# Interpret the graph of $\frac{ax+b}{cx+d}$ as a transformation of $y=\frac{1}{x}$

As part of a problem-set I'm self-studying, I'm trying to interpret the graph of $f(x)=\frac{ax+b}{cx+d}$ as a transformation of the graph of $y=\frac{1}{x}$, including determining what restrictions should be placed on $a, b, c,$ and $d$ for the interpretation to remain valid.

For context, the preceding problem was as follows: Sketch the graph of $f(x)=\frac{3x-7}{x-2}$ by applying transformations to the graph of $y=\frac{1}{x}$. We used polynomial long division to show $f(x)=3-\frac{1}{x-2}$, yielding the following transformations:

$\rightarrow$

$\rightarrow$$\rightarrow Applying the same approach to f(x)=\frac{ax+b}{cx+d}, I did some long division and came up with,$$f(x)=\frac{a}{c}+\frac{bc-ad}{c(cx+d)}=\frac{a}{c}+\left(\frac{bc-ad}{c}\right)\left(\frac{1}{cx+d}\right)$$I'm not totally confident in my interpretation of the transformations involved, but here's what I've got: Clearly, we must have c\ne 0, and the first transformation of y=\frac{1}{x} is y=(\frac{1}{c})(\frac{1}{x})=\frac{1}{cx}, which could be viewed as either a vertical or horizontal scaling. If c>1, it shrinks the graph, compressing by a factor of c. If 0<c<1, it grows by a factor of \frac{1}{c} (again, either horizontally or vertically, the graph is such that the two are equivalent). Trivially, if c=1, it remains as is. A similar trio of cases for negative values of c would apply similar results and a reflection (and like the scaling, the reflection could be applied either vertically or horizontally). I suspect it makes more sense to interpret the above as a horizontal transformation, which can then be followed by a horizontal shift of d units, yielding y=\frac{1}{cx+d}. If d>0, everything shifts to the left, if d<0 it shifts to the right, and if d=0 it all stays put. The next transformation scales everything vertically by \frac{bc-ad}{c}, and the final transformation applies a vertical shift of \frac{a}{c}. Does that sound about right? I didn't come up with any restrictions on a,b,c, and d, beyond c\ne 0. I feel a bit less than competent when handling horizontal graph transformations in general. Dealing with shifts is easy enough, and I can handle simple scalings, but when they all combine, and possibly involve reflections as well, I start struggling to keep my head above water. • Check your calculations for the long division: it should say f(x)=a/c + (bc-ad)/a(cx+d) (with an a instead of a c on the bottom). – John Gowers Mar 18 '13 at 22:42 • Good catch, @Donkey_2009. – Sammy Black Mar 18 '13 at 22:56 • @Donkey_2009 - Are you sure? I checked it a couple times, but still get the same. – ivan Mar 18 '13 at 23:08 • Yeah sorry, scratch that. What you wrote was right. – John Gowers Mar 19 '13 at 0:10 ## 2 Answers Everything looks correct except for the order in which the horizontal transformations are applied. The transformation$$ y = x \qquad \leadsto \qquad y = \frac{1}{cx + d}$$first shifts the graph left by$d$, then scales towards the$y$-axis by a factor of$c$, which means that the shift gets scaled as well. If you like, factor:$cx + d = c\left(x + \tfrac{d}{c}\right)$, which allows you to apply the transformations in the usual order: shrink by factor of$c$, then shift by$-\tfrac{d}{c}$. • That right there is exactly what always trips me up. How do you think that ($\frac{1}{cx+d}$) through so the order makes sense? Thanks, by the way, @SammyBlack – ivan Mar 19 '13 at 0:19 • At the moment, your suggestion of writing$cx+d$as$c\left(x+\frac{d}{c}\right)\$ seems like my best bet for understanding on an intuitive level what's going on here. – ivan Mar 19 '13 at 0:42

We should also put the restriction bc-ad ≠ 0 although this is not directly on a, b, c and d as you asked.

f(cx) should always be interpreted as a horizontal transformation (horizontal scaling)

Therefore, in the transformation f(cx), if 0 < c < 1, the graph is scaled by a factor of c (i.e. expanded by a factor of c) and if c > 1 the graph is scaled by a factor of 1/c (i.e. shrinked by a factor of c), one cannot say "it grows by a factor of 1/c" because it would be equivalent to saying "it shrinks by a factor of c" which is incorrect in your example.

You arrived to f(cx+d) = 1/(cx+d) from f(x) = 1/x by a series of steps

1. you multiplied x by c to get cx