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Thread: Transformation

  1. #1
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    Transformation

    hey MHF just wondering here, im suppose to do a transformation with y=-3(x-5)^2 from its original function..

    so i know that y=af[k(x-d)]+c

    but in here i have no idea if the -3 is "a" or it is a "k" thats been factored out.
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  2. #2
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    Re: Transformation

    Quote Originally Posted by Thenewguy View Post
    hey MHF just wondering here, im suppose to do a transformation with y=-3(x-5)^2 from its original function..

    so i know that y=af[k(x-d)]+c

    but in here i have no idea if the -3 is "a" or it is a "k" thats been factored out.

    What are you using for f(x) ?
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  3. #3
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    Re: Transformation

    Hello, Thenewguy!

    I'm suppose to do a transformation with $\displaystyle y\:=\:-3(x-5)^2$ from its original function..

    so I know that: .$\displaystyle y\:=\:a\,f[k(x-d)]+c$ .
    Who gave you this? . . . It's awful!

    but in here, I have no idea if the "-3" is "a" or "k" that's been factored out.

    You should have been given a better definition for Transformations.


    If $\displaystyle y \,=\,f(x)$ is the original function,

    . . then: .$\displaystyle a\!\cdot\!f(x-b) + c$ is the transformed function.

    $\displaystyle a$ is a scale factor in the vertical direction.
    . . If $\displaystyle |a| > 1$, the graph is "stretched" vertically.
    . . If $\displaystyle |a| < 1$, the graph is "squashed" vertically.
    . . If $\displaystyle a$ is negative, the graph is reflected over the $\displaystyle x$-axis.

    $\displaystyle b$ is the horizontal displacement.
    . . If $\displaystyle b$ is positive, the graph is moved $\displaystyle b$ units to the right.
    . . If $\displaystyle b$ is negative, the graph is moved $\displaystyle b$ units to the left.

    $\displaystyle c$ is the vertical displacement.
    . . If $\displaystyle c$ is positive, the graph is moved $\displaystyle c$ units upward.
    . . If $\displaystyle c$ is negative, the graph is moved $\displaystyle c$ units downward.


    Your problem has: .$\displaystyle y \:=\:-3(x-5)^2$

    The original function is: .$\displaystyle f(x) \,=\,x^2$
    . . an up-opening parabola with its vertex at the Origin.

    Code:
                |
          *     |     *
           *    |    *
             *  |  *
        - - - - * - - - - 
                |

    Since $\displaystyle b = 5$, the graph is moved 5 units to the right.

    Code:
                |
                |   *           *
                |    *         *
                |      *     *
          - - - + - - - - * - - - - 
                |         5

    Since $\displaystyle a = \text{-}3$, the graph is "stretched" vertically by a factor of 3
    . . and is reflected over the x-axis.

    Code:
                |
                |
                |
                |         5
          - - - + - - - - * - - - - 
                |       *   * 
                |      *     *
                |
                |     *       *
                | 
                |
                |    *         * 
                |
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  4. #4
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    Re: Transformation

    Quote Originally Posted by Soroban View Post
    Hello, Thenewguy!


    You should have been given a better definition for Transformations.


    If $\displaystyle y \,=\,f(x)$ is the original function,

    . . then: .$\displaystyle a\!\cdot\!f(x-b) + c$ is the transformed function.

    $\displaystyle a$ is a scale factor in the vertical direction.
    . . If $\displaystyle |a| > 1$, the graph is "stretched" vertically.
    . . If $\displaystyle |a| < 1$, the graph is "squashed" vertically.
    . . If $\displaystyle a$ is negative, the graph is reflected over the $\displaystyle x$-axis.

    $\displaystyle b$ is the horizontal displacement.
    . . If $\displaystyle b$ is positive, the graph is moved $\displaystyle b$ units to the right.
    . . If $\displaystyle b$ is negative, the graph is moved $\displaystyle b$ units to the left.

    $\displaystyle c$ is the vertical displacement.
    . . If $\displaystyle c$ is positive, the graph is moved $\displaystyle c$ units upward.
    . . If $\displaystyle c$ is negative, the graph is moved $\displaystyle c$ units downward.


    Your problem has: .$\displaystyle y \:=\:-3(x-5)^2$

    The original function is: .$\displaystyle f(x) \,=\,x^2$
    . . an up-opening parabola with its vertex at the Origin.

    Code:
                |
          *     |     *
           *    |    *
             *  |  *
        - - - - * - - - - 
                |

    Since $\displaystyle b = 5$, the graph is moved 5 units to the right.

    Code:
                |
                |   *           *
                |    *         *
                |      *     *
          - - - + - - - - * - - - - 
                |         5

    Since $\displaystyle a = \text{-}3$, the graph is "stretched" vertically by a factor of 3
    . . and is reflected over the x-axis.

    Code:
                |
                |
                |
                |         5
          - - - + - - - - * - - - - 
                |       *   * 
                |      *     *
                |
                |     *       *
                | 
                |
                |    *         * 
                |
    Actually, the graph is squashed when $\displaystyle \displaystyle |a| > 1$ and stretched when $\displaystyle \displaystyle |a| < 1$...
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  5. #5
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    Re: Transformation

    Quote Originally Posted by Prove It View Post
    Actually, the graph is squashed when $\displaystyle \displaystyle |a| > 1$ and stretched when $\displaystyle \displaystyle |a| < 1$ ...

    I disagree . . .

    $\displaystyle y = 3x^2$ will rise faster than $\displaystyle y = x^2$, resulting in a stretched parabola.

    And $\displaystyle y = \tfrac{1}{3}x^2$ will rise more slowly.

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  6. #6
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    Re: Transformation

    Quote Originally Posted by Soroban View Post

    I disagree . . .

    $\displaystyle y = 3x^2$ will rise faster than $\displaystyle y = x^2$, resulting in a stretched parabola.

    And $\displaystyle y = \tfrac{1}{3}x^2$ will rise more slowly.

    Yes, $\displaystyle \displaystyle 3x^2$ approaches $\displaystyle \displaystyle \infty$ much faster than $\displaystyle \displaystyle x^2$. Surely you can see how this equates to a SQUASHED parabola.

    Also, $\displaystyle \displaystyle \frac{1}{3}x^2$ approaches $\displaystyle \displaystyle \infty$ much slower than $\displaystyle \displaystyle x^2$. Surely you can see how this equates to a STRETCHED parabola.



    Check mate
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  7. #7
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    Re: Transformation

    to be precise:

    $\displaystyle y = 100x^2$ appears "squashed" along the x-axis, and "stretched" along the y-axis.

    $\displaystyle y = x^2/100$ appear "stretched" along the x-axis and "squashed" along the y-axis.

    however, since we typically assume the "x-scale" to remain constant, it makes more sense to use the y-axis change as a description of how the function is affected.
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  8. #8
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    Re: Transformation

    Quote Originally Posted by Thenewguy View Post
    hey MHF just wondering here, im suppose to do a transformation with y=-3(x-5)^2 from its original function..
    It is impossible to answer this without knowing what the "orginal function" was. The others answering this are assuming that the original function was $\displaystyle x^2$. Is that true?

    so i know that y=af[k(x-d)]+c

    but in here i have no idea if the -3 is "a" or it is a "k" thats been factored out.
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