<s>[INST] Sketch the curve $\frac{ax + b}{cx + d}$ I am working through a pure maths book out of interest. I am stuck on this problem:
If $a$, $b$, $c$ and $d$ are positive, sketch the curve $y = \frac{ax + b}{cx + d}$ when
i) $ad - bc \lt 0$
ii)  $ad - bc = 0$
iii)  $ad - bc \gt 0$
I can work out the basics. Thus
$x$-intercept:  $f(x) = 0 \implies \frac{ax + b}{cx + d} = 0 \implies x = \frac{-b}{a}$
so intercept is $[\frac{-b}{a}, 0]$
$y$-intercept: $f(0) = \frac{b}{d}$ at point $[0,\frac{b}{d}]$
horizontal asymptote: $\frac{a}{c}$
vertical asymptote: $cx + d = 0 \implies x = \frac{-d}{c}$
I can also see that $\frac{ax + b}{cx + d} = \frac{a}{c}\left(1 + \frac{\frac{b}{a}-\frac{d}{c}}{x + \frac{d}{c}}\right)$
but I cannot put all this together to see how one could sketch the graphs except in one case: where $ad -bc = 0$ because then $\frac{b}{a}-\frac{d}{c} = 0$ and the graph will be a horizontal line equal to $\frac{a}{c}$
I think not knowing the actual magnitudes makes it difficult for me to visualise.
 [/INST] From your last observation, $$\frac{ax+b}{cx+d} =\frac 1c \left( a +\frac{bc-ad}{cx+d} \right)$$
You can think of this as a series of transformations applied to the graph of $y=\frac 1x$.
Starting from it, first shift to the left by $d$. That gives $\frac {1}{x+d}$. Then squish parallel to the $x$ axis by a factor of $c$. This gives $\frac{1}{cx+d}$.
Now, scale along the $y$ axis by a factor of $bc-ad$ (here, the sign comes into play: if $bc-ad\lt 0$, you have to flip the graph over). Then, move the graph up by $a$ units and finally, squish down by a factor of $c$.
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