Having some trouble with combining functions in the form y=af(k(x-d))+c

Sep 2016
19
0
ottawa
Hi,

This is my first attempt at getting some help on this website so I apologize if I did anything incorrectly.

I recently started MRC3U Grade 11 University Prep Functions through an ILC course, and I've come across a question that has me stumped. We have to describe the transformations that must be applied to the base function to obtain the transformed function. The two functions are f(x)=x^2 and y=7f(-1/6(x-1))+1. I already know that the parameters are a=7 k=-1/6 d=1 c=1 (or at least I think those are correct). Where I'm stumped is combining the two functions. When putting f(x)=x^2 into the second function I'm not sure on where to place the exponent. Throughout the lesson text it doesn't go into detail on how to do this, but judging by a couple of examples shown my best guess would be the equation is y=7f(-1/6(x-1)^2)+1. My initial guess was y=7f(-1/6(x^2-1))+1 because wouldn't it make sense for the exponent to be placed on x, like in the base function?
I've spent literally the last couple of hours just trying to deal with this and it's driving me insane.
If anybody could clear this up for me I would be very grateful, and any tips for this going forward would also be greatly appreciated.
Thanks very much
Joe
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
$f(x)=x^2$

$y= 7f\left(-\dfrac 1 6 (x-1)\right) + 1$

$y=7\left(-\dfrac 1 6 (x-1)\right)^2 + 1$

$y = 7\left(\dfrac{1}{36}(x^2-2x+1)\right)+1$

$y= \dfrac{7}{36} x^2 - \dfrac{7}{18} x + \left( \dfrac {7}{36}+1 \right)$

$y = \dfrac{7}{36} x^2 - \dfrac{7}{18} x + \dfrac{43}{36}$

$y = \dfrac{1}{36}\left(7x^2 - 14 x + 43\right)$
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
parent function ... $f(x) = x^2$

transformation #1 ... $f(x-1) = (x-1)^2$, shift right

transformation #2 ... $f\bigg[-\dfrac{1}{6}(x-1)\bigg] = \dfrac{1}{36}(x-1)^2$, horizontal stretch

transformation #3 ... $7 \cdot f\bigg[-\dfrac{1}{6}(x-1)\bigg] = \dfrac{7}{36}(x-1)^2$, vertical stretch

transformation #4 ... $7 \cdot f\bigg[-\dfrac{1}{6}(x-1)\bigg]+1 = \dfrac{7}{36}(x-1)^2+1$, vertical shift
 

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Debsta

MHF Helper
Oct 2009
1,361
633
Brisbane
f(x)=x^2 and y=7f(-1/6(x-1))+1
Firstly you need to understand function notation.

f is just the name of the function. In this case, f is the function that maps x onto x^2.

If f(x) = x^2, then f(whatever is in brackets) = (whatever is in brackets)^2.

Eg:

f(a) = a^2

f(x+2) = (x+2)^2

f(elephant) = (elephant)^2

In other words, replace x with "whatever is in brackets".

f(-1/6(x-1)) = (-1/6(x-1))^2

So, 7f(-1/6(x-1))+1 = 7(-1/6(x-1))^2+1

Hope that helps. Pick up the bracket and put it instead of x.
 
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