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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:

f1$\rightarrow$f2

$\rightarrow$f3$\rightarrow$f4

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.

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  • $\begingroup$ 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). $\endgroup$ – John Gowers Mar 18 '13 at 22:42
  • $\begingroup$ Good catch, @Donkey_2009. $\endgroup$ – Sammy Black Mar 18 '13 at 22:56
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    $\begingroup$ @Donkey_2009 - Are you sure? I checked it a couple times, but still get the same. $\endgroup$ – ivan Mar 18 '13 at 23:08
  • $\begingroup$ Yeah sorry, scratch that. What you wrote was right. $\endgroup$ – John Gowers Mar 19 '13 at 0:10
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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 $-\tfrac{d}{c}$.

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  • $\begingroup$ 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 $\endgroup$ – ivan Mar 19 '13 at 0:19
  • $\begingroup$ 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. $\endgroup$ – ivan Mar 19 '13 at 0:42
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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
  2. you added d

What you do now is to reverse the process:

  1. substract d (which corresponds to a horizontal shift, to the right if d<0 and to the left if d>0)
  2. divide by c (which is a horizontal scaling by c or expansion by c if 0 < c < 1 and a horizontal scaling by 1/c or shrinking by a factor of c if c > 1)
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