MISSION · Algebra
Rational Expressions and Asymptotes
What happens to a function where its denominator hits zero
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The idea in one sentence
A rational expression is one polynomial divided by another, and everything interesting about it happens where the bottom is zero.
\[f(x) = \frac{x^2 - 4}{x - 2}\]Excluded values come first
Before simplifying anything, find where the denominator is zero. Those inputs are excluded from the domain — not because of a rule, but because dividing by zero has no meaning.
For \(\frac{x+1}{x^2-9}\), the bottom vanishes at \(x = 3\) and \(x = -3\), so both are excluded.
Do this before cancelling, because cancelling can hide the exclusion. That’s the one procedural habit this whole topic rests on.
A hole or an asymptote
Both come from a zero denominator, and which one you get depends on whether the same factor also appears on top.
A hole — the factor cancels. In \(\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2}\), the \((x-2)\) cancels and leaves \(x + 2\). The graph is the line \(y = x + 2\) with a single point missing at \(x = 2\). The function approaches 4 there but never arrives.
A vertical asymptote — the factor doesn’t cancel. In \(\frac{1}{x-2}\) nothing cancels, and as \(x\) nears 2 the values blow up towards \(\pm\infty\). The graph runs alongside the vertical line \(x = 2\) without ever touching it.
So: cancels means a hole, survives means an asymptote. Same condition, two very different pictures.
Horizontal asymptotes
Those describe the far ends instead, and comparing degrees settles it:
- Bottom degree bigger — the fraction shrinks towards 0, so \(y = 0\).
- Degrees equal — it settles at the ratio of leading coefficients.
- Top degree bigger — no horizontal asymptote; the graph runs off.
For \(\frac{3x^2+1}{x^2-4}\) the degrees match, so the graph flattens towards \(y = 3\) at both ends.
Worth knowing: a graph may cross a horizontal asymptote in the middle. The asymptote describes long-run behaviour, not a barrier. Vertical asymptotes genuinely can’t be crossed, because the function isn’t defined there at all.
Simplifying, and the trap
Cancelling is only legal for factors, never terms.
\[\frac{x^2 - 4}{x - 2} = x + 2 \quad \text{but} \quad \frac{x + 4}{x} \ne 4\]In the second, \(x\) is added, not multiplied — there’s nothing to cancel. Factor both top and bottom completely first; if a factor appears in both, it goes.
And carry the exclusion through. \(\frac{x^2-4}{x-2} = x + 2\) is only true for \(x \ne 2\). The simplified form has forgotten the hole, but the original function still has it.
Common mix-ups
- Cancelling across addition. Factors only.
- Simplifying before finding excluded values, which loses them.
- Assuming every zero denominator is an asymptote. Cancelling factors give holes.
- Treating a horizontal asymptote as unreachable. The graph may cross it.
Check yourself
For \(\frac{x^2 - 9}{x^2 - 2x - 3}\), find the excluded values and say what the graph does at each.
The bottom factors as \((x-3)(x+1)\), so \(x = 3\) and \(x = -1\) are excluded. The top is \((x-3)(x+3)\), so \((x-3)\) cancels: a hole at \(x = 3\), and a vertical asymptote at \(x = -1\).
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