MHF4U: Graphing Rational Functions

How It Works

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

1. Easy

   Sketch $f(x) = \frac{1}{x - 2}$: find its asymptotes, intercepts and the sign of each branch.

   Show solution

   1. The denominator is zero at $x = 2$, so the domain is $x \ne 2$ and $x = 2$ is a vertical asymptote. Nothing cancels, so there is no hole.
   2. The numerator is the constant $1$, never zero, so there is no $x$-intercept. The $y$-intercept is $f(0) = -\frac{1}{2}$.
   3. Degree $0$ on top, degree $1$ on the bottom: horizontal asymptote $y = 0$.
   4. Test points: $f(1) = -1$, so the left branch is below the $x$-axis and drops to $-\infty$ at the asymptote; $f(3) = 1$, so the right branch is above and climbs to $+\infty$.

   Answer: Vertical asymptote $x = 2$, horizontal asymptote $y = 0$, $y$-intercept $(0, -\frac{1}{2})$, no $x$-intercept; the left branch is negative, the right branch positive.

2. Easy

   Find every feature of $f(x) = \frac{x + 1}{x - 3}$ and sketch it.

   Show solution

   1. Vertical asymptote at $x = 3$; the numerator is $4$ there, so nothing cancels.
   2. $x$-intercept where $x + 1 = 0$: $(-1, 0)$. $y$-intercept $f(0) = \frac{1}{-3} = -\frac{1}{3}$.
   3. Degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients, $y = 1$.
   4. Sign check: $f(2) = -3$, negative, so the curve falls to $-\infty$ as $x \to 3^-$. $f(4) = 5$, positive, so it rises to $+\infty$ as $x \to 3^+$.
   5. Does it cross $y = 1$? $\frac{x + 1}{x - 3} = 1$ gives $x + 1 = x - 3$, impossible, so it never does.

   Answer: Vertical asymptote $x = 3$, horizontal asymptote $y = 1$, intercepts $(-1, 0)$ and $(0, -\frac{1}{3})$; negative just left of $x = 3$, positive just right.

3. Medium

   Sketch $f(x) = \frac{2x^2 - 8}{x^2 - 9}$.

   Show solution

   1. Factor: $f(x) = \frac{2(x - 2)(x + 2)}{(x - 3)(x + 3)}$. Nothing cancels. Domain $x \ne \pm 3$, vertical asymptotes $x = 3$ and $x = -3$.
   2. $x$-intercepts at $x = \pm 2$. $y$-intercept $f(0) = \frac{-8}{-9} = \frac{8}{9}$.
   3. Equal degrees: horizontal asymptote $y = \frac{2}{1} = 2$. Crossing check: $2x^2 - 8 = 2x^2 - 18$ has no solution, so the curve never meets $y = 2$.
   4. Sign check in the five intervals, using $x = -4, -2.5, 0, 2.5, 4$: $f(-4) = \frac{24}{7} > 0$, $f(-2.5) = \frac{4.5}{-2.75} < 0$, $f(0) > 0$, $f(2.5) < 0$, $f(4) > 0$.
   5. So the outer branches sit above $y = 2$, and the middle piece is an arch through $(-2, 0)$, $(0, \frac{8}{9})$ and $(2, 0)$ that falls to $-\infty$ at both asymptotes.

   Answer: Vertical asymptotes $x = \pm 3$, horizontal asymptote $y = 2$, intercepts $(-2, 0)$, $(2, 0)$ and $(0, \frac{8}{9})$. The outer branches lie above $y = 2$; the middle piece is an arch through the three intercepts that drops to $-\infty$ at both asymptotes.

4. Medium

   Sketch $f(x) = \frac{x^2 - x - 6}{x^2 - 4}$, including any hole.

   Show solution

   1. Factor: $f(x) = \frac{(x - 3)(x + 2)}{(x - 2)(x + 2)}$. Domain $x \ne 2, -2$.
   2. $(x + 2)$ cancels, so $x = -2$ is a hole. Simplified: $f(x) = \frac{x - 3}{x - 2}$, and at $x = -2$ that is $\frac{-5}{-4} = \frac{5}{4}$. Hole at $(-2, \frac{5}{4})$.
   3. $(x - 2)$ stays: vertical asymptote $x = 2$. Equal degrees: horizontal asymptote $y = 1$.
   4. Intercepts from the simplified form: $x$-intercept $(3, 0)$, $y$-intercept $f(0) = \frac{3}{2}$.
   5. Sign check: $f(1) = 2 > 0$, so the curve rises to $+\infty$ as $x \to 2^-$; $f(2.5) = -1 < 0$, so it falls to $-\infty$ as $x \to 2^+$, then climbs back through $(3, 0)$ toward $y = 1$.

   Answer: Hole $(-2, \frac{5}{4})$, vertical asymptote $x = 2$, horizontal asymptote $y = 1$, intercepts $(3, 0)$ and $(0, \frac{3}{2})$; positive left of $x = 2$, negative between $2$ and $3$.

5. Test level

   Sketch $f(x) = \frac{x^2 - 3x}{x + 1}$, including its oblique asymptote.

   Show solution

   1. Factor: $f(x) = \frac{x(x - 3)}{x + 1}$. Domain $x \ne -1$, vertical asymptote $x = -1$, no hole.
   2. $x$-intercepts $(0, 0)$ and $(3, 0)$; the $y$-intercept is the origin too.
   3. The numerator is one degree higher, so divide: $x^2 - 3x = (x + 1)(x - 4) + 4$. Oblique asymptote $y = x - 4$; the remainder term $\frac{4}{x + 1}$ shrinks to $0$ at both ends.
   4. Crossing check: $f(x) = x - 4$ would need $\frac{4}{x + 1} = 0$, impossible, so the curve never touches the line. The remainder term is positive for $x > -1$ and negative for $x < -1$, so the right branch sits above the line and the left branch below it.
   5. Near the asymptote: $f(-0.5) = \frac{1.75}{0.5} > 0$, so the curve rises to $+\infty$ as $x \to -1^+$; $f(-1.5) = \frac{6.75}{-0.5} < 0$, so it falls to $-\infty$ as $x \to -1^-$.

   Answer: Vertical asymptote $x = -1$, oblique asymptote $y = x - 4$, intercepts $(0, 0)$ and $(3, 0)$. The right branch comes down from $+\infty$, dips below the axis between its two intercepts, then rises along $y = x - 4$ from above; the left branch lies below the line and drops to $-\infty$ at the asymptote.

Learn It on Martian Lab

Martian Lab teaches this in the Intercepts & Asymptotes of Rational Functions lesson of its MHF4U Rational Functions unit. Watch it, then take the Rational Functions practice tests to see if it stuck.

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