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Algebra 2
Chapter 5 Notes
Quadratic Functions
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:
1. 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
2a
4. Axis of symmetry is one vertical line, x = ─ b
2a
Example: Graph y = 2 x2 – 8 x + 6
Lowest or highest point of
the quadratic function
a=2,b=─8,c=6
Since a > 0 , parabola opens up
X- coordinate
=
─
Y- coordinate
=
2 (2)2 – 8 (2) + 6
b
2a
Vertex,
(─8)
2 (2)
─
=8
4
=2
=
─ 2
}
Vertex
(x,y)
(2,─ 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
Since a < 0 ,
parabola opens
down
h = −3
k =
Vertex =
4
(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
a = −1
2
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 )
p = −2
q =
4
X –intercepts = ( 4 , 0 ) and (− 2 , 0 )
(− 2 , 0 )
●
●
x=1
a = −1
Since a > 0 ,
parabola opens
down
Axis of symmetry : x = 1 , which is halfway
between the x-intercepts
Plot 2 pts on one side of axis of symmetry
y = −1 ( 1 + 2 ) ( 1 – 4 )
Y=9
(4 , 0 )
Graphing Quadratic Equations
Name of
Form
Equation Form
Standard
y = ax2 + bx + c
Vertex
y = a (x – h) 2 + k
Intercept y = a (x – p) (x – q)
How do you find the xcoordinate of the Vertex
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 is (h, k)
then substitute another value for x to
get another point
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
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
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
Factors of 28 =
( 28 • 1 )
( 14 • 2 )
Example 2 : Factoring a trinomial of the form, ax2 + b x + c
Factor: 3x2 − 17 x + 10
Factors of 10 =
( 10 • 1 )
Factors of 3 =
(3•1)
“What are the factors of 28 that
combine to make a difference of − 12?”
(7•4)
“What are the factors of 10 and 3 that
combine to add up to − 17, when
multiplied together?”
(5•2)
3 • − 5 + 1 • − 2 = − 17
[ 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)
where + is >
What are the factors of 18
that have difference of + 7 ?
ax2 – bx – c
( – )( + )
where – is >
z2 – 6z – 27
(z–9)(z +3)
where – is >
What are the factors of 27
that have difference of – 6 ?
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 )
a2 + 2ab + b2 = ( a + b ) 2
x2 + 12x + 36 = ( x + 6 ) 2
a2 – 2ab + b2 = ( a – b ) 2
x2 – 8x + 16 = ( x – 4 ) 2
Perfect Square Trinomial
{
4 x2 – 25 =
(2x) 2 – (5) 2 =
(2 x + 5 ) (2 x – 5)
Difference of 2
Squares
9 y2 + 24 y + 16 =
(3y) 2 + 2 (3y)(4) + 42 =
(3y + 4) 2
Perfect Square
Trinomial
49 r2 – 14r + 1 =
(7r) 2 – 2 (7r) (1) =
( 7r – 1) 2
Perfect Square
Trinomial
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 + 45 = 3t – 5
2t2 – 20t + 50 = 0
t2 – 10t + 25 = 0
(t – 5) 2 = 0
t–5=0
t=5
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)
NOTES: Page 60, Section 5.3
x = x½
Radical sign
Solving Quadratic Functions
r is a square root of s if r2 = s
3 is a square root of 9 if 32 = 9
Radican:
r
Radical
3
Since (3)2 = 9 and (-3)2 = 9, then 2 square roots of 9 are: 3 and – 3
Therefore,
±
r or ±
9
Examples of Perfect Squares
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
( a > 0 , b > 0)
Quotient Property
ab =
a
=
b
4
6 •
15 =
6
•
9 is 3x3
16 is 4x4
25 is 5x5
36 is 6x6
49 is 7x7
64 is 8x8
81 is 9x9
100 is 10x10
36
b
4
=
9
a
b
Examples:
24 =
a
4 is 2x2
=
2
6
90
=
9
•
10
=
3
4
9
10
=
4 •
9
NOTES: Page 60a, Section 5.3
Solving Quadratic Functions
A Square Root expression is considered simplified if
1. No radican has a Perfect Square other than 1
2. There is no radical in the denominator
“Rationalizing the denominator”
Examples
7
7 =
16
16
=
7
7
2
4
=
7
2
2
2
=
2
Solve:
Solve:
2 x2 + 1 = 17
1 ( x + 5)2 =
3
2 x2 = 16
x2 =
8
X =±
4
X =± 2
7
( x + 5)2 = 21
• ±
2
2
( x + 5)2 = ±
21
x+5 = ±
21
X = –5 ±
21
{
X =–5 +
21
and
X = –5 –
14
21
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 =
−1
and i2 = − 1
Property of the square root of a negative number:
If r = + real number, then
−r =
−5 =
(i
r )2 = − 1 • r = − r
(i
5 )2 = − 1 • 5 = − 5
−1• r = −1 •
r
= i
r
−1• 5 = −1 •
5
= i
5
Solving Quadratic Equation
3 x2 + 10 = − 26
x =
3 x2 = − 36
x = i
x2 = − 12
x2 = − 12
x = − 12
−1
12
4•3
x = ±2 i
3
Imaginary Number
i =
−1
Imaginary Number Squared
and
2
i
=−1
What is the Square
Root of – 25?
? = − 25
=
=
−1
25
i
±5
NOTES: Page 61a, Section 5.4
( Real number + imaginary number )
Complex Numbers ( a + b i )
Real Numbers
Imaginary Numbers
(a+bi)
(a+0i)
−1
5
2
(2+3i)
(5−5i)
Pure Imaginary Numbers
3
( 0 + b i ) , where b ≠ 0
∏
2
(−4i)
(6i)
NOTES: Page 62, Section 5.4
Imaginary
Complex Numbers: Plot
(−3+2i)
●
Real
●
(2 − 3 i )
NOTES: Page 62a, Section 5.4
Complex Numbers: Add and Subtract
a) ( 4 − i ) + ( 3 − 2 i ) = 7 − 3 i
b) ( 7 − 5 i ) − ( 1 − 5 i ) = 6 + 0 i
c) 6 − ( − 2 + 9 i ) + ( − 8 + 4 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)
(7−4i )(−1+2i )=
− 7 + 4 i + 14 i − 8 i 2
− 7 + 18 i − 8 (−1)
− 7 + 18 i + 8
1 + 18 i
b)
(6+3i )(6−3i )=
36 + 18 i − 18 i − 9 i 2
36 + 0 i − 9 (−1)
36 + 0 + 9
45
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 #.”
5 + 3i
•
1 − 2i
1 + 2i
1 + 2i
Complex Conjugates
( a + b i ) • ( a − b i ) = REAL #
( 6 + 3 i ) • ( 6 − 3 i ) = REAL #
=
5 + 10 i + 3 i + 6 i2
1 + 2 i – 2 i – 4 i2
=
5 + 13 i + 6 (– 1 )
1 – 4 (– 1 )
=
– 1 + 13 i
5
=
– 1 + 13 i
5
5
[ standard form ]
NOTES: Page 62, Section 5.4
Complex Numbers: Absolute Value
Imaginary
(−1+5i)
●
Z =a+bi
│Z│=
(3 + 4 i )
●
a2 + b2
Absolute Value of a complex number
is a non-negative real number.
Real
●
a) │ 3 + 4 i │ =
32 + 42 =
b) │ − 2 i │ = │ 0 + 2 i │ =
c) │− 1 + 5 i │=
− 12 + 52 =
25
= 5
02 + ( − 2 )2 = 2
26
≈ 5.10
(−2i)
NOTES: Page 64, Section 5.5
Completing the Square
x
b
2
x
b
x2
bx
x
b
2
x
b
2
x2
b x
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
Example 1
x2 − 7 x + c “What is ½ of the linear coefficient SQUARED?”
c = [ ½ (− 7 ) ] 2 = ( − 7 ) 2 = 49
2
4
x2 − 7 x + 49
4
= ( x − 7 )2
2
Perfect Square Trinomial = the Square of a Binomial
Example 2
x2 + 10 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
x2 + 10 x − 3 = 0
x2 + 10 x
= +3
x2 + 10 x + 25 = + 3 + 25
( x + 5 )2 = 28
( x + 5 )2 =
28
x+5 =
4
7
x = –5 ±2
7
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 – 6 x + 12 = 0
3
x2 – 2 x + 4 = 0
x2 – 2 x
=–4
x2 – 2 x + 1 = – 4 + 1
( x – 1 )2 = – 3
( x – 1 )2 =
(x–1) =
x = +1±i
–3
–1
3
3
As + 4 isn’t [ ½ (– 2) ]2 , move 4 to other side of =
What is [ ½ (– 2) ]2 = (– 1)2 = 1 ?
NOTES: Page 65a, Section 5.5
Writing Quadratic Functions in Vertex Form
y = a ( x − h )2 + k
y = x2 – 8 x + 11
11 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 + 16 = ( x2 – 8 x + 16 ) + 11
What is [ ½ ( – 4 ) ]2 = (– 4 )2 = 16
y + 16 = ( x – 4 )2 + 11
– 16
– 16
y = ( x – 4 )2 – 5
( x2 – 8 x + 16 ) = ( x – 4 )2
(x,y) =(4,–5)
NOTES: Page 66, Section 5.6
The Quadratic Formula and the Discriminant
Quadratic Equation
x = −b ±
b2 − 4 ac
2a
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
a
Complete the square
( + to both sides of = )
x2 + b x + ( b )2 = − c + ( b )2 = b2 − 4 a c
a
( 2a )
a ( 2a )
4 a2
Binomial Squared
(x + b )2 = b2 − 4 a c
2a
4 a2
= − c
a
Square Root both sides of =
(x + b )2 =
2a
b2 − 4 a c
4 a2
Squared Root undoes Squared
x+ b = ±
2a
b2 − 4 a c
2a
Solve for x by − b to both sides
2a
x = − b ±
2a
b2 − 4 a c
2a
=
−b ±
[ combine both terms]
b2 − 4 a c
2a
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
NOTES: Page 67, Section 5.6
x = −b ±
The Quadratic Formula and the Discriminant
b2 − 4 ac
2a
Ex 1: Solving a quadratic equation with 2 real solutions
a
x2
+bx+c=0
2 x2 + 1 x = 5
2 x2 + 1 x − 5 = 0
x= −b ±
b2 − 4 a c
2a
x= −1 ±
12 − 4 (2 ) ( −5 )
2(2)
x= −1 ±
41
4
NOTES: Page 67, Section 5.6
x = −b ±
The Quadratic Formula and the Discriminant
b2 − 4 ac
2a
Ex 2: Solving a quadratic equation with 1 real solutions
a
x2
+bx+c=0
1 x2 − 1 x = 5 x − 9
1 x2 − 6 x + 9 = 0
x= −b ±
b2 − 4 a c
2a
x = − (−6) ±
(−6) 2 − 4 (1 ) ( 9)
2(1)
x=
6 ±
0
2
X = 3
NOTES: Page 67, Section 5.6
x = −b ±
The Quadratic Formula and the Discriminant
b2 − 4 ac
2a
Ex 3: Solving a quadratic equation with 2 imaginary
solutions
a
x2
+bx+c=0
−1 x2 + 2 x = 2
−1 x2 + 2 x − 2 = 0
x= −b ±
b2 − 4 a c
2a
x = − (2) ±
(2) 2 − 4 (−1 ) (− 2)
2 (−1 )
x = − (2) ±
−4
2
x = −2 ± 2i
−2
x = − 2(1 ± I )
−2
x = 1±i
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 ±
x2 − 6 x + 10 = 0
x2 − 6 x + 9 = 0
x2 − 6 x + 8 = 0
(− 6 )2 − 4 (1) (10 ) = − 4
(− 6 )2 − 4 (1) (9) = 0
(− 6 )2 − 4 (1) (8) = 4
b2 − 4 a c
2a
x = − (− 6 ) ±
−4
2 (1)
x = − (− 6 ) ± 2 i = 3 ± i
2 (1)
x = − (− 6 ) ±
0
2 (1)
x = − (− 6 ) ± 0 = 3
2 (1)
x = − (− 6 ) ±
2 (1)
+4
x = − (− 6 ) ± 2 = 2 or 4
2 (1)
Quadratic Equation in Standard Form:
a x2 + b x + c = 0
3 x2 – 11 x – 4 = 0
x = + 11 ±
(11)2 − 4 (3) (– 4)
2 (3)
x = + 11 ±
121+ 48
2 (3)
x = + 11 ±
6
169
x = + 11 ± 13 = 24 , – 2 = 4 , – 1
6
6
6
3
x = −b ±
b2 − 4 ac
2a
Sum of Roots: – b
a
4+–1 =
11
3
3
Product of Roots: c
a
4•–1 = –4
3
3
Solve this Quadratic Equation: a x2 + b x + c = 0
x2 + 2 x – 15 = 0
Factoring
x 2 + 2 x – 15 = 0
Quadratic Formula
x = −b ±
(x–3)(x+5)=0
x – 3 = 0 or x + 5 = 0
x = 3 or x = – 5
Completing the Square
x 2 + 2 x – 15 = 0
x2+ 2x
( x + 1 ) 2 = 16
(x+1) =
x 2 + 2 x – 15 = 0
x = – 2 ±
16
x = – 1 ± 4 = 3 or – 5
(– 2 )2 − 4 (1) (– 15)
2 (1)
x = – 2 ±
4 + 60
2
= + 15
x 2 + 2 x + 1 = + 15 + 1
b2 − 4 ac
2a
x = – 2 ±
64
2
x = – 2 ± 8 = 3 or – 5
2
NOTES: Page 68a, Section 5.6
The Quadratic Formula and the Discriminant
IMMAGINARY
x2 − 6 x + 10 = 0 = 3 ± i
No intercept
One intercept
x2 − 6 x + 9 = 0 = 3
Two intercepts
x2 − 6 x + 8 = 0 = 2 or 4
●
● ●
REAL
y > a x2 + b x + c [graph of the line is a dash]
NOTES: Page 69, Section 5.7
Graphing & Solving Quadratic Inequalities
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]
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
Vertex (standard form) = − b = − (− 2 ) = 1
2a
2 (1 )
y = 1 x2 − 2 x − 3
y = 1 (1)2 − 2 (1) − 3 = − 4
Vertex = ( 1 , − 4 )
Line of symmetry = 1
●
●
●
●
x
Y
1
−4
3
0
1
0
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
NOTES: Page 69a, Section 5.7
Graphing & Solving Quadratic Inequalities
x
y
−1
2
2 1
4
−2
0
1
0
y < − x2 − x + 2
y < − ( x2 + x − 2 )
y <−(x−1)(x+2)
y < − x2 − x + 2
y < − ( − 1 )2 − (− 1 ) + 2
2
2
y < −1 +1 +2
4
2
y < 2 1
4
y
●
●
●
●
x
x
y
0
−4
2
0
−2
0
y ≥ x2 − 4
y ≥ (x−2)(x+2)
●
x = −b = 0 = 0
2a 2
y ≥ x2 − 4
y ≥ ( 0 )2 − 4
y ≥ −4