hey MHF just wondering here, im suppose to do a transformation with y=3(x5)^2 from its original function..
so i know that y=af[k(xd)]+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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hey MHF just wondering here, im suppose to do a transformation with y=3(x5)^2 from its original function..
so i know that y=af[k(xd)]+c
but in here i have no idea if the 3 is "a" or it is a "k" thats been factored out.
Hello, Thenewguy!
Quote:
I'm suppose to do a transformation with $\displaystyle y\:=\:3(x5)^2$ from its original function..
so I know that: .$\displaystyle y\:=\:a\,f[k(xd)]+c$ . 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: .$\displaystyle a\!\cdot\!f(xb) + c$ 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: .$\displaystyle y \:=\:3(x5)^2$
The original function is: .$\displaystyle f(x) \,=\,x^2$
. . an upopening 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 xaxis.
Code:


 5
   +     *    
 * *
 * *

 * *


 * *

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 :)
to be precise:
$\displaystyle y = 100x^2$ appears "squashed" along the xaxis, and "stretched" along the yaxis.
$\displaystyle y = x^2/100$ appear "stretched" along the xaxis and "squashed" along the yaxis.
however, since we typically assume the "xscale" to remain constant, it makes more sense to use the yaxis change as a description of how the function is affected.
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(xd)]+c
but in here i have no idea if the 3 is "a" or it is a "k" thats been factored out.