To sketch a rational function, combine the asymptotes and holes with intercepts and a sign chart, the same tool you used for Polynomial and Rational Inequalities.
Key idea
Steps: factor and note the domain; find holes and vertical asymptotes; find the horizontal asymptote; find the $x$-intercepts (zeros of the simplified numerator) and the $y$-intercept $f(0)$; test the sign in each interval; then draw. A graph never crosses a vertical asymptote, but it can cross a horizontal one. Solve $f(x) = c$, where $y = c$ is the asymptote, to find where.
Why it works
A horizontal asymptote only describes what happens as $x \to \pm\infty$. In the middle of the graph, nothing stops the function from taking that value. A vertical asymptote is different: the function is undefined there, so the graph can't touch it.
Worked examples
Worked example: The whole picture
Describe the graph of $f(x) = \frac{x^2 + 6x + 8}{x^2 + 2x - 3}$.
- Factor: $\frac{(x + 2)(x + 4)}{(x - 1)(x + 3)}$. Nothing cancels, so there are no holes.
- Vertical asymptotes $x = 1$ and $x = -3$. Equal degrees: horizontal asymptote $y = 1$.
- $x$-intercepts $-2$ and $-4$; $y$-intercept $f(0) = \frac{8}{-3} = -\frac{8}{3}$.
- Signs: test $x = -5, -3.5, -2.5, 0, 2$ to get $+, -, +, -, +$.
Answer: Two vertical asymptotes, horizontal asymptote $y = 1$, intercepts $(-4, 0)$, $(-2, 0)$, $\left(0, -\frac{8}{3}\right)$.
Worked example: Crossing the horizontal asymptote
Where does that graph cross $y = 1$?
- Solve $\frac{x^2 + 6x + 8}{x^2 + 2x - 3} = 1$: $x^2 + 6x + 8 = x^2 + 2x - 3$.
- $4x = -11$, so $x = -\frac{11}{4}$.
Answer: At $x = -\frac{11}{4}$
Common mistake
Thinking a graph can never cross its horizontal asymptote. It can, as the second example shows. The rule that holds is about vertical asymptotes: those are never crossed, because the function is undefined there.