1. Algebra 2
Chapter 5 Notes
Quadratic Functions

2. Graphing Quadratic Equations
Axis of Symmetry,
The vertical line through the vertex
5.1
Graphing Quadratic Equations
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. Vertex Form of a Quadratic Equations
5.1
Vertex Form of a Quadratic Equations
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. Intercept Form of a Quadratic Equation
5.1
Intercept Form of a Quadratic Equation
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. 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 2
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 + Outer + Inner + Last ]
5.1
FOIL Method
FOIL Method for changing intercept form or vertex form to standard form:
[ First + Outer + Inner + Last ]
( x + 3 ) ( x + 5 )
= x2 + 5 x + 3 x + 15
= x2 + 8 x + 15

7. Solving Quadratic Equations by Factoring
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. 5.2 [ First + Outer + 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. Solving Quadratic Equations by Factoring Special Factoring Patterns
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. Solving Quadratic Equations Factoring Monomials First
5.1
Solving Quadratic Equations
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. Finding Zeros of Quadratic Functions
5.1
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. Solving Quadratic Equations
5.3
Solving Quadratic Equations
Radical sign
Radican:
Radical
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 ± r
x = x ½ 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. Solving Quadratic Functions “Rationalizing the denominator”
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. 5.4 Complex Numbers − 5 = − 1 • 5 = − 1 • 5 = i 5
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
Complex Numbers
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
5.4
Imaginary Number
Imaginary Number
i = − 1
and
i2 = − 1
Imaginary Number Squared

16. What is the Square Root of – 25?
5.4
Imaginary Number
What is the Square Root of – 25?
? = − 25
= −
= i ± 5

17. ( Real number + imaginary number ) Pure Imaginary Numbers
5.4
Complex Numbers
( Real number + imaginary number )
Complex Numbers ( a + b i )
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. 5.4
Plot Complex Numbers
Imaginary
( − i ) ●
Real ●
(2 − 3 i )

19. Complex Numbers: Add, Subtract, Multiply
5.4
Complex Numbers: Add, Subtract, Multiply
( 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 Numbers: Divide and Complex Conjugates
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. Complex Numbers: Absolute Value
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. 5.5 Completing the Square RULE: x2 + b x + c, where c = ( ½ 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. Examples of Completing the Square
5.1
Examples of Completing the Square
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. Completing the Square
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. Quadratic Functions in Vertex Form
5.1
Quadratic Functions in Vertex Form
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. Quadratic Formula
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. Quadratic Formula
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
Quadratic Formula
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
Quadratic Formula
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
Quadratic Formula
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. Quadratic Formula 5.6 The Quadratic Formula and the Discriminant
EQUATION
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. 5.6 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. Using the Discriminant
5.6
The Quadratic Formula and the Discriminant
Using 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. ● ● Graphing & Solving Quadratic Inequalities
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]
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. Graphing & Solving Quadratic Inequalities
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 ●