Transformation

**Thenewguy** · October 23rd 2011, 09:09 AM

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.

**SammyS** · October 23rd 2011, 01:11 PM

Originally Posted by Thenewguy

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) ?

**Soroban** · October 23rd 2011, 01:59 PM

Hello, Thenewguy!

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

so I know that: . $y\:=\:a\,f[k(x-d)]+c$ . 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 $y \,=\,f(x)$ is the original function,

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

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

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

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

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

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

Code:

            |
      *     |     *
       *    |    *
         *  |  *
    - - - - * - - - - 
            |

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

Code:

            |
            |   *           *
            |    *         *
            |      *     *
      - - - + - - - - * - - - - 
            |         5

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

Code:

            |
            |
            |
            |         5
      - - - + - - - - * - - - - 
            |       *   * 
            |      *     *
            |
            |     *       *
            | 
            |
            |    *         * 
            |

**Prove It** · October 23rd 2011, 05:59 PM

Originally Posted by Soroban

Hello, Thenewguy!

You should have been given a better definition for Transformations.

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

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

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

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

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

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

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

Code:

            |
      *     |     *
       *    |    *
         *  |  *
    - - - - * - - - - 
            |

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

Code:

            |
            |   *           *
            |    *         *
            |      *     *
      - - - + - - - - * - - - - 
            |         5

Since $a = \text{-}3$ , 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 $\displaystyle |a| > 1$ and stretched when $\displaystyle |a| < 1$ ...

**Soroban** · October 23rd 2011, 08:01 PM

Originally Posted by Prove It

Actually, the graph is squashed when $\displaystyle |a| > 1$ and stretched when $\displaystyle |a| < 1$ ...

I disagree . . .

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

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

**Prove It** · October 23rd 2011, 10:09 PM

Originally Posted by Soroban

I disagree . . .

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

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

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

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

Check mate

**Deveno** · October 23rd 2011, 10:35 PM

to be precise:

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

$y = x^2/100$ 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.

**HallsofIvy** · November 4th 2011, 06:42 AM

Originally Posted by Thenewguy

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 $x^2$ . 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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