphs of higher degree equations, like y = x^3 and y = x^4.
The patterns will help you sketch graphs of polynomials.
***
These points satisfy the CUBIC:
3
f(x) = x
x |f(x)
---+----
-1 | -1
0 | 0
1 | 1
-3 | -27
-2 | -8
2 | 8
3 | 27
The points on the graph satisfy the function:
3
f(x) = x
Notice the value of f(x) at:
x = 4 f(4) = 64
Find the value of f(x) at:
x = 5 f(5) = 125
x = 8 f(8) = 512
The function is CONSTANTLY INCREASING as x increases.
***
These points satsify:
3
f(x) = (-1/2)x
x |f(x)
---+----
-2 | 4
-1 | 1/2
0 | 0
1 | -1/2
2 | -4
***
You just saw one pattern of cubic graphs. They are either constantly increasing or constantly decreasing.
Now you will compare graphs of cubics and graphs of parabolas.
***
Graphs of quadratic functions are SYMETRICAL about a vertical line.
You can plot points to determine how cubic functions are symetrical.
***
These points satisfy:
3
f(x) = 2x
x |f(x)
---+----
-2 | -16
-1 | -2
0 | 0
1 | 2
2 | 16
When the f(x)-values are opposites for positive and negative x-values, the graph is SYMMETRICAL about the origin.
***
Before continuing, use your own words to describe the function:
3
f(x) = x
And it's graph.
The function is a CONSTANTLY INCREASING function. That is, as x increases, f(x) increases. The graph has SYMMETRY with respect to the origin.
***
Next, you will see how the values for a, h, and k affect the graph of:
3
a(x - h) + k
***
Notice the values of h and k in the equation of this graph.
3
f(x) = (x - 5) + 2
h = 5 k = 2
***
Describe in your own words how a, h, and k affect the graph of a cubic equation.
The value of "a" changes the STEEPNESS and direction of the graph.
The value of "h" translates the graph RIGHT OR LEFT.
The value of "k" translates the graph UP OR DOWN.
***
Before continuing, use your own words to describe the number of possible x-intercepts of a cubic function.
The graph of a cubic function may intersect the x-axis in 1, 2, or 3 points.
***
You just saw some characteristic patterns in the graphs of cubics.
Graphs of quartics and other higher degree functions also have unique patterns.
***
These points satisfy the QUARTIC:
4
f(x) = x
x |f(x)
---+----
-2 | 16
-1 | 1
0 | 0
1 | 1
2 | 16
These points have symmetry with respect to the y-axis.
For x > 0, the function is increasing. For x < 0 the function is decreasing.
***
Now you will see how the values for a, h, and k affect the graph of:
4
y = a(x - h) + k
***
Use your own words to describe the number of possible x-intercepts for the graph of a quartic function of the form:
4
f(x) = a(x - h) + k
The graph crosses the x-axis in 0, 1, or 2 points.
***
The graphs you just saw are special cases of quartic functions.
Now you will graph a variety of other fourth degree functions.
***
Here's the graph of:
4 2
f(x) = -x + 8x + 1
Intersecting the x-axis in 2 points.
***
Before continuing, use your own words to describe the number of possible x-intercepts of a quartic function.
The graph of a quartic function may cross the x-axis in 0, 1, 2, 3, or 4 points.
***
Higher degree functions follow the patterns set by cubic and quartic functions.
***
Before continuing, use your own words to describe the number of x-intercepts on the graph of a higher degree function.
- The graph of a function of degree n intersects the x-axis in at most n points.
- If n is even, the graph might not cross the x-axis at all.
- If n is odd, the graph has at least one x-intercept.
WORKSPACE:
ANS:
-27 , -8 , 8 , 27 , 125 , 512 , -1/2 , -4 , #1 , #1 , #4 , #1 , 0 , 2 , 16 , 2 , 3 , -2 , -1,-3 , 2,0 , 1 , 1 , 0 , 1 , 16 , 1,2 , -1/4,1 , 3,-3 , 4 , 4 , 2 , 3 , 5 , 2 , 6 , 1 , 7 , n , one , #2 , #2 , #3 , #4 , #4
EVALUATE AND GRAPH POLYNOMIALS: Transforming Polynomials
- Transform polynomial equations algebraically and graphically
NOTES:
Here are the graphs of y = x + 4 and y = 3.
[shows graph]
By SUBSTITUTION x + 4 = 3.
TRANSFORMING the equation gives x = -1.
Here are the graphs of y = x and y = -1.
[shows another graph]
These two lines intersect at x = -1, the solution of x + 4 = 3.
Similar transformations can be performed on CUBIC and QUARTIC equations.
***
Describe in your own words what happens to the solution set of two equations when you transform each side.
Performing algebraic operations on both sides of an equation transforms it into equivalent equations.
Although the equations change in form, the graphs show that the solutions are preserved.
WORKSPACE:
ANS:
1 , 4x^2 , 3 , x , 1 , -3 , 3/5 , preserved , 5x^2 , -7 , 5x + 7 , 1 , -7/5
EVALUATE AND GRAPH POLYNOMIALS: Problem Set 1
NOTES:
Stereo amplifiers have "characteristic equations." For any root, a + bi, "a" must be less than or equal to zero, or there will be distortion.
Find the value of "a" for the amplifier with the characteristic equation:
2
x - 6x + 25 = 0
***
Two hoses working together fill a pool in 5 hours. The larger hose fills the pool in 3 hours less than the smaller hose working alone.
How long does it take the larger hose alone to do the work?
***
A cyclist goes 10 miles at a certain speed. She then goes 5 more miles at a new speed, 5 mph slower. The total trip takes 2 hours.
What was the first speed?
WORKSPACE:
ANS:
3 , 8.72 , 10
The below is continued. Current Date: May 30, 2006
FACTORS AND ROOTS OF POLYNOMAILS: Cubic Equations
- Build polynomials by multiplying factors
- Find factors of polynomial funtions by looking at their graphs
NOTES:
In the last activity, you saw that graphs of cubic equations cross the x-axis in one, two, or three places.
In this activity you will BUILD EQUATIONS of cubics that intersect the x-axis in one, two, or three places.
***
To build a cubic equation, MULTIPLY:
(x - 2)(x - 2)(x - 2)
To form:
3 2
y = x - 6x + 12x - 8
3
Here's the graph of y = (x - 2). 2 is a root of MULTIPLICITY 3.
***
To build a cubic equation, multiply (x + 3)(x + 3)(x + 3) to form:
3 2
y = x + 9x + 27x + 27
The graph intersects the x-axis once. A real root of the equation is -3.
It's multiplicity is 3.
***
To build a cubic equation, multiply x(x - 2)(x - 2) to form:
3 2
y = x - 4x + 4x
The graph intersects the x-axis twice. 2 is a double root. The other real root is 0.
***
To build a cubic equation, multiply x(x - 2)(x + 1) to form:
3 2
y = x - x - 2x
Where will the graph cross the x-axis? -1, 0, and 2.
***
To build a cubic equation using IMAGINARIES, multiply (x - i)(x + i)(x + 3) to form:
3 2
y = x + 3x + x + 3
The graph crosses the x-axis at the real root -3. From the factors you know that the two imaginary roots are i and -i.
***
To build a cubic equation using imaginaries, multiply (x + 2i)(x - 2i)(x + 2) to form:
3 2
y = x + 2x + 4x + 8
The imaginary roots are 2i and -2i. From the graph the one real root is -2.
***
To build a cubic equation using imaginaries, multiply (x + i)(x - i)(x - i) to form:
3 2
y = x - x + x - 1
The imaginary roots are i and -i. From the graph the one real root is 1.
***
Describe in your own words how you can build cubic equations.
Multiply factors to form a polynomial in expanded form.
The factors indicate where the graph crosses the x-axis.
Every cubic equation has three roots. Possible combinations include:
- 1 real root of multiplicity 3.
- 1 real root and 1 double real root.
- 3 distinct real roots.
- 1 real root and 2 imaginary roots.
***
Given factors, you have built cubic equations.
You also can build a cubic equation using x-intercepts from the graph.
***
From the graph, you see the roots are -5, -2, and 3.
[shows picture of graph, all x-intercepts are at those points]
So, the factors are (x + 5), (x + 2), and (x - 3).
Multiplying the factors gives the cubic equation:
3 2
y = x + 4x - 11x - 30
***
In your own words, describe how you can write a cubic equation from its graph.
1. Read the roots from the graph.
2. Form factors from the roots.
3. Multiply factors to write the equation.
***
You have built polynomial functions by multiplying factors.
y = (x + 1)(x + 2)(x + 3)
You also can build the GRAPH of a polynomial function by multiplying the GRAPHS OF THE FACTORS.
Let's graph these LINEAR FACTORS:
(x + 1)(x + 2)(x + 3)
y = x + 1
1
y = x + 2
2
y = x + 3
3
At the point x = 0...
1 * 2 * 3 = 6
y = (x + 1)(x + 2)(x + 3)
3 2
y = x + 6x + 11x + 6
Notice the arena in which the graphs of all three factors are positive. The graph of the product also is positive.
All y-values for the graph of the product between -3 < x < -2 are positive since (-) * (-) * (+) = (+).
***
Let's graph these linear factors:
x(x + 3)(x - 1)
y = x
1
y = x - 1
2
y = x + 3
3
At the point x = 2...
2 * 1 * 5 = 10
y = x(x - 1)(x + 3)
3 2
y = x + 2x - 3x
Notice that for -3 < x < 0, (-) * (-) * (+) = (+).
When x > 1, the y-values are positive.
***
Let's graph a linear factor and a quadratic factor.
2
(x - 1)(x + x + 1)
y = x - 1
1
2
y = x + x + 1
2
At the point x = -2...
-3 * 3 = -9
2
y = (x - 1)(x + x + 1)
3
y = x - 1
The graph of the cubic crosses the x-axis at the point where the linear graph crosses the x-axis.
***
You have build the graph of a cubic function by multiplying graphs of factors.
y = (x + 3)(x - 2)(x - 1)
You can also build the graph of a cubic by adding graphs of terms.
3
y = x + 4
***
You can add the terms of a cubic to find the graph of the cubic.
3
y = x + 4
3
Here is the graph of y = x.
Here is the graph of y = 4.
At x = 0, 0 + 4 = 4.
3
Here is the graph of y = x + 4.
Notice that it is the sum of any pair of points on the other two graphs.
WORKSPACE:
ANS:
27x , -3 , 3 , 4x , 0 , x^2 , -1,0,2 , 2x^2 , -2 , -1 , 1 , 3 , 1 , 3 , 2 , #1 , #4 , factors , 0 , 0 , 0 , -6 , 10 , 0 , 4 , 6 , 0 , -20 , -2 , -1 , 0 , 7 , 3 , 5 ,