Transformation

• Oct 23rd 2011, 09:09 AM
Thenewguy
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.
• Oct 23rd 2011, 01:11 PM
SammyS
Re: Transformation
Quote:

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) ?
• Oct 23rd 2011, 01:59 PM
Soroban
Re: Transformation
Hello, Thenewguy!

Quote:

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       - - - + - - - - * - - - -             |      *  *             |      *    *             |             |    *      *             |             |             |    *        *             |
• Oct 23rd 2011, 05:59 PM
Prove It
Re: Transformation
Quote:

Originally Posted by Soroban
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 $\displaystyle \displaystyle |a| < 1$...
• Oct 23rd 2011, 08:01 PM
Soroban
Re: Transformation
Quote:

Originally Posted by Prove It
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.

• Oct 23rd 2011, 10:09 PM
Prove It
Re: Transformation
Quote:

Originally Posted by Soroban

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.

http://i22.photobucket.com/albums/b3.../checkmate.jpg

Check mate :)
• Oct 23rd 2011, 10:35 PM
Deveno
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.
• Nov 4th 2011, 06:42 AM
HallsofIvy
Re: Transformation
Quote:

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 $\displaystyle x^2$. Is that true?

Quote:

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.