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Math Help - Graph of the Function

  1. #1
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    Graph of the Function

    Graph the function f below by starting with the graph of y = x^2 and using transformations (shifting, compressing, stretching and/or reflection.

    HINT: If needed, write f in the form f(x) = a(x - h)^2 + k.

    f(x) = (2/3)x^2 + (4/3)(x) - 1


    I need someone to show me how to graph such a function f.
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  2. #2
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    Quote Originally Posted by symmetry View Post
    Graph the function f below by starting with the graph of y = x^2 and using transformations (shifting, compressing, stretching and/or reflection.

    HINT: If needed, write f in the form f(x) = a(x - h)^2 + k.

    f(x) = (2/3)x^2 + (4/3)(x) - 1


    I need someone to show me how to graph such a function f.
    Hello,

    first transcribe the equation of your function into the given general form:

    f(x)=\frac{2}{3}x^2+\frac{4}{3}x-1
    f(x)=\frac{2}{3}(x^2+2x)-1
    f(x)=\frac{2}{3}((x^2+2x+1)-1)-1
    f(x)=\frac{2}{3}(x+1)^2-\frac{2}{3}-1
    f(x)=\frac{2}{3}(x+1)^2-\frac{5}{3}

    Now you can describe the steps of transformation of the normal parabola:

    1. shift 1 unit to the left because of (x+1)
    2. compress the graph by the factor \frac{2}{3} because the factor is smaller than 1
    3. move down by \frac{5}{3}

    I've attached a diagram which contains these three steps. The order of transformation is: black - blue - green - red

    EB
    Attached Thumbnails Attached Thumbnails Graph of the Function-transf_parab.gif  
    Last edited by earboth; January 25th 2007 at 03:28 AM.
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    ok

    What can I say? I am overwhelmed by your kindness and willingness to help me with tough math questions.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    Now you can describe the steps of transformation of the normal parabola:

    1. shift 1 unit to the left because of (x+1)
    2. compress the graph by the factor \frac{2}{3} because the factor is smaller than 1
    I know what you are saying here, but theoretically you need to compress the graph first, then shift 1 unit to the left. Since you are compressing the entire x-axis the location of the vertex would also shift, unless the vertex is at x = 0. (In other words the way you have done it puts the vertex at x = -2/3, not x = -1.)

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    ... Since you are compressing the entire x-axis the location of the vertex would also shift, unless ...
    -Dan
    Hello, Dan,

    thanks for your reply.

    The factor \frac{2}{3} scales down only all y-values of coordinates of points. Geometrically it is a dilation(?) of the complete coordinate-plane perpendicular to the x-axis. All points forming the x-axis are fixed points(?) and stay therefore unmoved.

    Therefore it doesn't matter here if I first compress the graph and move afterwards or if I move the graph first and compress afterwards.

    EB
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