Post on 17-Dec-2015
Algebra 2
Chapter 5 Notes
Quadratic Functions
NOTES: Page 56, Section 5.1
Graphing a Quadratic Function
Quadratic Function in standard form: y = a x2 + b x + cQuadratic functions are U-shaped, called “Parabola.”
●Vertex,
Lowest or highest point of the quadratic function
Axis of Symmetry,The vertical line through
the vertex
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│> 13. x-coordinate of vertex = ─ b
2 a 4. Axis of symmetry is one vertical line, x = ─ b
2 a
Example: Graph y = 2 x2 – 8 x + 6 a = 2 , b = ─ 8 , c = 6Since 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
NOTES: Page 57, Section 5.1
Vertex & Intercept Forms of a Quadratic Function
Vertex form: y = a ( x – h ) 2 + k
●
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
x =
− 3
● ●
(− 3 , 4 )
(− 1 , 2 )(− 5 , 2 )
NOTES: Page 57a, Section 5.1
Vertex & Intercept Forms of a Quadratic Function
Intercept form: y = a ( x – p ) ( x – q )
●
Example 2: Graph y = −1 ( x + 2 ) ( x – 4 )
a = − 1Since 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
x =
1
● ●
(1 , 9 )
(4 , 0 )(− 2 , 0 )
y = −1 ( 1 + 2 ) ( 1 – 4 )Y = 9
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 + kVertex 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
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
“What are the factors of 28 that combine to make a difference of − 12?”
Factors of 28 = ( 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 = ( 10 • 1 ) ( 5 • 2 )3 • − 5 + 1 • − 2 = − 17
Factors of 3 = ( 3 • 1 )
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 ?
[ First + Outter + Inner + Last ]
( x + 3 ) ( x + 5 )
= x2 + 5 x + 3 x + 15
= x2 + 8 x + 15
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 x2 + 12x + 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 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) = 0x – 3 = 0 or x + 6 = 0x = 3 or x = – 6
2t2 – 17t + 45 = 3t – 52t2 – 20t + 50 = 0t2 – 10t + 25 = 0(t – 5) 2 = 0t – 5 = 0t = 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
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
r
3
r 9 Examples of Perfect Squares4 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
Product Property ab = a b
===
=
==
=
• 36 = •4 9
Quotient Property ab
b
a 49
4
9Examples:
24 4 6 62
3 10156 90 9 10• •
( a > 0 , b > 0)
x = x ½
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.
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 12. There is no radical in the denominator
Examples
716
= = = =416
7
2
•
77 72
2
2 14
2
Solve:
2 x2 + 1 = 17
2 x2 = 16
x2 = 8
X = ± 4
X =
“Rationalizing the denominator”
± 2
± 2 2
Solve:
13
( x + 5)2 = 7
( x + 5)2 = 21
( x + 5)2 = 21 x + 5 = 21
X = – 5 21
±
±
±
X = – 5 21
X = – 5 21
+
–and{
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 = − 1 • r = − 1 • r = i r
− 5 = − 1 • 5 = − 1 • 5 = i 5
( i r )2 = − 1 • r = − r
( i 5 )2 = − 1 • 5 = − 5 Solving Quadratic Equation
3 x2 + 10 = − 26
3 x2 = − 36
x2 = − 12
x2 = − 12
x = − 12
x = − 1 12
x = i 4 • 3
x = ± 2 i 3
i = − 1
and i2 = − 1
Imaginary Number
Imaginary Number Squared
? = − 25
What is the Square Root of – 25?
= − 1 25 = i ± 5
NOTES: Page 61a, Section 5.4
Complex Numbers ( a + b i )
Real Numbers Imaginary Numbers
Pure Imaginary Numbers
( a + 0 i )
( 0 + b i ) , where b ≠ 0
( a + b i )
( Real number + imaginary number )
( 2 + 3 i ) ( 5 − 5 i )
( − 4 i ) ( 6 i )
− 1 52
3
∏ 2
NOTES: Page 62, Section 5.4
Complex Numbers: Plot Imaginary
Real
●
●( − 3 + 2 i )
(2 − 3 i )
NOTES: Page 62a, Section 5.4
Complex Numbers: Add and Subtract
a) ( 4 − i ) + ( 3 − 2 i ) = 7 − 3 ib) ( 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 − 4 i ) ( − 1 + 2 i ) =
− 7 + 4 i + 14 i − 8 i 2
− 7 + 18 i − 8 (−1) − 7 + 18 i + 8 1 + 18 i
b) ( 6 + 3 i ) ( 6 − 3 i ) =
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
5 + 3 i1 − 2 i
Complex Conjugates( a + b i ) • ( a − b i ) = REAL #( 6 + 3 i ) • ( 6 − 3 i ) = REAL #
• 1 + 2 i 1 + 2 i
= 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 ]
CONJUGATE means to“Multipy by same real # and same imaginary # but with opposite sign
to eliminate the imaginary #.”
NOTES: Page 62, Section 5.4
Complex Numbers: Absolute ValueImaginary
Real
●
●( − 1 + 5 i )
(3 + 4 i )
● ( − 2 i )
a) │ 3 + 4 i │ = 32 + 42 = 25 = 5
b) │ − 2 i │ = │ 0 + 2 i │ = 02 + ( − 2 )2 = 2 c) │− 1 + 5 i │= − 12 + 52 = 26 ≈ 5.10
Z = a + b i
│ Z │ = a2 + b2
Absolute Value of a complex number is a non-negative real number.
NOTES: Page 64, Section 5.5
Completing the Square
x
x
x2 bx
b
b2
x
x
b2
b2
x2
b x2
b x2
( 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
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 1
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 = 0x2 + 10 x = + 3x2 + 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 As + 4 isn’t [ ½ (– 2) ]2 , move 4 to other side of = x2 – 2 x = – 4x2 – 2 x + 1 = – 4 + 1 What is [ ½ (– 2) ]2 = (– 1)2 = 1 ?( x – 1 )2 = – 3
( x – 1 )2 = – 3
( x – 1 ) = – 1 3
x = + 1 ± i 3
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 ( x2 – 8 x + 16 ) = ( x – 4 )2 – 16 – 16
y = ( x – 4 )2 – 5 ( x , y ) = ( 4 , – 5 )
NOTES: Page 66, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac2a
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 ) 4 a2
Binomial Squared (x + b )2 = b2 − 4 a c 2a 4 a2
Square Root both sides of = (x + b )2 = b2 − 4 a c 2a 4 a2
Squared Root undoes Squared x + b = ± b2 − 4 a c 2a 2 a
Solve for x by − b to both sides 2a x = − b ± b2 − 4 a c
2a 2 a= − b ± b2 − 4 a c 2a
NOTES: Page 66a, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac2a
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
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 ± 12 − 4 (2 ) ( −5 ) 2 ( 2 )
2 x2 + 1 x − 5 = 0x = − 1 ± 41 4
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac2a
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 = 0x = 6 ± 0 2
X = 3
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac2a
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 = 0x = − (2) ± − 4 2
x = − 2 ± 2 i −2
x = − 2(1 ± I ) −2 x = 1 ± i
NOTES: Page 67, Section 5.6
The Quadratic Formula and the Discriminant x = − b ± b2 − 4 ac2a
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 2 (1)
x2 − 6 x + 9 = 0 (− 6 )2 − 4 (1) (9) = 0
x = − (− 6 ) ± 0 2 (1)
x = − (− 6 ) ± 0 = 3 2 (1)
x2 − 6 x + 8 = 0 (− 6 )2 − 4 (1) (8) = 4
x = − (− 6 ) ± + 4 2 (1)
x = − (− 6 ) ± 2 = 2 or 4 2 (1)
x = − b ± b2 − 4 ac 2a
Quadratic Equation in Standard Form:
a x2 + b x + c = 03 x2 – 11 x – 4 = 0
Sum of Roots: – b a
4 + – 1 = 11 3 3
Product of Roots: c a
4 • – 1 = – 4 3 3
x = + 11 ± (11)2 − 4 (3) (– 4) 2 (3)
x = + 11 ± 121+ 48 2 (3)
x = + 11 ± 169 6
x = + 11 ± 13 = 24 , – 2 = 4 , – 1 6 6 6 3
x 2 + 2 x – 15 = 0
x 2 + 2 x = + 15
x 2 + 2 x + 1 = + 15 + 1
( x + 1 ) 2 = 16
( x + 1 ) = 16
x = – 1 ± 4 = 3 or – 5
Solve this Quadratic Equation: a x2 + b x + c = 0 x2 + 2 x – 15 = 0
x 2 + 2 x – 15 = 0
x = – 2 ± (– 2 )2 − 4 (1) (– 15) 2 (1)
x = – 2 ± 4 + 60 2
x = – 2 ± 64 2
x = – 2 ± 8 = 3 or – 5 2
x = − b ± b2 − 4 ac 2a
Quadratic Formula
x 2 + 2 x – 15 = 0
( x – 3 ) ( x + 5 ) = 0
x – 3 = 0 or x + 5 = 0
x = 3 or x = – 5
Factoring
Completing the Square
NOTES: Page 68a, Section 5.6
The Quadratic Formula and the Discriminant
REAL
IMMAGINARY
x2 − 6 x + 10 = 0 = 3 ± i
x2 − 6 x + 9 = 0 = 3
x2 − 6 x + 8 = 0 = 2 or 4
No intercept
One intercept
Two intercepts
●● ●
NOTES: Page 69, Section 5.7
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]
Example 1: y > 1 x2 − 2 x − 3 0 = (x − 3 ) ( x + 1 )
So, either (x − 3 ) = 0 or ( x + 1 ) = 0Then x = 3 or x = − 1
Vertex (standard form) = − b = − (− 2 ) = 1 2a 2 (1 )
y = 1 x2 − 2 x − 3y = 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 shadey > 1 x2 − 2 x − 30 > 1 (1)2 − 2 (1) − 30 > 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
0 − 4
2 0
−2 0● ●
●
●●
●
y
x
y ≥ x2 − 4y ≥ ( x − 2 ) ( x + 2 )
x = −b = 0 = 0 2a 2
y ≥ x2 − 4y ≥ ( 0 )2 − 4y ≥ − 4
x y
− 12
2 1 4
− 2 0
1 0
y < − x2 − x + 2y < − ( x2 + x − 2 )y < − ( x − 1 ) ( x + 2 )
y < − x2 − x + 2y < − ( − 1 )2 − (− 1 ) + 2 2 2y < − 1 + 1 + 2 4 2y < 2 1 4