Sketch the curve $\frac{ax + b}{cx + d}$

Question

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.

Answer 1

Assuming $c \not=0$, you can write $$y = \frac ac + \frac{bc-ad}{c^2(x+\frac dc)}$$

so $y$ is not defined when $x=-\frac dc$ and

• if $ad=bc$ you have $y$ otherwise being constant $\frac ac$

• if $ad > bc$ you have a hyperbola with horizontal asymptote $y=\frac ac$ and vertical asymptote $x=-\frac dc$ using the second and fourth quadrants

• if $ad < bc$ you have a hyperbola with horizontal asymptote at $y=\frac ac$ and vertical asymptote at $x=-\frac dc$ using the first and third quadrants

Meanwhile if $c=0$ and $d \not=0$, you have the straight line $y=\frac ad x+\frac bd$, with the slope affected by the sign of $\frac ad$, which is equivalent to the sign of $ad-bc$

Answer 2

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 $c$.