Thread: Further Transformation issues with y=af(k(x-d)+c

I am given a base function of f=x²

and given a transformed equation of y = 7f((-1/6)(x-1))+1

I am being asked to describe the transformations that must be applied to the graph of the base function f(x) to obtain the transformed function, which I understand, and then I am ask to write the transformed equation in simplified form. This is where I'm hung up.

I guess the next line would be

y=7((-1/6)(x-1)²+1This is where I get stuck. How do I multiply -1/6 by x-1? What do I do with the square? Also, I don't know how to write fractions on the site yet, so -1/6 is negative one sixt to clear up any confusion. I've placed it in parenthesis to try to indicate that.

Hi OverlyFrustrated, this seems to be what you are required to do.



From here on you can expand using the rules of algebra and fractions and you'll find out the simplified form. The fact that you are asking about multiplying -1/16 with (x-1), and then saying that -1/6 is negative one sixth, suggests to me that you are unsure about the basics of fractions. For example



Multiplying fractions works like this



In your case you have


Finally, the way to write math in this forum is to write [..TEX..] before the math expression and [../TEX..] after the math (remove the dots). Fractions are written as \frac{}{}, for example, \frac{x-1}{16}