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How do we sketch rational functions by finding their intercepts and asymptotes?

Sketch rational functions, identifying vertical, horizontal and oblique asymptotes.

Sketching rational functions using intercepts, vertical asymptotes, horizontal and oblique asymptotes via division, and sign analysis, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this dot point is asking

This dot point pulls together everything in the sketching strand. A rational function combines zeros, asymptotes and sign behaviour, and a clean sketch comes from working through a fixed checklist rather than plotting random points.

The checklist for a rational function

Domain and vertical asymptotes. Set $Q(x) = 0$ . Each zero of the denominator that is not cancelled by the numerator gives a vertical asymptote.
Intercepts. The $y$ intercept is the value at $x = 0$ . The $x$ intercepts are where $P(x) = 0$ (and $Q(x) \ne 0$ ).
End behaviour. Compare degrees of $P$ and $Q$ to find horizontal or oblique asymptotes.
Sign analysis. Use the zeros and asymptotes to split the number line into intervals and test the sign in each.

End behaviour rules

Finding an oblique asymptote

When the numerator degree is one more than the denominator, divide $P(x)$ by $Q(x)$ . The quotient is a linear expression $y = mx + c$ , which is the oblique asymptote, and the remainder term vanishes as $x \to \pm\infty$ .

Sketching a rational function with a horizontal asymptote

Sketch $y = \dfrac{2x}{x - 1}$ .

Step 1: Find the vertical asymptote

Set the denominator equal to zero to locate where the function is undefined. Each genuine zero of the denominator that is not cancelled by the numerator produces a vertical asymptote.

The denominator is zero at $x = 1$ . The numerator $2x$ is not zero there, so there is a vertical asymptote at $x = 1$ .

Step 2: Find the intercepts

Setting $x = 0$ gives the $y$ -intercept, and setting the numerator to zero (excluding denominator zeros) gives any $x$ -intercepts. Both intercepts are needed to anchor the sketch.

At $x = 0$ : $y = \tfrac{0}{-1} = 0$ , so the graph passes through the origin. This single point is both the $x$ -intercept and the $y$ -intercept.

Step 3: Determine the end behaviour

Compare the degrees of numerator and denominator to find the horizontal asymptote. When the degrees are equal, the asymptote is the ratio of the leading coefficients.

Both numerator and denominator have degree $1$ , so the horizontal asymptote is:

y = \frac{2}{1} = 2.

Step 4: Analyse the sign and direction near each asymptote

Checking the sign of the function on each side of the vertical asymptote tells us whether each branch runs to $+\infty$ or $-\infty$ , and knowing whether the curve approaches the horizontal asymptote from above or below completes the shape.

Just left of $x = 1$ : the denominator is small and negative while the numerator is near $2$ , so $y \to -\infty$ . Just right of $x = 1$ : the denominator is small and positive, so $y \to +\infty$ . For large positive $x$ , $y$ approaches $2$ from above; for large negative $x$ , it approaches $2$ from below.

Final answer: the curve has two branches separated by the vertical asymptote $x = 1$ , passes through the origin, and both branches approach the horizontal asymptote $y = 2$ .

Sketching a rational function with an oblique asymptote

Sketch $y = \dfrac{x^2 + 1}{x}$ .

Step 1: Find the vertical asymptote and intercepts

Set the denominator to zero to find the vertical asymptote. Then check for $x$ -intercepts by solving $x^2 + 1 = 0$ ; since this has no real solutions ( $x^2 + 1 > 0$ always), there are none. The function is also undefined at $x = 0$ so there is no $y$ -intercept.

Vertical asymptote: $x = 0$ . No $x$ -intercepts and no $y$ -intercept.

Step 2: Determine the end behaviour by polynomial division

When the numerator degree is exactly one more than the denominator degree, long division reveals an oblique asymptote. Dividing gives a linear quotient plus a remainder term that vanishes as $x \to \pm\infty$ .

\frac{x^2 + 1}{x} = x + \frac{1}{x}.

The oblique asymptote is $y = x$ , and the remainder $\tfrac{1}{x}$ vanishes at both extremes.

Step 3: Locate turning points by differentiating

Writing the function in divided form makes differentiation straightforward. Setting the derivative to zero finds the local extrema, which determine the precise shape of each branch.

Differentiating $x + \tfrac{1}{x}$ :

\frac{dy}{dx} = 1 - \frac{1}{x^2} = 0 \quad\Longrightarrow\quad x^2 = 1 \quad\Longrightarrow\quad x = \pm 1.

At $x = 1$ : $y = 1 + 1 = 2$ (local minimum on the positive branch). By the odd symmetry of $x + \tfrac{1}{x}$ , there is a local maximum at $(-1, -2)$ on the negative branch.

Step 4: Describe the final shape

For $x > 0$ the term $\tfrac{1}{x} > 0$ means the curve lies above the asymptote $y = x$ ; for $x < 0$ the term $\tfrac{1}{x} < 0$ means the curve lies below $y = x$ . Combined with the turning points, this gives the full picture.

Final answer: the graph has a vertical asymptote at $x = 0$ and oblique asymptote $y = x$ , with a local minimum at $(1, 2)$ and a local maximum at $(-1, -2)$ . Both branches hug the line $y = x$ for large $|x|$ .

Sign and shape

After placing intercepts and asymptotes, a sign table over the intervals between them tells you which side of each asymptote the curve sits on. This is usually enough to sketch without finding every turning point, though setting the derivative to zero pins down any local maxima or minima precisely.

Locating turning points by calculus

For a rational function with an oblique asymptote, the cleanest route to turning points is to write the function in divided form $f(x) = (mx + c) + \dfrac{r}{x - a}$ and differentiate. The TASC questions above all reduce to $f'(x) = 1 - \dfrac{r}{(x - a)^2} = 0$ , giving $(x - a)^2 = r$ and two symmetric critical points either side of the vertical asymptote. The branch above the asymptote carries a local minimum and the branch below carries a local maximum, which is a useful sanity check on the sign of the second derivative.

Why this matters

Rational functions appear throughout the course, from partial fractions in integration to modelling problems with limiting behaviour. A systematic checklist, intercepts then asymptotes then sign then calculus for turning points, produces a correct sketch quickly and avoids the guesswork of point plotting.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20227 marksConsider

f(x) = \dfrac{x^2 - 5x + 7}{x - 2}

. (a) Determine the locations of any zeros and critical points of

f(x)

. (b) Sketch the graph of

y = f(x)

Show worked answer →

(a) Zeros: set the numerator to zero. $x^2 - 5x + 7 = 0$ has discriminant $25 - 28 = -3 < 0$ , so there are no real zeros. There is a vertical asymptote at $x = 2$ (denominator zero). Divide to find the oblique asymptote: $x^2 - 5x + 7 = (x - 2)(x - 3) + 1$ , so $f(x) = x - 3 + \dfrac{1}{x - 2}$ and the oblique asymptote is $y = x - 3$ . For critical points, $f'(x) = 1 - \dfrac{1}{(x - 2)^2} = 0$ gives $(x - 2)^2 = 1$ , so $x = 1$ or $x = 3$ . Then $f(1) = -3$ (local maximum) and $f(3) = 1$ (local minimum). (4 marks)

(b) Sketch the two branches separated by the vertical asymptote $x = 2$ , approaching the oblique asymptote $y = x - 3$ at both ends. The left branch passes through the local maximum $(1, -3)$ ; the right branch passes through the local minimum $(3, 1)$ . There are no $x$ -intercepts and the $y$ -intercept is $f(0) = \dfrac{7}{-2} = -3.5$ . (3 marks)

TCE 20237 marksConsider

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

. (a) Determine the intercepts and critical points of

f(x)

. (b) Sketch the graph of

y = f(x)

Show worked answer →

(a) The $y$ -intercept is $f(0) = \dfrac{3}{-1} = -3$ . For $x$ -intercepts, $x^2 + 3 = 0$ has no real solution, so there are none. There is a vertical asymptote at $x = 1$ . Polynomial division gives $x^2 + 3 = (x - 1)(x + 1) + 4$ , so $f(x) = x + 1 + \dfrac{4}{x - 1}$ with oblique asymptote $y = x + 1$ . Critical points: $f'(x) = 1 - \dfrac{4}{(x - 1)^2} = 0$ gives $(x - 1)^2 = 4$ , so $x = 3$ or $x = -1$ . Then $f(3) = 6$ (local minimum on the right branch) and $f(-1) = -2$ (local maximum on the left branch). (5 marks)

(b) Draw the vertical asymptote $x = 1$ and oblique asymptote $y = x + 1$ . The left branch rises to the local maximum $(-1, -2)$ and passes through the $y$ -intercept $(0, -3)$ ; the right branch dips to the local minimum $(3, 6)$ . (2 marks)

TCE 20219 marksConsider

f(x) = \dfrac{x^2 - 4x + 20}{x - 2}

. (a) Determine the intercepts and critical points of

f(x)

. (b) Find the behaviour of

f(x)

x \to \pm\infty

. (c) Sketch the graph of

y = f(x)

Show worked answer →

(a) The $y$ -intercept is $f(0) = \dfrac{20}{-2} = -10$ . For $x$ -intercepts, $x^2 - 4x + 20 = 0$ has discriminant $16 - 80 < 0$ , so none. There is a vertical asymptote at $x = 2$ . Since $x^2 - 4x + 20 = (x - 2)^2 + 16$ , $f(x) = (x - 2) + \dfrac{16}{x - 2}$ . Then $f'(x) = 1 - \dfrac{16}{(x - 2)^2} = 0$ gives $(x - 2)^2 = 16$ , so $x = 6$ or $x = -2$ . Then $f(6) = 8$ (local minimum) and $f(-2) = -8$ (local maximum). (4 marks)

(b) As $x \to \pm\infty$ , $\dfrac{16}{x - 2} \to 0$ , so $f(x)$ approaches the oblique asymptote $y = x - 2$ . (1 mark)

(c) Sketch the vertical asymptote $x = 2$ and oblique asymptote $y = x - 2$ . The left branch peaks at the local maximum $(-2, -8)$ ; the right branch bottoms at the local minimum $(6, 8)$ . (4 marks)