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Algebra 2 Chapter 5 Notes Quadratic FunctionsPowerPoint Presentation

Algebra 2 Chapter 5 Notes Quadratic Functions

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The vertical line through the vertex

NOTES: Page 56, Section 5.1

Graphing a Quadratic Function

Quadratic Function in standard form: y = ax2 + bx + 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

2 a

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

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

●

●

(− 5 , 2 )

(− 1 , 2 )

x =− 3

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)

(− 2 , 0 )

(4 , 0 )

●

●

x =1

y = −1 ( 1 + 2) ( 1– 4)

Y = 9

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 28that combine to make a difference of − 12?”

Example 2 : Factoring a trinomial of the form, ax2 + b x + c

Factor: 3x2 − 17 x + 10

“What are the factors of 10and 3 that combine to add up to − 17, when multiplied together?”

Solving Quadratic Equations by Factoring

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.

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

x = x ½

r

3

r

9

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

•

36

4

•

9

b

=

( a > 0 , b > 0)

Quotient Property

a

b

4

9

a

4

=

=

b

9

Examples:

=

6

=

2

6

24

4

=

=

90

•

=

10

6

•

15

9

10

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

14

2

7

7

7

2

7

16

=

=

=

=

4

16

2

2

2

Solve:

2 x2 + 1 = 17

2 x2 = 16

x2 = 8

X = ± 4

X =

Solve:

1

3

( x + 5)2 = 7

( x + 5)2 = 21

( x + 5)2 = 21

x + 5 = 21

X = – 5 21

±

•

±

2

2

X = – 5 21

± 2

±

+

{

and

±

X = – 5 21

–

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 theImaginary 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

Complex Numbers (a + b i)

( Real number + imaginary number )

Imaginary Numbers

Real Numbers

( a+b i)

( a+0 i)

( 2+3 i)

( 5−5 i)

− 1

5

2

Pure Imaginary Numbers

3

( 0 + b i ) , where b ≠ 0

∏

2

( − 4 i)

( 6 i)

Complex Numbers: Add and Subtract

- ( 4− i ) + ( 3− 2 i ) = 7− 3 i
- ( 7− 5 i ) − ( 1− 5 i ) = 6 + 0 i
- 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 ) =

b) ( 6+ 3 i) ( 6− 3 i ) =

− 7 + 4 i+ 14 i− 8 i2

36+ 18 i− 18 i− 9 i2

− 7 + 18 i− 8 (−1)

− 7 + 18 i+ 8

1+ 18 i

36 + 0 i− 9 (−1)

36 + 0 + 9

45

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 + 3 i

1− 2 i

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 ]

=

Complex Numbers: Absolute Value

Imaginary

( − 1 + 5 i )

Z =a + b i

│ Z │ =a2 + b2

●

(3 +4 i )

●

Absolute Value of a complex number is a non-negative real number.

Real

( −2 i )

●

│ 3 + 4 i│ = 32 + 42 = 25 = 5

│ −2 i│ = │ 0 + 2 i│ = 02 + ( − 2 )2 = 2

c) │− 1 + 5 i│= − 12 + 52 = 26 ≈ 5.10

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

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 − 3half 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

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 = – 4

x2 – 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 ± i3

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

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 ± (11)2− 4 (3) (– 4) 2 (3)

x= + 11 ± 121+ 482 (3)

x= + 11 ± 1696

x= + 11 ± 13 = 24 , – 2 = 4 , – 16 6 6 3

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

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

x= − b ± b2− 4 ac

2a

x 2 + 2 x – 15 = 0

x= – 2 ± (– 2)2− 4 (1) (– 15) 2 (1)

x= – 2 ± 4 + 602

x= – 2 ± 642

x= – 2 ± 8= 3 or – 5 2

Completing the Square

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

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 = 2or4

REAL

●

●

●

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 2 (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

●

●

●

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

●

Graphing & Solving Quadratic Inequalities

y

●

x

●

●

●

●

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 ≥ x2 − 4

y ≥ ( x − 2 ) ( x + 2 )

x = −b = 0 = 0

2a 2

y ≥ x2− 4

y ≥ (0)2 − 4

y ≥ − 4

●

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