Rational Functions and How to Sketch Them

You already know how to read a cubic or a reciprocal graph by its silhouette, and you can cancel a common factor from an algebraic fraction. Put those two skills together and you can sketch a whole new family: the rational functions.

A rational function is nothing more exotic than one polynomial divided by another:

f(x) = \dfrac{p(x)}{q(x)}, \qquad q(x) \neq 0.

The reciprocal y = \dfrac{1}{x} is the simplest interesting one (p(x)=1, q(x)=x). Swap in bigger polynomials and you get curves that rocket off to infinity at certain x-values and flatten out towards a level line far away. The wonderful thing is that you can predict exactly where both of those things happen just by looking at the top and the bottom — no forty-point table required.

This page is a recipe. By the end you'll be able to take any \dfrac{p(x)}{q(x)} and read off its vertical asymptotes, its horizontal (or slant) asymptote, its holes, and its intercepts — then sketch the thing in under a minute.

It feels like a contradiction — wild in one place, tame in another — but it's just the fraction doing two different jobs at two different scales.

Near a bad spot: if the denominator q(x) shrinks towards 0 while the numerator stays away from 0, you're dividing a normal number by something microscopic. Divide 7 by 0.001 and you get 7000; by 0.000001, seven million. The output explodes — that's a vertical asymptote.

Far away: when x is huge, only the highest power on the top and the bottom matters (the little terms are rounding errors beside it). So the whole fraction settles down to the ratio of those leading terms — a flat horizontal line. Same formula, two regimes: violent where the bottom vanishes, sleepy where the top and bottom race each other to infinity and tie.

Vertical asymptotes: where the bottom hits zero

A fraction goes haywire exactly when its denominator is 0 — you cannot divide by nothing. So the candidates for trouble are the zeros of q(x). At (almost) every such x the curve shoots up to +\infty on one side and plunges to -\infty on the other, hugging an invisible vertical line it never touches: a vertical asymptote.

I said almost every zero, and that word is the whole subtlety of this lesson — we'll meet the exception (a hole) in a moment. First, the clean case. Take

f(x) = \dfrac{2x+1}{x-3}.

The denominator is zero when x-3=0, i.e. x=3, and the numerator there is 2(3)+1 = 7 \neq 0. Seven divided by an ever-tinier number: the curve has a vertical asymptote at x=3.

Horizontal asymptote: let the powers fight it out

Now zoom way out. For large x, a polynomial is dominated by its top term, so compare the degree (highest power) of the top with that of the bottom. Three cases, and that's the whole story:

Back to f(x) = \dfrac{2x+1}{x-3}: top and bottom are both degree 1, with leading coefficients 2 and 1. Equal degrees, so the horizontal asymptote is y = \dfrac{2}{1} = 2. (You can see why: for enormous x, \dfrac{2x+1}{x-3} \approx \dfrac{2x}{x} = 2.)

Worked example 1 — sketch f(x) = \dfrac{2x+1}{x-3}

Let's run the full recipe on our example and turn it into a picture. Five quick questions:

Mark the two asymptotes as a dashed cross-hairs, drop in those two intercepts, and the shape draws itself — a reciprocal-style curve threaded between the lines x=3 and y=2. The chart below plots it live: drag the sliders to change the leading coefficient a (which sets the horizontal asymptote y=a) and the root c (which sets the vertical asymptote x=c), and watch both dashed lines and the whole curve move with them. Setting a = 2 and c = 3 gives exactly our worked example.

Worked example 2 — the trap: f(x) = \dfrac{x^2-1}{x-1}

Here the denominator is zero at x=1, so you might reach for a vertical asymptote there. Stop. Check the numerator first — it factorises:

f(x) = \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1}.

The factor (x-1) appears on both top and bottom, so it cancels:

f(x) = x+1 \quad \text{for all } x \neq 1.

Away from x=1 this is just the straight line y = x+1. But we are not allowed to put x=1 into the original fraction (that would be \tfrac{0}{0}), so the graph is the line y=x+1 with a single point punched out at x=1. Since the line would pass through (1,\, 1+1) = (1,2), the graph is the line y=x+1 with an open circle — a hole — at (1,2).

A hole is a removable discontinuity: a factor that cancelled between top and bottom leaves a single missing point, not an asymptote. The curve doesn't blow up there — it simply skips one dot. In the picture below, the line is drawn solid and the hole marked at (1,2).

This is the single most common mistake with rational functions. A zero of the denominator gives a vertical asymptote only if the factor does NOT cancel with the numerator. If it cancels, you get a hole instead — no blow-up, just one missing point.

The rule: always fully factor and cancel first. Then the un-cancelled zeros of the denominator are your vertical asymptotes, and every factor that cancelled marks a hole. Skip the cancelling step and you'll draw an asymptote where there's really just a pinprick — or miss a hole entirely.

To sketch f(x) = \dfrac{p(x)}{q(x)}, factor top and bottom, cancel, then read off:

Then there's no horizontal or slant asymptote — the curve runs away to infinity following a parabola or higher curve. For example \dfrac{x^3}{x-1} behaves like x^2 for large x (polynomial long division gives x^2 + x + 1 + \tfrac{1}{x-1}), so it hugs the parabola y = x^2 + x + 1 far out. Same idea, one rung up: the quotient of the division is the asymptotic curve, whatever its degree. When the top beats the bottom by just one, that quotient is a line — the slant asymptote — which is why "degree + 1" is the special case worth naming.