1. A detailed summary of all of the information within chapter 5

2. Table of Contents Review- Exponents…………………………………………….…….1 slide
Practice- Exponents…………………………….… slide
Lesson 5.1: Graphing Quadratic Equations ……………….……..8 slides
Lesson 5.2: Solving Quadratic Equations with factoring…………5 slides
Lesson 5.3: Solving Quadratic Equations with square roots……2 slides
Lesson 5.4: Complex Numbers……………………………….……8 slides
Lesson 5.5: Completing the Square……………………….………4 slides
Lesson 5.6: The Quadratic Formula and the Discriminant………6 slides
Lesson 5.7: Graphing and Solving Quadratic Inequalities………2 slides
Lesson 5.8: Modeling with Quadratic Functions………………….5 slides
Review………………………………………………………………...5 slides

3. X Review- Exponents 2 Base
An Exponent is a multiple determining how many times the base is multiplied. X 2
Exponent
Base

4. Practice- Exponents What is the value of 32 ? 32 = 3∙3 = 9
Remember that a Negative Exponent refers to “1/” the exponent.
What is the value of 3-4?
3-4 = 1/34 = 1/3∙3∙3∙3 = 1/81

5. Lesson 5.1: Graphing Quadratic Functions
A standard quadratic function is written in the form
Y=ax2+bx+c
And appears in the shape of a parabola, or U-Shape Curve. Parabolas are symmetrical over the Y-axis, because exponents always create positive numbers.
Graph of Y=X2

6. Lesson 5.1: Graphing Quadratic Functions
Most Quadratic Equations (Such as x2+4x+3) have a zero, root, or solution.
x2+4x+3=0 (Which is equal to (x+1)(x+3)=0)
Roots
X=-1,-3 Which are solutions.
They are also Zeros, where (x,0) occurs.

7. Lesson 5.1: Graphing Quadratic Functions
The highest or lowest point on the U-Shaped Curve is known as the Vertex.
When an equation is written as Y=ax2+bx+c, the vertex’s X-value is located at –b/2a.

8. Lesson 5.1: Graphing Quadratic Functions
What is the vertex of Y=2x2-3x+4?
-b/2a = -(-3)/2(2) = ¾
Y=2(3/4)2 -3(3/4) +4 = 2(9/16) – 9/4 +4;
9/8 – 9/4 + 4 = 9/8 – 18/8 +4 = -9/8 + 32/8 = 23/8
So, the Vertex of this Equation is (3/4, 23/8).

9. Lesson 5.1: Graphing Quadratic Equations
After plotting the Vertex (3/4, 23/8), plot nearby points and connect them to form a parabola.
Y=2X2-3X+4
Points based on Equation:
(0,4)
(-1,9)
(1,3)
(2,6)

10. Lesson 5.1: Graphing Quadratic Equations
Besides Standard Form, there are two other common Quadratic Expressions. Both are graphed in the same way, but provide information that can help you.
Vertex Form
Intercept Form
Y=a(x-h)2 +k Y=a(x-m)(x-n)
Where the intercepts, or zeros, (Y=0) can easily be identified as (m, 0) and (n, 0), and the vertex X-value is the number between the two points.
Where the vertex can easily be identified as (h,k).

11. Lesson 5.1: Graphing Quadratic Equations
Graph Y=2(x-1)2 +4.
1. Find Vertex (h,k): (1,4)
2. Graph nearby points; (0,6) (2,5) (-1, 12) (3,12)
3. Connect the points.

12. Lesson 5.1: Graphing Quadratic Equations
Graph Y=3(x-3)(x-7).
Find the Y-Intercepts: (3,0) (7,0)
Find the X-value in-between, which is the vertex: (7+3)= 10; 10/2=5
Plot Vertex, (5,-12) and other points (4,-9), (6,-9)
Connect the points to draw the graph.

13. Lesson 5.2: Solving Quadratic Equations with Factoring
Solving a quadratic equation involves changing the changing the standard quadratic equation
Y=ax2+bx+c
To a binomial term (intercept form) which will have two roots.
Y=(x-m)(x-n)

14. Lesson 5.2: Solving Quadratic Equations with Factoring
Factor 2x2+x-3.
To do this, first set up two sets of parentheses.
( )( )
Then, examine a and c. Are either of them prime? If they are prime, there only factors are 1 and itself. Fill in the parentheses as necessary.
Somewhere in the parentheses, there is a 2x, an x, a 1 or -1, and a 3 or -3, Because 3 and 2 are prime.
Now that you know the factors (2x, x, 3, 1), you need to organize them so that the two numbers multiply to c, and the x values add to b.

15. Lesson 5.2: Solving Quadratic Equations with Factoring
Try out some combinations using F.O.I.L.
Does (2x-3)(x-1) work?
FIRST: 2x∙x=2x2
INSIDE: ∙x=-3x
LAST: ∙-1=3
OUTSIDE: 2x∙-1=-2x

16. Lesson 5.2: Solving Quadratic Equations with Factoring
Combine like terms to get 2x2-5x+3; not the solution. But if 3 were positive, would it work?
(2x+3)(x-1)
Use F.O.I.L.
You’ll get 2x2+x-3; you’ve found the solution.

17. Lesson 5.2: Solving Quadratic Equations with Factoring
Here’s another method.
Factor 6x2+x-2.
First, find factors of 6: 2/3, 3/2, -2/-3, -3/-2
Then, find factors of -2: -2/1, 1/-2, 2/-1, -1/2
Which pair would make sense? You want a positive x; therefore, you should use a pair of 2 and -1. 2∙3=6; so use the pair of positive 3.
You’ll get (2x+2)(3x-1). Using FOIL, change the digits to match the equation. The correct combination is (2x-1)(3x+2).

18. Lesson 5.3: Solving Quadratic Equations with Square Roots
Solve 2x2+5=41.
This isn’t too hard; start out by solving it normally:
2x2=36
x2=18
x=√18, OR
x=-√18.
However, remember that exponents always create positive numbers. That means there are two solutions here:

19. Lesson 5.3: Solving Quadratic Equations with Square Roots
Solve (2x-5)2=81.
Again, it will solve normally; but don’t forget that there are two solutions when it comes to exponents.
(2x-5)2=y+81
2x-5=9 –OR- 2x-5=-9
2x=14 –OR- 2x=-4
x=7, -2
Both solutions work perfectly well; it’s almost exactly like absolute value equations.

20. Lesson 5.4: Complex Numbers
Solve x2+15=-21.
This problem doesn’t make sense; you would need a negative square root. How can a square root be negative? It can’t. But, the imaginary number i is used to represent negative square roots.
Remember that i=√-1.
With the number i, unreal solutions can be simplified.
x2+15=-21
x2=-36
x=√-1∙36
x=i√36
x=6i; the solution

21. Lesson 5.4: Complex Numbers
Complex Numbers are graphed on a real-imaginary plane, where the Y-axis represents the number of i, and the X-Axis represents the number of real numbers.

22. Lesson 5.4: Complex Numbers
Where does our solution, 6i, go on this plane?
On the Y-axis, because 6i has no real numbers, only i’s.

23. Lesson 5.4: Complex Numbers
How would you add (4-i) + (3+2i)?
Simply combine like terms.
(4+3) + (-i+2i)
7+i
Remember that i is always goes behind the real number, in this case 7 goes before the i.

24. Lesson 5.4: Complex Numbers
How do you multiply Complex numbers?
To multiply Complex Numbers, remember that i2=-1. Other than that fact, the multiplication is standard.
What is 4i(6-i)?
4∙6(i) – 4(i2)
24i-4(-1)
24i+4
Solution: 4+24i

25. Lesson 5.4: Complex Numbers
How do you divide Complex Numbers?
Dividing complex numbers is more difficult than other operations. To divide Complex Numbers, it is similar to normal division except you must multiply by a complex conjugate (similar to an inverse).
A complex conjugate of a+bi is a-bi.
To practice this, we will solve -7+6i ∕9-4i.

26. Lesson 5.4: Complex Numbers
-7+6i∕9-4i
Find C.C.: 9+4i
(9-4i)(9+4i)=97
-7+6i/97
Solution: -7/97 + 6i/97

27. Lesson 5.4: Complex Numbers
Finally, how do you find the absolute value of a complex number?
To find the absolute value of a complex number, ignore the i and square all the terms, then find the square root. (This is similar to a distance formula; your finding the distance from 0). Remember absolute value is always positive.
What is the absolute value of 4-8i?
Remember, absolute values are always positive.
(4)2 + (-8)2
16+64=80; Find √80
Simplifies to 4√5
The absolute value of 4-8i is 4√5.

28. Lesson 5.5: Completing the Square
The Ultimate Goal of completing a square is to change a trinomial, such as this:
x2+20x+100
To a binomial squared, such as this:
(x+10)2

29. Lesson 5.5: Completing the Square
Find the value of c to make x2+15x+c a perfect square trinomial.
The easiest way to do this is to use the formula x2+bx+(b/2)2 = (x + b/2)2.
x2+15x+(15/2)2=(x+15/2)2
x2+15x+(225/4) = (x+15/2)2
Perfect Square Trinomial
Resulting square of a binomial

30. Lesson 5.5: Completing the Square
It is very important not to forget the formula, x2 + bx +(b/2) = (x+(b/2)2, because it can be used for many different problems.

31. Lesson 5.5: Completing the Square
Solve x2+3x-1=0.
Remember (b/2)2
(3/2)2=9/4
Add one to both sides, x2 +3x = 1
Add 9/4 to both sides, x2 +3x +9/4 = 13/4
Regroup to a squ. Binomial, (x+3/2)2=13/4
Square Root both sides, x+3/2=√13/2 OR x+3/2=- √13/2
x= √13/2 -3/2, - √13/2 -3/2
Always, Always… Two Solutions!

32. Lesson 5.6: The Quadratic Formula and the Discriminant
There is a formula that can be used to find the exact value of a quadratic equation Y-intercept (When X=0).
Graph of Y=x2+3x-10
Normally when graphing Quadratic Equations, you’re Y-intercept is difficult to determine, in terms of exact value.

33. Lesson 5.6: The Quadratic Formula and the Discriminant
Using ax2+bx+c, the value of the X-intercepts are:
X= (-b +-√(b2-4ac) / 2a

34. Lesson 5.6: The Quadratic Formula and the Discriminant
So, on the equation we previously graphed, Y= x2+3x-10, the X-intercepts are:
~And~
X= (-(3) +√((3)^2-4(1)(-10)) / 2(1)
X= -3 +√9+40 / 2(1)
X= -3 +√49 / 2
X=4/2
X=2
X= (-(3) -√((3)^2-4(1)(-10)) / 2(1)
X= -3 -√9+40 / 2(1)
X= -3 -√49 / 2
X= -10/2
X=-5
Final Solution: X=2,-5
Like completing the square, it’s all about remembering the basic equation.

35. Lesson 5.6: The Quadratic Formula and the Discriminant
A part of the Quadratic Formula can be used to tell how many solutions a Quadratic Equation has. It could have two real solutions, one real solution, or two imaginary solutions (i.e. no real solution). It is called the Discriminant.
Blue: Two Solutions
Green: One Solution
Red: Two Imaginary

36. Lesson 5.6: The Quadratic Formula and the Discriminant
The Discriminant formula is b2-4ac.
If:
>0, then the equation has two real solutions.
=0, then the equation has one real solution.
<0, then the equation has two imaginary solutions.

37. Lesson 5.6: The Quadratic Formula and the Discriminant
How many solutions are there for:
(x-3)2
x2-6x+8
x2-6x+10
Change to standard form: x2-6x+9
(-6)2 -4(1)(9) = = 0; One Real Solution
(-6)2 -4(1)(8) = = 4 = 4>0; Two Real Solutions
(-6)2 -4(1)(10) = = -4 = -4<0; Two Imaginary Solutions

38. Lesson 5.7: Graphing and Solving Quadratic Inequalities
To graph a Quadratic Inequality:
Graph the Equation, as if Y=x2…
Remember to dash the line for < or >, shade for ≤ or ≥.
Use a Test Point to determine whether to shade inside or outside the parabola.

39. Lesson 5.7: Graphing and Solving Quadratic Inequalities
Graph Y ≥ -2x2-x+3.
Find the Vertex (-b/2a): /-4 = -(1/4); (-1/4, 25/8)
Plot the graph as normal: (0,3) (-1,2) (1,0) (-2,-3)
≥ Means you should have a solid line (Draw that)
Test (0,0). (0,0) is NOT a solution (0≥3?), so shade OUTSIDE the parabola.
The Graph is Complete.
Systems of Inequalities work in the exact same way.

40. Lesson 5.8: Modeling with Quadratic Functions
Sometimes with Quadratics you will already have a graph, and instead need to find out the equation of the graph.
A good memory of the different forms of equations (Intercept, Standard, Vertex) can help you find out the equation quickly.

41. Lesson 5.8: Modeling with Quadratic Functions
What is the equation for this graph?
You know the two intercepts, so try graphing it in intercept form.
Y=a(x-m)(x-n)
Y=a(x+2)(x-1)
Test (-1,-6); -6=a(1)(-2)
-6=-2a
a=3
(-2, 0)
(1,0)
Therefore, the equation of this graph is Y=3(x+2)(x-1)
(-1,-6)

42. Lesson 5.8: Modeling with Quadratic Functions
What is the equation for this graph?
You know the Vertex, so try using Vertex Form this time.
Y=a(x-h)2 +k
Y=a(x-1)2+0
Test (-1,-3); -3=a(-1-1)2
-3=a(4)
a=-3/4
Vertex: (1,0)
(-1,-3)
So, the equation of this graph is Y=-3/4(x-1)2

43. Lesson 5.8: Modeling with Quadratic Functions
What is the equation of this graph?
You can’t use Intercept- or Vertex-Form here… but because there are three points, you can make three equations with standard form.
Y=ax2+bx+c
(-4,0)
(0,1)
(0,1): 1=a(0)2+b(0)+c; 1=c
(-4,0): 0=a(-4)2+b(-4)+c; 0=16a-4b+c
(-5,-4): -4=a(-5)2+b(-5)+c; -4=25a-5b+c
(-5,-4)

44. Lesson 5.8: Modeling with Quadratic Functions
With these three equations, you can find the value of a, b, and c.
c=1
16a-4b+c=0
25a-5b+c=-4
a= -3/4
b= -11/4
c=1
So, the final equation is
Y=(-3/4)x2-(11/4)x+1

45. Chapter 5 Review Factor 8y2-28y-60.
Start off by removing like terms (In this case, 4):
2y2-7y-15
Find the factors of the numbers:
2: 1,2 or -1,-2
-7: -7,1 or -1,7
-15: 3,-5 5,-3 or 15,-1 1,-15
By trial-and-error, find the correct combination:
(2y+3)(y-5) = 2y2-7y (2y-15)(y+1) = 2y2-13y-15
(2y-5)(y+3) = 2y2+y-15

46. Chapter 5 Review y+4 y Recall the formula for area of a trapezoid:
Find y if the area of the trapezoid is 71units2.
y+4 y
Recall the formula for area of a trapezoid:
A=((b1+b2)/2) *h
Add in the information given:
((y+9y-10)/2) *(y+4)
(5y-5) *(y+4)
5y2+15y-14 = 71
Solve with square roots:
9y-10
5y2+15y-14 = 71 5y2+15y=85
y2+5y=17 y2+5y+6.25=23.25
(y+2.5)2=23.25
y+2.5=√23.25
y= √
y=~7.32

47. Chapter 5 Review
How many natural numbers between 1 and 50 have a value of 3 when the number’s square is added to eight times the number and subtracted by 2?
This is a word problem. It is important in word problems to convert the words into an algebraic expression.
the number’s square is added to eight times the number and subtracted by 2
have a value of 3
x2+8x-2
= 3
Knowing the solution needs to be between 1 and 50 is not useful here until the solution is found!

48. Chapter 5 Review Simplify: x2+8x-2=3 Simplify: x2+8x-5=0
Recall the quadratic formula:
X= (-b +-√(b2-4ac) / 2a
x= (-8+-√(82-4(1)(-5) /2(1)
x= (-8+-√(84) /2
x= ~ or ~

49. Chapter 5 Review Recall the list of possible answers.
.58258<1; not a solution, because 1<x<50
<1; not a solution, because 1<x<50
So, the answer is that there are no whole numbers that fit the requirement.