How to find the equation, intercepts, and intersection points of two rational graphs

Key concept: When two rational functions are involved, the cleanest move is to read the asymptotes first, then match any shifted form to the graph. After that, the intersection equation usually becomes much simpler than it looks at first glance.

Step by step

Step 1: Read the structure of each function

We have

$f(x)=\frac{4}{3-x}=-\frac{4}{x-3}$

and

$g(x)=\frac{1}{x+p}+q.$

For a rational function of the form $\frac{a}{x-h}+k$:

• the vertical asymptote is $x=h$
• the horizontal asymptote is $y=k$

So for $f(x)$:

• vertical asymptote: $x=3$
• horizontal asymptote: $y=0$

Its graph is a rectangular hyperbola shifted right by 3, and because of the minus sign it sits in the opposite orientation from $\frac{1}{x}$.

Step 2: Sketch $y=f(x)$

Key features to plot:

• asymptote $x=3$
• asymptote $y=0$
• one point to the left, one to the right

Try a couple of values:

$f(2)=\frac{4}{1}=4$ $f(4)=\frac{4}{-1}=-4$

So the graph passes through $(2,4)$ and $(4,-4)$.

That is enough to sketch the curve accurately around the asymptotes.

Step 3: Determine $p$ and $q$

Match the asymptotes of $g(x)=\frac{1}{x+p}+q$ to the graph shown.

• vertical asymptote comes from $x+p=0$, so $x=-p$
• horizontal asymptote is $y=q$

From the graph of $g(x)$, read those asymptotes directly:

• if the vertical asymptote is $x=3$, then $-p=3$, so $p=-3$
• if the horizontal asymptote is $y=0$, then $q=0$

So the values are:

$p=-3,\qquad q=0$

Step 4: Solve $f(x)=g(x)$

Substitute the values into $g(x)$:

$g(x)=\frac{1}{x-3}$

So solve

$\frac{4}{3-x}=\frac{1}{x-3}$

Since

$3-x=-(x-3),$

we can rewrite

$\frac{4}{3-x}=-\frac{4}{x-3}$

Then the equation becomes

$-\frac{4}{x-3}=\frac{1}{x-3}$

Multiply both sides by $x-3$:

$-4=1$

That is impossible.

So there are no solutions.

Final answers

• (a) Sketch $f(x)=\frac{4}{3-x}$ with asymptotes $x=3$ and $y=0$.
• (b) $p=-3$, $q=0$
• (c) No solution

Pitfall alert

A common slip is to confuse the vertical asymptote of $\frac{1}{x+p}+q$ with $x=p$. It is actually $x=-p$. Another easy mistake is to forget that $\frac{4}{3-x}$ can be rewritten as $-\frac{4}{x-3}$, and that sign changes the whole orientation of the sketch.

Try different conditions

If the graph of $g(x)$ were shifted up or down, only $q$ would change; the vertical asymptote would still give $p$. If the coefficient in front of the fraction were changed, the asymptotes would stay the same, but the branches would get stretched or flipped. The intersection step would still begin by rewriting both equations into the same denominator form.

Further reading

vertical asymptote, horizontal asymptote, rational function