Algebra 2 Chapter 5 Notes Quadratic Functions Algebra 2 Chapter 5 Notes Quadratic Functions

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


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


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


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


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


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 )


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.


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{


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


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


Complex Numbers: Plot Imaginary

Real

●

●( − 3 + 2 i )

(2 − 3 i )


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 #.”


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.


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


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


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 )


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



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



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



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




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


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

●● ●


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


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

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Algebra 2 Chapter 5 Notes Quadratic Functions Algebra 2 Chapter 5 Notes Quadratic Functions