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Graphing Parabolas

Using the VertexAxis of Symmetry

& y-Intercept

By: Jeffrey Bivin

Lake Zurich High School

jeff.bivin@lz95.org

Last Updated: October 15, 2007

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Graphing Parabolas

y = x2 + 4x - 7

With your graphing calculator, grapheach of the following quadraticequations and identify the vertex andaxis of symmetry.

y = 2x2 + 10x + 4

y = -3x2 + 5x + 9

Jeff Bivin -- LZHS

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Graph the following parabola

y = x2 + 4x - 7

axis of symmetry:

vertex:

2!x

)11,2(

)7,0( y-intercept:

Jeff Bivin -- LZHS

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Graph the following parabola

y = 2x2 + 10x + 4

axis of symmetry:

vertex:

25!x

),( 21725

)4,0(y-intercept:

25

!x

),(217

25

)4,0(

Jeff Bivin -- LZHS

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Graph the following parabola

y = -3x2 + 5x + 9

axis of symmetry:

vertex:

6

5!x

),( 1213365

)9,0(y-intercept:

6

5!x

),( 1213365

)9,0(

Jeff Bivin -- LZHS

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Graphing Parabolas

y = x2 + 4x - 7

Now look at the coefficients of theequation and the value of the axis ofsymmetry especially a and b

y = ax2 + bx + c

y = 2x2 + 10x + 4

y = -3x2 + 5x + 9

2!x

25

!x

6

5!x

Jeff Bivin -- LZHS

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Graphing Parabolas

y = ax2 + bx + c

Vertex:

Axis of symmetry: ab

x 2

!

)(,22 ab

ab f

Jeff Bivin -- LZHS

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Graph the following parabola

y = x2 + 4x - 7

axis of symmetry:

vertex:

224

)1(24

2!!!!

abx

)11,2(

)7,0( y-intercept:

117)2(4)2()2( 2 !!f

Jeff Bivin -- LZHS

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Graph the following parabola

y = 2x2 + 10x + 4

axis of symmetry:

vertex:

25

410

)2(210

2 !!!!

abx

),(217

25

)4,0(y-intercept:

217

252

25

25 4)(10)(2)( !!f

25!x

Jeff Bivin -- LZHS

),(217

25

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Graph the following parabola

y = -3x2 + 5x + 9

axis of symmetry:

vertex:

6

56

5)3(2

52

!!!!

abx

),(12

1336

5

)9,0(y-intercept:

12133

6

52

6

56

5 9)(5)(3)( !!f

6

5!x

),( 1213365

)9,0(

Why did this parabola opendownward instead of upward as

did the previous two?

Jeff Bivin -- LZHS

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Graph the following parabolay = x2 + 6x - 8

Axis of symmetry:

Vertex:

3)1(2 62 !!! abx

178)3(6)3()3(2

!!f

)17,3(

)8,0( y-intercept:Jeff Bivin -- LZHS

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Graph the following parabolay = -2x2 + 7x + 12

Axis of symmetry:

Vertex:

47

47

)2(27

2 !!!!

abx

8145472

4747 12)(7)(2)( !!f

),(8

14547

)12,0(y-intercept:

),(8

14547

47

!x

Jeff Bivin -- LZHS

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Graphing Parabolas

In Vertex Form

Jeff Bivin -- LZHS

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Graphing Parabolas

y = x2

With your graphing calculator, grapheach of the following quadraticequations and identify the vertex andaxis of symmetry.

y = (x - 5)2 - 4

y = -3(x + 2)2 + 5

y = (x - 3)2 + 1

Jeff Bivin -- LZHS

0!x)0,0(

5!x)4,5(

2!x)5,2(

3!x)1,3(

vertex axis of sym.

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Graph the following parabola

y = (x - 5)2 - 4

axis of symmetry:

vertex:

5!x

)4,5(

)21,0(y-intercept:

Jeff Bivin -- LZHS

05 !x

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Graph the following parabola

y = -3(x + 2)2 + 5

axis of symmetry:

vertex:

2!x

)5,2(

)7,0( y-intercept:

Jeff Bivin -- LZHS

02 !x

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Graph the following parabola

y = (x - 3)2 - 1

axis of symmetry:

vertex:

3!x

)1,3(

),0(8

19y-intercept:

Jeff Bivin -- LZHS

03 !x

),0(8

19

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Graphing Parabolas

In Intercept Form

Jeff Bivin -- LZHS

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Graph the following parabola

y = (x 4)(x + 2)

x-intercepts:

vertex:

)8,0( y-intercept:Jeff Bivin -- LZHS

04 !x 02!x

)0,4( )0,2(

axis of symmetry:

2

24 !x

1!x)9,1(

9)3)(3()21)(41( !!!y

8)2)(4()20)(40( !!!y

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Graph the following parabola

y = (x - 1)(x - 9)

x-intercepts:

vertex:

)9,0(y-intercept:Jeff Bivin -- LZHS

01!x 09!x

)0,1( )0,9(

axis of symmetry:

2

91 !x

5!x)16,5(

16)4)(4()95)(15( !!!y

9)9)(1()90)(10( !!!y

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Graph the following parabola

y = -2(x + 1)(x - 5)

x-intercepts:

vertex:

)10,0(y-intercept:Jeff Bivin -- LZHS

01!x 05!x

)0,1( )0,5(

axis of symmetry:

2

51 !x

2!x)18,2(

18)3)(3(2)52)(12(2 !!!y

10)5)(1(2)50)(10(2 !!!y

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Convert to standard form

y = -2(x + 1)(x - 5)

Jeff Bivin -- LZHS

y = -2(x2 5x + 1x 5)

y = -2(x2 4x 5)

y = -2x2 + 8x + 10

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Now graph from standard form.

y = -2x2 + 8x + 10

Axis of symmetry:

Vertex:

248)2(2 82 !!!! abx

1810)2(8)2(2)2(2

!!f

)18,2(

)10,0(y-intercept:Jeff Bivin -- LZHS

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A taxi service operates between two airports transporting200 passengers a day. The charge is \$15.00. The ownerestimates that 10 passengers will be lost for each \$2increase in the fare. What charge would be most profitablefor the service? What is the maximum income?

Jeff Bivin -- LZHS

Income = Price Quantity

f(x) = ( 15 + 2x ) ( 200 10x )

Define the variable

x = number of \$2

price increases15 + 2x = 0 200 10x = 0

25.3781)25.6(10200)25.6(215)25.6( !!f

2x = -15Vertex is:

25.3781,25.6

So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = \$27.50

f(x) = income

200 = 10x

215!x x!20

25.6:4

252

240

215

!!!

xsymmetryofaxis

Maximumincome:

VERTEX

27.50 137.50

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A taxi service operates between two airports transporting200 passengers a day. The charge is \$15.00. The ownerestimates that 10 passengers will be lost for each \$2increase in the fare. What charge would be most profitablefor the service? What is the maximum income?

Jeff Bivin -- LZHS

Income = Price Quantity

f(x) = ( 15 + 2x ) ( 200 10x )

Define the variable

x = number of \$2

price increasesf(x) = 3000 150x + 400x 20x2

f(x) = 20x2 + 250x + 3000

VERTEX

a

bx2

!

)20(2250

!x

25.6!x

f(6.25) = 20(6.25)2 + 250(6.25) + 3000

f(6.25) = 3781.25 Vertex is: 25.3781,25.6

So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = \$27.50

f(x) = income

Maximum income = f(x) = \$3781.25