# Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.

• Aug 14th 2012, 07:50 AM
OverlyFrustrated
Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.
Hi everyone. This is my first post here, and I hope I am doing everything right. Please feel free to correct me if I've done anything wrong. Thanks in advance for taking the time to read through my frustrating issues.

Okay, so I'm doing transformations using y=af[K(x-d)]+c.

I know that K>1 = horizontal compression of a function and 0<k, & K<1 = horizontal stretch of a function.

I've been given the base function f(x)=x, and have been told that it is being stretched horizontally by a factor of 1/2, and having some other transformations applied to it. I understand everything else that's been provided in the answer, but I don't understand what's going on with K.

The answer the text provides states that, since the graph is being stretched horizontally by a factor of 1/2, then K=2. How does K=2? Shouldn't it equal 1/2? Also, if K=2, then the graph is being compressed horizontally rather than stretched, correct? What is going on here? Is my text book wrong? Or am I not getting something?

Secondly,

it's occurring to me that I don't know how to work with fractions that include variables. Another question I have done involving transformations looked like

Y=(-1/5(1/x)-1)-3

and the textbook reduced it to (5/x-1) -3, and I have absolutely no idea how that happened.

Thank you so much for your help, I'm pulling my hair out over here trying to understand what the hell is going on.
• Aug 14th 2012, 08:35 AM
emakarov
Re: Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.
Welcome to the forum. First note that variable names are case-sensitive in mathematics, so k and K may very well denote different entities. In this case, since all other variables are lowercase, it would be reasonable to use k everywhere instead of K.

Quote:

Originally Posted by OverlyFrustrated
The answer the text provides states that, since the graph is being stretched horizontally by a factor of 1/2, then K=2. How does K=2? Shouldn't it equal 1/2? Also, if K=2, then the graph is being compressed horizontally rather than stretched, correct?

Your text probably considers that "stretching by k" by definition means moving a point (x, y) into (kx, y). For k > 1 this is indeed stretching; for k < 1 this is actually compressing. However, in both cases the term "stretching by k" is used. If this is so, then stretching a graph by k amounts to changing f(x - d) to f((x - d) / k). In your case, stretching by 1/2 (which is compressing around the y-axis) amounts to changing x to 2x.

Quote:

Originally Posted by OverlyFrustrated
Y=(-1/5(1/x)-1)-3

and the textbook reduced it to (5/x-1) -3, and I have absolutely no idea how that happened.

Please recall the order of operations and use parentheses properly. $-\frac{1}{5}\frac{1}{x}-1-3$ is indeed different from $\frac{5}{x}-1-3$.
• Aug 14th 2012, 04:20 PM
OverlyFrustrated
Re: Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.
Thanks a bunch, the first bit really helps to clear things up.

As far as the second bit, how do I write like you did, with the numbers over each other?

The response you've given was not, unfortunately, what I'd intended to portray. x-1 is the denominator in the second case.
• Aug 15th 2012, 02:55 AM
emakarov
Re: Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.
You can write $$\frac{5}{x-1}$$ to produce $\frac{5}{x-1}$. See also the LaTeX Help subforum. However, every expression can be unambiguously rendered in plain text using parentheses.

$-\frac{1}{5}\frac{1}{x}-1\ne\frac{5}{x-1}$

Edit: You can also click the "Reply with Quote" button to see the source code for a post.
• Aug 20th 2012, 11:25 AM
OverlyFrustrated
Re: Transformation of the form y=af(k(x-d)+c - very lost. Also help with fractions.
Thank you very much for your help!