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Math Help - Horizontal stretch by a factor of 5

  1. #1
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    Horizontal stretch by a factor of 5

    Will you please perform horizontal stretch by a factor of 5 of the parent function y=3x^2+5x

    Thank you very much
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by oceanmd View Post
    Will you please perform horizontal stretch by a factor of 5 of the parent function y=3x^2+5x

    Thank you very much
    A horizontal stretch means we effect the x-axis.

    To stretch the x-axis by a factor of 5, we "paradoxically" alter x by a factor of 1/5:
    y' = 3 \left ( \frac{x}{5} \right ) ^2 + 5 \left ( \frac{x}{5} \right )

    y' = \frac{3}{25}x^2 + x

    -Dan
    Attached Thumbnails Attached Thumbnails Horizontal stretch by a factor of 5-stretch.jpg  
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  3. #3
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    Dan,

    thank you very much, you confirmed what I did, the only difference is that when I graphed both parabolas, the vertex for foth functions is at (0,0). Why have you graphed them differently?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by oceanmd View Post
    Dan,

    thank you very much, you confirmed what I did, the only difference is that when I graphed both parabolas, the vertex for foth functions is at (0,0). Why have you graphed them differently?
    The vertex for y=3x^2+5x can be found by completing the square:
    y=3x^2+5x

    y = 3 \left ( x^2 + \frac{5}{3}x \right )

    y = 3 \left ( x^2 + \frac{5}{3}x + \frac{25}{36} - \frac{25}{36} \right )

    y = 3 \left ( x^2 + \frac{5}{3}x + \frac{25}{36} \right ) - 3 \cdot \frac{25}{36}

    y = 3 \left ( x + \frac{5}{6} \right )^2 - \frac{25}{12}

    So the vertex is at \left ( -\frac{5}{6}, -\frac{25}{12} \right ) as my graph shows.

    -Dan
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  5. #5
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    Dan,

    Thanks, it is just hard to keep everything in mind. Axis od symmetry formula -b/2a will give x=-5/6

    Thank you so much
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