1. Algebra 2
Chapter 5 Notes
Quadratic Functions

2. } ● X- coordinate = 2 a 2 (2) = 8 4 = 2 Vertex ( x , y ) ( 2 , ─ 2 )
Axis of Symmetry,
The vertical line through the vertex
NOTES: Page 56, Section 5.1
Graphing a Quadratic Function
Quadratic Function in standard form: y = a x2 + b x + c
Quadratic functions are U-shaped, called “Parabola.” ●
Graph of a Quadratic Function:
If parabola opens up, then a > 0 [POSITIVE VALUE]
If parabola opens down, then a < 0 [NEGATIVE VALUE]
2. Graph is wider than y = x2 , if│a│< 1
Graph is narrower than y = x2 , if │a│> 1
3. x-coordinate of vertex = ─ b
2 a
4. Axis of symmetry is one vertical line, x = ─ b
Vertex,
Lowest or highest point of the quadratic function
Example: Graph y = 2 x2 – 8 x + 6
a = 2 , b = ─ 8 , c = 6
Since a > 0 , parabola opens up
X- coordinate =
─ b
2 a
─ (─8)
2 (2)
= 8 4
= 2 }
Vertex
( x , y )
( 2 , ─ 2 )
Y- coordinate
2 (2)2 – 8 (2) + 6
─ 2

3. (− 3 , 4 ) ● a = − 1 2 h − 3 k 4 ( h , k ) (− 3 , 4 ) ● ● (− 5 , 2 )
NOTES: Page 57, Section 5.1
Vertex & Intercept Forms of a Quadratic Function
Vertex form: y = a ( x – h ) 2 + k
(− 3 , 4 ) ●
Example 1: Graph y = −1 ( x + 3 ) 2 + 4 2 a =
− 1 2
Since a < 0 , parabola opens down h
− 3 k 4
Vertex
( h , k )
(− 3 , 4 )
Axis of symmetry : x = − 3
Plot 2 pts on one side of axis of symmetry ● ●
(− 5 , 2 )
(− 1 , 2 )
x = − 3

4. NOTES: Page 57a, Section 5.1
Vertex & Intercept Forms of a Quadratic Function
Intercept form: y = a ( x – p ) ( x – q ) ●
(1 , 9 )
Example 2: Graph y = −1 ( x + 2 ) ( x – 4 ) a =
− 1
Since a > 0 , parabola opens down p
− 2 q 4
X –intercepts = ( 4 , 0 ) and (− 2 , 0 )
Axis of symmetry : x = 1 , which is halfway between the x-intercepts
Plot 2 pts on one side of axis of symmetry
(− 2 , 0 )
(4 , 0 ) ● ●
x = 1
y = −1 ( ) ( 1 – 4 )
Y = 9

5. How do you find the x-coordinate of the Vertex
Graphing Quadratic Equations
Name of Form
Equation Form
How do you find the x-coordinate of the Vertex
Standard
y = ax2 + bx + c
x = – b 2a
Then substitute x into equation to get y of the vertex, then substitute another value for x to get another point
Vertex
y = a (x – h) 2 + k
Vertex is (h, k)
then substitute another value for x to get another point
Intercept
y = a (x – p) (x – q)
x = midpoint between p and q,
then substitute x into equation to get y of the vertex, then substitute another value for x to get another point

6. [ First + Outter + Inner + Last ]
NOTES: Page 57b, Section 5.1
FOIL Method for changing intercept form or vertex form to standard form:
[ First + Outter + Inner + Last ]
( x + 3 ) ( x + 5 )
= x2 + 5 x + 3 x + 15
= x2 + 8 x + 15

7. So, the sum of ( m + n ) must = b and the product of m n must = c
NOTES: Page 58, Section 5.2
Solving Quadratic Equations by Factoring
Use factoring to write a trinomial as a product of binomials
x2 + b x + c = ( x + m ) ( x + n )
= x2 + ( m + n ) x + m n
So, the sum of ( m + n ) must = b and the product of m n must = c
Example 1 : Factoring a trinomial of the form, x2 + b x + c
Factor: x2 − 12 x − 28
“What are the factors of 28 that combine to make a difference of − 12?”
Factors of =
( 28 • 1 )
( 14 • 2 )
( 7 • 4 )
Example 2 : Factoring a trinomial of the form, ax2 + b x + c
Factor: 3x2 − 17 x + 10
“What are the factors of 10 and 3 that combine to add up to − 17, when multiplied together?”
Factors of =
( 10 • 1 )
( 5 • 2 )
3 • − • − 2 = − 17
Factors of 3 =
( 3 • 1 )

8. [ First + Outter + Inner + Last ]
( x + 3 ) ( x + 5 )
= x2 + 5 x + 3 x + 15
= x2 + 8 x + 15
Signs of Binomial Factors for Quadratic Trinomials
The four possibilities when the quadratic term is +
ax2 + bx + c
( ) ( )
x2 + 8x + 15
( x + 5 ) ( x + 3 )
What are the factors of 15 that add to + 8 ?
ax2 – bx + c
( – ) ( – )
y2 – 7y + 12
( y – 4 ) ( y – 3 )
What are the factors of 12 that add to – 7 ?
ax2 + bx – c
( ) ( – )
where + is >
r2 + 7r – 18
( r + 9 ) ( r – 2 )
What are the factors of 18 that have difference of + 7 ?
ax2 – bx – c
( – ) ( )
where – is >
z2 – 6z – 27
( z – 9 ) ( z + 3 )
What are the factors of 27 that have difference of – 6 ?

9. Special Factoring Patterns
NOTES: Page 58a, Section 5.2
Solving Quadratic Equations by Factoring
Special Factoring Patterns
Name of pattern
Pattern
Example
Difference of 2 Squares
a2 – b2 = ( a + b ) ( a – b )
x2 – 9 = ( x + 3) ( x – 3 )
Perfect Square Trinomial {
a2 + 2ab + b2 = ( a + b ) 2
x x + 36 = ( x + 6 ) 2
a2 – 2ab + b2 = ( a – b ) 2
x2 – 8x + 16 = ( x – 4 ) 2
4 x2 – 25 =
(2x) 2 – (5) 2 =
(2 x + 5 ) (2 x – 5)
Difference of 2 Squares
9 y y + 16 =
(3y) (3y)(4) + 42 =
(3y + 4) 2
Perfect Square Trinomial
49 r2 – 14r + 1 =
(7r) 2 – 2 (7r) (1) =
( 7r – 1) 2

10. Factoring Monomials First
NOTES: Page 59, Section 5.2
Factoring Monomials First
5x2 – 20 =
5 ( x2 – 4) =
5 (x + 2) (x – 2)
6p2 – 15p + 9 =
3 (2p2 – 5 p + 3) =
3 ( 2p – 3) ( p – 1 )
2u2 + 8 u =
2 u ( u + 4)
4 x2 + 4x + 4 =
4 ( x2 + x + 1)
Zero Product Property: If A • B = 0, the A = 0 or B= 0
With the standard form of a quadratic equation written as ax2 + bx + c = 0,
if you factor the left side, you can solve the equation.
Solve Quadratic Equations
x2 + 3x – 18 = 0
(x – 3) (x + 6) = 0
x – 3 = 0 or x + 6 = 0
x = 3 or x = – 6
2t2 – 17t = 3t – 5
2t2 – 20t + 50 = 0
t2 – 10t = 0
(t – 5) 2 = 0
t – 5 = 0
t = 5

11. • • (– 2 , 0 ) ( 3 , 0 ) NOTES: Page 59a, Section 5.2
Finding Zeros of Quadratic Functions
x – intercepts of the Intercept Form: y = a (x – p ) ( x – q)
p = (p , 0 ) and q = (q , 0)
Example:
y = x2 – x – 6
y = ( x + 2 ) ( x – 3 ), then Zeros of the function are p = – 2 and q = 3. • •
(– 2 , 0 )
( 3 , 0 )

12. Examples of Perfect Squares
NOTES: Page 60, Section 5.3
Solving Quadratic Functions
r is a square root of s if r2 = s
3 is a square root of 9 if 32 = 9
Since (3)2 = 9 and (-3)2 = 9, then 2 square roots of 9 are: 3 and – 3
Therefore, ± or ±
Radical sign
Radican:
Radical
x = x ½ r 3 r 9
Examples of Perfect Squares
4 is 2x2
9 is 3x3
16 is 4x4
25 is 5x5
36 is 6x6
49 is 7x7
64 is 8x8
81 is 9x9
100 is 10x10
Square Root of a number means:
What # times itself = the Square Root of a number?
Example: 3 • 3 = 9, so the Square Root of 9 is 3.
Product Property ab = a • b 36 = 4 • 9
( a > 0 , b > 0)
Quotient Property a b a 4 9 4 = = b 9
Examples: 24 = 4 6 = 2 6 = = 6 • 15 90 9 • 10 = 3 10

13. “Rationalizing the denominator”
NOTES: Page 60a, Section 5.3
Solving Quadratic Functions
A Square Root expression is considered simplified if
No radican has a Perfect Square other than 1
There is no radical in the denominator
Examples
“Rationalizing the denominator” 7 2 14 7 7 7 16 7 2 = = = = 4 16 2 2 2
Solve:
2 x2 + 1 = 17
2 x2 = 16
x2 =
X = ±
X =
Solve: 1 3
( x + 5)2 = 7
( x + 5)2 = 21
( x + 5)2 =
x + 5 =
X = – • 2
± 2 2
X = – + {
and
X = – –

14. − 5 = − 1 • 5 = − 1 • 5 = i 5 NOTES: Page 61, Section 5.4
Complex Numbers
Because the square of any real number can never be negative, mathematicians had
to create an expanded system of numbers for negative number
Called the Imaginary Unit “ i “
Defined as i = − and i2 = − 1
Property of the square root of a negative number:
If r = + real number, then − r = − 1 • r = − 1 • r = i r
− 5 = − 1 • 5 = − 1 • = i
( i r )2 = − 1 • r = − r
( i )2 = − 1 • 5 = − 5
Solving Quadratic Equation
3 x = − 26
3 x2 = − 36
x2 = − 12
x2 = − 12
x = − 12
x = −
x = i • 3
x = ± 2 i 3

15. Imaginary Number Squared
and i2 = − 1
Imaginary Number Squared

16. What is the Square Root of – 25?
? = − 25
= −
= i ± 5

17. ( Real number + imaginary number ) Pure Imaginary Numbers
NOTES: Page 61a, Section 5.4
Complex Numbers ( a + b i )
( Real number + imaginary number )
Imaginary Numbers
Real Numbers
( a + b i )
( a + 0 i )
( i )
( 5 − 5 i )
− 1 5 2
Pure Imaginary Numbers 3
( 0 + b i ) , where b ≠ 0 ∏ 2
( − 4 i )
( 6 i )

18. Imaginary Real NOTES: Page 62, Section 5.4 Complex Numbers: Plot●
Real ●
(2 − 3 i )

19. NOTES: Page 62a, Section 5.4
Complex Numbers: Add and Subtract
( 4 − i ) + ( 3 − 2 i ) = 7 − 3 i
( 7 − 5 i ) − ( 1 − 5 i ) = i
6 − ( − i ) + ( − i ) = 0 − 5 i = − 5 i
Complex Numbers: Multiply
a) 5 i ( − 2 + i ) = − 10 i + 5 i2 = − 10 i + 5 ( − 1) = − 5 − 10 i
b) ( 7 − 4 i ) ( − i ) =
b) ( i ) ( 6 − 3 i ) =
− i i − 8 i 2
i − 18 i − 9 i 2
− i − 8 (−1)
− i + 8 i
i − 9 (−1) 45

20. Complex Conjugates ( a + b i ) • ( a − b i ) = REAL #
NOTES: Page 62b, Section 5.4
Complex Numbers: Divide and Complex Conjugates
CONJUGATE means to
“Multipy by same real # and same imaginary # but with opposite sign to eliminate the imaginary #.”
Complex Conjugates
( a + b i ) • ( a − b i ) = REAL #
( i ) • ( 6 − 3 i ) = REAL # i
1 − 2 i i
i + 3 i + 6 i2
i – 2 i – 4 i2 = •
i + 6 (– 1 )
1 – 4 (– 1 ) =
– i 5 =
– i =
[ standard form ]

21. Imaginary Real NOTES: Page 62, Section 5.4
Complex Numbers: Absolute Value
Imaginary
( − i )
Z = a + b i
│ Z │ = a2 + b2 ●
( i ) ●
Absolute Value of a complex number is a non-negative real number.
Real ●
( − 2 i )
│ i │ = = = 5
│ − 2 i │ = │ i │ = ( − 2 )2 = 2
c) │− i │= − = ≈

22. RULE: x2 + b x + c, where c = ( ½ b )2
NOTES: Page 64, Section 5.5
Completing the Square b 2 x b x x x
b x 2 x2 bx x2 b 2 b 2
b x 2
( b )2
( 2 )
RULE: x2 + b x + c, where c = ( ½ b )2
In a quadratic equation of a perfect square trinomial,
the Constant Term = ( ½ linear coefficient ) SQUARED.
x2 + b x + ( ½ b )2 = ( x + ½ b )2
Perfect Square Trinomial = the Square of a Binomial

23. Example 1
x2 − 7 x + c “What is ½ of the linear coefficient SQUARED?”
c = [ ½ (− 7 ) ] 2 = ( − 7 ) 2 = 49
x2 − 7 x + 49 4
= ( x − 7 )2 2
Perfect Square Trinomial = the Square of a Binomial
Example 2
x x − 3 “Is − 3 half of the linear coefficient SQUARED?”
[ if NOT then move the − 3 over to the other side of = , then replace it with the number that is half of the linear coefficient SQUARED ]
c = [ ½ (+ 10 ) ] 2 = ( 5 ) 2 = 25
x x − 3 = 0
x x = + 3
x x =
( x + 5 )2 = 28
( x + 5 )2 =
x =
x = – 5 ±

24. NOTES: Page 65, Section 5.5
Completing the Square where the coefficient of x2 is NOT “ 1 “
3 x2 – 6 x + 12 = 0 3
x2 – 2 x = As + 4 isn’t [ ½ (– 2) ]2 , move 4 to other side of =
x2 – 2 x = – 4
x2 – 2 x = – What is [ ½ (– 2) ]2 = (– 1)2 = 1 ?
( x – 1 )2 = – 3
( x – 1 ) = –
x = ± i

25. NOTES: Page 65a, Section 5.5
Writing Quadratic Functions in Vertex Form y = a ( x − h )2 + k
y = x2 – 8 x doesn’t work here, so move 11 out of the way and replace the constant “c” with a # that makes a perfect square trinomial
y = ( x2 – 8 x + 16 ) What is [ ½ ( – 4 ) ]2 = (– 4 )2 = 16
y = ( x – 4 ) ( x2 – 8 x + 16 ) = ( x – 4 )2
– – 16
y = ( x – 4 )2 – 5 ( x , y ) = ( 4 , – 5 )

26. NOTES: Page 66, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a
Quadratic Equation
a x2 + b x + c = 0
Divide by a to both sides of =
x2 + b x + c = 0
a a
− c to both sides a
x2 + b x = − c
a a
Complete the square ( + to both sides of = )
x2 + b x + ( b )2 = − c + ( b )2 = b2 − 4 a c [ combine both terms]
a ( 2a ) a ( 2a ) a2
Binomial Squared
(x + b )2 = b2 − 4 a c
2a a2
Square Root both sides of =
(x + b )2 = b2 − 4 a c
2a a2
Squared Root undoes Squared
x + b = ± b2 − 4 a c
2a a
Solve for x by − b to both sides 2a
x = − b ± b2 − 4 a c
2a a
= − b ± b2 − 4 a c

27. NOTES: Page 66a, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a
Number and type of solutions of a quadratic equation determined by the DISCRIMINANT
If b2 − 4 a c > 0
Then equation has 2 real solutions
If b2 − 4 a c = 0
Then equation has 1 real solutions
If b2 − 4 a c < 0
Then equation has 2 imaginary solutions

28. Ex 1: Solving a quadratic equation with 2 real solutions
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a
Ex 1: Solving a quadratic equation with 2 real solutions
a x2 + b x + c = 0
x = − b ± b2 − 4 a c
2 a
2 x2 + 1 x = 5
x = − 1 ± − 4 (2 ) ( −5 )
2 ( 2 )
2 x2 + 1 x − 5 = 0
x = − 1 ± 4

29. Ex 2: Solving a quadratic equation with 1 real solutions
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a
Ex 2: Solving a quadratic equation with 1 real solutions
a x2 + b x + c = 0
x = − b ± b2 − 4 a c
2 a
1 x2 − 1 x = 5 x − 9
x = − (−6) ± (−6) 2 − 4 (1 ) ( 9)
2 ( 1 )
1 x2 − 6 x + 9 = 0
x = ± 2
X = 3

30. Ex 3: Solving a quadratic equation with 2 imaginary solutions
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac 2a
Ex 3: Solving a quadratic equation with 2 imaginary solutions
a x2 + b x + c = 0
x = − b ± b2 − 4 a c
2 a
−1 x2 + 2 x = 2
x = − (2) ± (2) 2 − 4 (−1 ) (− 2)
2 (−1 )
−1 x2 + 2 x − 2 = 0
x = − (2) ± − 4 2
x = − 2 ± 2 i −2
x = − 2(1 ± I )
x = 1 ± i

31. NOTES: Page 68, Section 5.6
The Quadratic Formula and the Discriminant
EAUATION
DISCRIMINANT
SOLUTIONS
a x2 + b x + c = 0
b2 − 4 a c
x = − b ± b2 − 4 a c
2 a
x2 − 6 x + 10 = 0
(− 6 )2 − 4 (1) (10 ) = − 4
x = − (− 6 ) ± − 4
2 (1)
x = − (− 6 ) ± 2 i = 3 ± i
x2 − 6 x + 9 = 0
(− 6 )2 − 4 (1) (9) = 0
x = − (− 6 ) ±
x = − (− 6 ) ± = 3
x2 − 6 x + 8 = 0
(− 6 )2 − 4 (1) (8) = 4
x = − (− 6 ) ±
x = − (− 6 ) ± = 2 or 4

32. x = − b ± b2 − 4 ac 2a a x2 + b x + c = 0 3 x2 – 11 x – 4 = 0
Quadratic Equation in Standard Form:
a x2 + b x + c = 0
3 x2 – 11 x – 4 = 0
x = − b ± b2 − 4 ac 2a
x = ± (11)2 − 4 (3) (– 4) (3)
x = ± (3)
x = ±
x = ± = 24 , – 2 = 4 , –
Sum of Roots: – b a
4 + – 1 =
Product of Roots: c a
4 • – 1 = – 4

33. Factoring Quadratic Formula Completing the Square
Solve this Quadratic Equation: a x2 + b x + c = 0
x x – 15 = 0
Factoring
Quadratic Formula
x x – 15 = 0
( x – 3 ) ( x + 5 ) = 0
x – 3 = 0 or x + 5 = 0
x = 3 or x = – 5
x = − b ± b2 − 4 ac 2a
x x – 15 = 0
x = – 2 ± (– 2 )2 − 4 (1) (– 15) (1)
x = – 2 ±
x = – 2 ±
x = – 2 ± = 3 or –
Completing the Square
x x – 15 = 0
x x = + 15
x x =
( x + 1 ) 2 = 16
( x + 1 ) =
x = – 1 ± 4 = 3 or – 5

34. ● ● ● NOTES: Page 68a, Section 5.6
The Quadratic Formula and the Discriminant
IMMAGINARY
x2 − 6 x + 10 = 0 = 3 ± i
No intercept
x2 − 6 x + 9 = 0 = 3
One intercept
Two intercepts
x2 − 6 x + 8 = 0 = 2 or 4
REAL ● ● ●

35. ● ● y > a x2 + b x + c [graph of the line is a dash]
y ≥ a x2 + b x + c [graph of the line is solid]
y < a x2 + b x + c [graph of the line is a dash]
y ≤ a x2 + b x + c [graph of the line is solid]
NOTES: Page 69, Section 5.7
Graphing & Solving Quadratic Inequalities
Vertex (standard form) = − b = − (− 2 ) = 1
2a (1 )
y = 1 x2 − 2 x − 3
y = 1 (1)2 − 2 (1) − 3 = − 4
Vertex = ( 1 , − 4 )
Line of symmetry = 1
Example 1: y > 1 x2 − 2 x − 3
0 = (x − 3 ) ( x + 1 )
So, either (x − 3 ) = 0 or ( x + 1 ) = 0
Then x = 3 or x = − 1 x Y 1
− 4 3 ● ● ●
Test Point (1,0) to determine which side to shade
y > 1 x2 − 2 x − 3
0 > 1 (1)2 − 2 (1) − 3
0 > 1 − 2 − 3
0 > − 4 This test point is valid, so graph this side ●

36. NOTES: Page 69a, Section 5.7
Graphing & Solving Quadratic Inequalities y x y
− 1 2
2 1 4
− 2 1 x y
− 4 2 −2 ● x ● ● ● ●
y < − x2 − x + 2
y < − ( x2 + x − 2 )
y < − ( x − 1 ) ( x + 2 )
y < − ( − 1 )2 − (− 1 ) + 2
y < −
y < 2 1 4
y ≥ x2 − 4
y ≥ ( x − 2 ) ( x + 2 )
x = −b = 0 = 0 2a
y ≥ ( 0 )2 − 4
y ≥ − 4 ●