Graphing Parabolas

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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