A-Level Further Maths 9231 / Topic 1.2 / AS & A Level
Find the lines a curve never quite reaches.
A rational function is one polynomial divided by another. Its graph is steered by straight lines it gets closer and closer to: vertical lines where the denominator is zero, and a horizontal or slanted line far out to the left and right. Find those lines first and the sketch almost draws itself.
For \(y = \dfrac{P(x)}{Q(x)}\): each real root of \(Q(x) = 0\) that is not also a root of \(P(x) = 0\) gives a vertical asymptote. If \(\deg P < \deg Q\) the horizontal asymptote is \(y = 0\); if \(\deg P = \deg Q\) it is \(y = \) (ratio of leading coefficients); if \(\deg P = \deg Q + 1\), divide to get \(y = mx + c + \dfrac{r}{Q(x)}\), and \(y = mx + c\) is the oblique asymptote. The sign of the remainder term tells you which side the curve is on.
Fig. 1: y = (x² + 3x + 1)/(x + 2) = x + 1 - 1/(x + 2). The dashed lines x = -2 and y = x + 1 are its asymptotes. The curve crosses the x-axis where x² + 3x + 1 = 0 and the y-axis at 1/2
Section 01
Vertical and horizontal asymptotes.
An asymptote is a straight line that the curve approaches more and more closely as \(x\) or \(y\) becomes very large. For a rational function \(y = \dfrac{P(x)}{Q(x)}\) there are two places to look: where the denominator is zero (vertical asymptotes) and what happens as \(x \to \pm\infty\) (horizontal or oblique asymptotes).
Vertical: solve \(Q(x) = 0\). At each real root \(x = a\), check \(P(a) \ne 0\); then \(x = a\) is a vertical asymptote and \(y \to \pm\infty\) there. A quadratic denominator can give two, one or none (if \(Q(x) = 0\) has no real roots).
Horizontal, \(\deg P < \deg Q\): the denominator wins for large \(|x|\), so \(y \to 0\): the asymptote is \(y = 0\).
Horizontal, \(\deg P = \deg Q\): divide top and bottom by the highest power of \(x\); only the leading coefficients survive, so the asymptote is \(y = \dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q}\).
Worked example Linear over linear
Find the asymptotes of \(y = \dfrac{2x + 1}{x - 3}\).
Denominator zero: \(x - 3 = 0\), so \(x = 3\). The numerator there is \(7 \ne 0\), so \(x = 3\) is a vertical asymptote.
Equal degrees (both 1): divide top and bottom by \(x\): \(y = \dfrac{2 + \frac{1}{x}}{1 - \frac{3}{x}}\).
As \(x \to \pm\infty\), \(\dfrac{1}{x} \to 0\) and \(\dfrac{3}{x} \to 0\), so \(y \to \dfrac{2}{1} = 2\).
Answer: \(x = 3\) and \(y = 2\)
Worked example Quadratic over quadratic
Find the asymptotes of \(y = \dfrac{3x^{2} - 2}{x^{2} - 5x + 6}\).
\(x^{2} - 5x + 6 = (x - 2)(x - 3)\), which is zero at \(x = 2\) and \(x = 3\).
Numerator at \(x = 2\): \(12 - 2 = 10 \ne 0\); at \(x = 3\): \(27 - 2 = 25 \ne 0\). Both are vertical asymptotes.
Equal degrees (both 2): the horizontal asymptote is the ratio of the \(x^{2}\) coefficients, \(\dfrac{3}{1}\).
Answer: \(x = 2\), \(x = 3\) and \(y = 3\)
Your turn
Find any asymptotes of \(y = \dfrac{x - 4}{x^{2} + 9}\).
\(x^{2} + 9 = 0\) has no real roots, so there is no vertical asymptote. The numerator has lower degree than the denominator, so the only asymptote is \(y = 0\).
Examiner trap
Always check the numerator at each root of the denominator. In \(y = \dfrac{x^{2} - 4}{x^{2} - x - 2} = \dfrac{(x - 2)(x + 2)}{(x - 2)(x + 1)}\) the factor \(x - 2\) cancels: \(x = 2\) is a gap in the curve, not an asymptote. The only vertical asymptote is \(x = -1\) (and \(y = 1\) is horizontal).
Section 02
Oblique asymptotes by division.
When the numerator's degree is one more than the denominator's (in this syllabus, quadratic over linear), \(y\) grows without limit as \(x \to \pm\infty\), but it settles close to a slanted straight line. Divide the numerator by the denominator: the quotient is the line, and the remainder over the denominator tends to zero.
Write \(\dfrac{ax^{2} + bx + c}{x - p} = mx + k + \dfrac{r}{x - p}\). As \(x \to \pm\infty\), \(\dfrac{r}{x - p} \to 0\), so \(y\) gets closer and closer to \(mx + k\). The oblique asymptote is \(y = mx + k\); the vertical asymptote is still \(x = p\).
Worked example The curve in Fig. 1
Find the asymptotes of \(y = \dfrac{x^{2} + 3x + 1}{x + 2}\).
Match the \(x\) term: \(q - 2 = -1\), so \(q = 1\), and \((x - 1)(2x + 1) = 2x^{2} - x - 1\).
Then \(2x^{2} - x + 3 = (x - 1)(2x + 1) + 4\), so \(y = 2x + 1 + \dfrac{4}{x - 1}\).
Vertical asymptote \(x = 1\) (numerator there is \(4 \ne 0\)).
Answer: \(x = 1\) and \(y = 2x + 1\)
Your turn
Write \(y = \dfrac{x^{2} + 1}{x + 1}\) in the form \(mx + k + \dfrac{r}{x + 1}\) and state both asymptotes.
\(x^{2} + 1 = (x + 1)(x - 1) + 2\), so \(y = x - 1 + \dfrac{2}{x + 1}\). The asymptotes are \(x = -1\) and \(y = x - 1\).
Examiner trap
The oblique asymptote is the whole quotient, not just the leading term. For \(\dfrac{x^{2} + 3x + 1}{x + 2}\), dividing only \(x^{2}\) by \(x\) gives \(y = x\), which is parallel to the true asymptote \(y = x + 1\) but wrong. Do the full division and check it by expanding.
Stage 1 · Learn
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What is the vertical asymptote of \(y = \dfrac{3x - 1}{2x + 5}\)?
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What is the horizontal asymptote of \(y = \dfrac{4x^{2} - 1}{2x^{2} + x + 3}\)?
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What is the oblique asymptote of \(y = \dfrac{x^{2} + 5x + 2}{x + 3}\)?
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Which curve has no vertical asymptote?
Section 03
Which side does the curve approach from?
Knowing the asymptotes is not enough for a sketch: you also need to know which side of each line the curve lies on. The form \(y = (\text{line}) + \dfrac{r}{Q(x)}\) answers both questions by the sign of the fraction.
Large \(|x|\): if \(\dfrac{r}{Q(x)}\) is positive, the curve is just above the horizontal or oblique asymptote; if negative, just below. Check \(x \to +\infty\) and \(x \to -\infty\) separately: the sign often changes.
Near \(x = a\): put in a value just to the right (\(x \to a^{+}\)) and just to the left (\(x \to a^{-}\)) and decide the sign of \(y\): it goes to \(+\infty\) on one side or both.
Add the intercepts and you have a first sketch; the next lesson adds turning points.
Worked example Approaches for Fig. 1
For \(y = x + 1 - \dfrac{1}{x + 2}\), describe how the curve approaches each asymptote.
\(x \to +\infty\): \(x + 2 > 0\), so \(-\dfrac{1}{x + 2} < 0\): the curve is just below \(y = x + 1\).
\(x \to -\infty\): \(x + 2 < 0\), so \(-\dfrac{1}{x + 2} > 0\): the curve is just above \(y = x + 1\).
\(x \to -2^{+}\): \(x + 2\) is small and positive, so \(-\dfrac{1}{x + 2} \to -\infty\) and \(y \to -\infty\).
\(x \to -2^{-}\): \(x + 2\) is small and negative, so \(y \to +\infty\).
Answer: Right branch: from \(-\infty\) beside \(x = -2\) up to just below \(y = x + 1\). Left branch: from just above \(y = x + 1\) up to \(+\infty\) beside \(x = -2\) (as in Fig. 1)
Worked example A horizontal asymptote
For \(y = \dfrac{2x + 1}{x - 3}\), find whether the curve lies above or below \(y = 2\) for large positive and large negative \(x\), and its behaviour near \(x = 3\).
Answer: Above \(y = 2\) on the right, below it on the left; \(y \to +\infty\) as \(x \to 3^{+}\) and \(y \to -\infty\) as \(x \to 3^{-}\)
Your turn
For \(y = 2x + 1 + \dfrac{4}{x - 1}\), describe the approach to \(y = 2x + 1\) as \(x \to \pm\infty\) and the behaviour near \(x = 1\).
\(x \to +\infty\): \(\dfrac{4}{x - 1} > 0\), above the line; \(x \to -\infty\): below. \(y \to +\infty\) as \(x \to 1^{+}\) and \(y \to -\infty\) as \(x \to 1^{-}\).
Examiner trap
Do not judge the side from the sign of the remainder \(r\) alone. In \(-\dfrac{1}{x + 2}\) the top is negative, yet for \(x < -2\) the denominator is negative too and the fraction is positive. Decide the sign of the whole fraction for the values of \(x\) you are looking at.
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State the equations of the asymptotes of the curve \(y = \dfrac{2x^{2} + 3x - 1}{x + 1}\).
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For \(y = \dfrac{x^{2} + x + 1}{x - 1}\), how does the curve approach its oblique asymptote as \(x \to +\infty\)?
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The curve \(y = \dfrac{x^{2} + ax + 4}{x - 2}\) has oblique asymptote \(y = x + 5\). Find \(a\).
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Find all the asymptotes of \(y = \dfrac{x^{2} - 9}{x^{2} - x - 6}\).
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