# Transformations Across functions

• September 9th 2008, 03:06 PM
mike_302
Transformations Across functions
So our lesson in class was about transforming any function, f(x) using generalization (stating: y=f(x), y= -3f(1/2(x-2))+3 etc.)

The homework question is the one thing that may have been the most basic, but I think that because it was so easy, we dismissed discussing it: How do you take points that were in f(x) and find their corresponding points in the transformed function?

Example:

You have the points (2,3), (4,6), (8,9) on y=f(x). Find their corresponding point on y= -2f(2x)+3 .

My first guess would simply be to plug x and y in to the transformed function but I don't want to assume in this case. I'd just like to double check. Confirmation?
• September 9th 2008, 03:46 PM
reinanyc317
The inverse function, no matter how you calculate it, or what it means geometrically, is just a regular function after all. To find points on it, plug the x in the equation to find the corresponding y.
• September 9th 2008, 03:57 PM
mike_302
sorry. I might be getting confused with the word "inverse" but I'm not talking of inverse functions. I'm just referring to a transformation (any transformation) of f(x) .... So I'm told that a point (a,b) is on the graph of f(x). Then I'm told to find what the equivalent point of (a,b) is on a given (albeit a randomly chosen) function, given (for example) y=2f(1/2(x+4))+2 ... And I have asked a few people since originally posting this question and we cannot decide... What do we do?

We cannot simply put 'a' in for x and 'b' in for y... so???

Haha, sorry if this is confusing :S this is me doing my best to explain this situation.
• September 9th 2008, 08:02 PM
reinanyc317
of course you can't keep the (a,b) as is. Otherwise they won't satisfy the equation of the new function that you got by transformation.

what i meant is that you plug the a in the new equation, and you find an answer that should be different than be.

ie, for the original function, the point (a,b) belonged to it. on the new function, the equivalent point is (a,z), where you find z by plugging a into the new function.

• September 10th 2008, 12:12 PM
mike_302
well, I had the class today, and, in case anyone else reads this, apparnetly what we were supposed to do is look at the transformations on f(x) and simply apply those trasformations to a and b just by thinking about it. Makes sense to me.

Thanks for the help though! appreciated
• September 10th 2008, 12:37 PM
Shyam
Quote:

Originally Posted by mike_302
So our lesson in class was about transforming any function, f(x) using generalization (stating: y=f(x), y= -3f(1/2(x-2))+3 etc.)

The homework question is the one thing that may have been the most basic, but I think that because it was so easy, we dismissed discussing it: How do you take points that were in f(x) and find their corresponding points in the transformed function?

Example:

You have the points (2,3), (4,6), (8,9) on y=f(x). Find their corresponding point on y= -2f(2x)+3 .

My first guess would simply be to plug x and y in to the transformed function but I don't want to assume in this case. I'd just like to double check. Confirmation?

for $y=f(x), \;points\;are (2,3),\;(4,6)\;(8,9),\;$

The transformed function $y=-3f \left(\frac{1}{2}(x-2) \right)+3$

The transformed point for (2,3) will be = (2{2} +2, -3{3} +3) = (6, -6)

The transformed point for (4,6) will be = (2{4} +2, -3{6} +3) = (10, -15)

The transformed point for (8,9) will be = (2{8} +2, -3{9} +3) = (18, -24)

In general, for any function $y = a f[k(x - p)] + q$, the transformed point for point (x, y) will be

$=\left( \frac{1}{k}x+p,\;ay+q \right)$

where,
$a =$ vertical stretch/compression.
$k=$ horizontal stretch/compression
$p=$ horizontal shift
$q=$ vertical shift