A-Level Further Maths 9231 / Topic 1.2 / AS & A Level
Find the features, then draw the curve.
A good sketch of a rational function is not a plot of points. It is a short list of facts: where the asymptotes are, where the curve meets the axes, where it turns and what it does near each asymptote. The discriminant then tells you exactly which values of \(y\) the curve can and cannot reach, and the finished sketch solves equations and inequalities for you.
To sketch \(y = \dfrac{P(x)}{Q(x)}\): asymptotes, intercepts, turning points (quotient rule), then behaviour near each asymptote. To find the values \(y\) can take, write \(y\,Q(x) = P(x)\) as a quadratic in \(x\); real \(x\) needs \(b^{2} - 4ac \ge 0\), an inequality in \(y\). The boundary values of \(y\) are the \(y\)-coordinates of the turning points.
Fig. 1: y = x(x + 3)/(x - 1), with vertical asymptote x = 1, oblique asymptote y = x + 4, intercepts at O and (-3, 0), a maximum at (-1, 1) and a minimum at (3, 9). No part of the curve has 1 < y < 9
Section 01
A checklist for every sketch.
Detailed plotting is not required; the examiner wants the significant features, each in the right place and each labelled. Work through the same checklist every time, and only then join the pieces up.
Vertical asymptotes: where the denominator is zero (and the numerator is not).
Horizontal or oblique asymptote: compare degrees, dividing if the numerator's degree is one more than the denominator's.
Intercepts: put \(x = 0\) for the \(y\)-intercept; put the numerator equal to zero for the \(x\)-intercepts.
Turning points: \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{u}{v}\right) = \dfrac{v u' - u v'}{v^{2}}\); only the numerator can be zero, so solve \(v u' - u v' = 0\).
Behaviour near the asymptotes: the sign of \(y\) just either side of each vertical asymptote, and whether the curve approaches the other asymptote from above or below as \(x \to \pm\infty\).
Worked example Quadratic over linear
Sketch \(y = \dfrac{x(x + 3)}{x - 1}\), showing its asymptotes, intercepts and turning points.
Vertical asymptote \(x = 1\). Dividing, \(x^{2} + 3x = (x - 1)(x + 4) + 4\), so \(y = x + 4 + \dfrac{4}{x - 1}\) and the oblique asymptote is \(y = x + 4\).
Intercepts: \(y = 0\) when \(x = 0\) or \(x = -3\); the curve passes through \(O\) and \((-3, 0)\).
Zero at \(x = -1\) (\(y = 1\)) and \(x = 3\) (\(y = 9\)). The gradient changes from \(+\) to \(-\) at \(x = -1\) and from \(-\) to \(+\) at \(x = 3\): a maximum and a minimum.
Near \(x = 1\): \(\dfrac{4}{x - 1} \to +\infty\) as \(x \to 1^{+}\) and \(\to -\infty\) as \(x \to 1^{-}\). As \(x \to +\infty\) the curve is just above \(y = x + 4\); as \(x \to -\infty\) just below it.
Answer: Two branches as in Fig. 1: maximum \((-1, 1)\), minimum \((3, 9)\), through \(O\) and \((-3, 0)\), asymptotes \(x = 1\) and \(y = x + 4\)
Worked example Quadratic over quadratic
Find the asymptotes, intercepts and turning point of \(y = \dfrac{x^{2} - 4}{x^{2} - 9}\), and describe its shape.
Vertical asymptotes \(x = 3\) and \(x = -3\). Equal degrees, leading coefficients \(1\) and \(1\): horizontal asymptote \(y = 1\).
Intercepts: \((2, 0)\), \((-2, 0)\) and \(\left(0, \dfrac{4}{9}\right)\).
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x(x^{2} - 9) - 2x(x^{2} - 4)}{(x^{2} - 9)^{2}} = \dfrac{-10x}{(x^{2} - 9)^{2}}\), zero only at \(x = 0\); the gradient goes from \(+\) to \(-\), so \(\left(0, \dfrac{4}{9}\right)\) is a maximum.
Write \(y = 1 + \dfrac{5}{x^{2} - 9}\). For \(|x| > 3\), \(y > 1\): the outer branches lie above \(y = 1\) and rise to \(+\infty\) at the asymptotes.
For \(-3 < x < 3\), \(y < 1\): the middle branch comes up from \(-\infty\), crosses at \(x = -2\), peaks at \(\left(0, \frac{4}{9}\right)\), crosses at \(x = 2\) and falls to \(-\infty\).
Answer: Three branches, symmetrical in the \(y\)-axis; maximum \(\left(0, \frac{4}{9}\right)\); asymptotes \(x = \pm 3\), \(y = 1\)
Examiner trap
A curve never crosses a vertical asymptote, but it CAN cross a horizontal or oblique one at a finite \(x\). Check by solving \(f(x) = \) asymptote: \(\dfrac{x^{2} - x - 6}{x^{2} - 1} = 1\) gives \(x = -5\), so that curve crosses \(y = 1\) at \((-5, 1)\).
Section 02
The discriminant gives the range.
Which values does \(y\) actually take? Fix a value of \(y\) and ask whether some real \(x\) produces it. Multiply out \(y = f(x)\) into a quadratic in \(x\) whose coefficients contain \(y\); a real \(x\) exists exactly when that quadratic's discriminant is not negative.
Clear the fraction: \(y\,Q(x) = P(x)\).
Collect as \(Ax^{2} + Bx + C = 0\), where \(A\), \(B\), \(C\) may involve \(y\).
Real \(x\): \(B^{2} - 4AC \ge 0\), then solve this inequality in \(y\).
Bonus: at each boundary value of \(y\) the quadratic has a repeated root, which is the \(x\)-coordinate of a turning point.
Worked example The range of the Fig. 1 curve
Find the set of values taken by \(y = \dfrac{x(x + 3)}{x - 1}\), and hence the turning points.
\(y(x - 1) = x^{2} + 3x\), so \(x^{2} + (3 - y)x + y = 0\).
Real \(x\): \((3 - y)^{2} - 4y \ge 0\), so \(y^{2} - 10y + 9 \ge 0\), i.e. \((y - 1)(y - 9) \ge 0\).
For real \(x\): \(y^{2} - 4(3y - 5) \ge 0\), so \(y^{2} - 12y + 20 \ge 0\).
\((y - 2)(y - 10) \ge 0\), so \(y \le 2\) or \(y \ge 10\).
Hence no real \(x\) gives \(2 < y < 10\). (The values 2 and 10 are reached at \(x = 1\) and \(x = 5\).)
Answer: \(y \le 2\) or \(y \ge 10\), so \(y\) cannot lie between 2 and 10
Your turn
Find the set of values of \(y = \dfrac{x}{x^{2} + 1}\).
\(yx^{2} - x + y = 0\); real \(x\) needs \(1 - 4y^{2} \ge 0\), so \(-\frac{1}{2} \le y \le \frac{1}{2}\).
Examiner trap
When \(A\) contains \(y\), the equation is only a quadratic if \(A \ne 0\). For \(y = \dfrac{x^{2} - 4}{x^{2} - 9}\): \((y - 1)x^{2} + (4 - 9y) = 0\) gives \((y - 1)(9y - 4) \ge 0\), but \(y = 1\) makes the equation \(-5 = 0\). The range is \(y \le \frac{4}{9}\) or \(y > 1\), with a strict inequality.
Stage 1 · Learn
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What are the vertical asymptotes of \(y = \dfrac{x^{2} + 1}{x^{2} - x - 6}\)?
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Where are the turning points of \(y = \dfrac{x(x + 3)}{x - 1}\), whose derivative is \(\dfrac{(x - 3)(x + 1)}{(x - 1)^{2}}\)?
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Setting \(y = \dfrac{x(x + 3)}{x - 1}\) gives \(x^{2} + (3 - y)x + y = 0\). Which condition gives the set of values of \(y\)?
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What is the set of values taken by \(y = \dfrac{x + 1}{x^{2} + 3}\)?
Section 03
Sketch plus algebra for inequalities.
Once the sketch is drawn, an equation \(f(x) = k\) is the question 'where does the horizontal line \(y = k\) meet the curve?' In Fig. 1, \(\dfrac{x(x + 3)}{x - 1} = k\) has two roots if \(k < 1\) or \(k > 9\), one repeated root if \(k = 1\) or \(k = 9\), and none if \(1 < k < 9\). Inequalities work the same way, with algebra to find the exact crossing points.
Never multiply an inequality by a denominator such as \(x - 1\) whose sign you do not know.
Multiply by its square \((x - 1)^{2}\), which is positive, then factorise with \((x - 1)\) as a factor.
If the denominator is positive for every \(x\) (such as \(x^{2} + 1\)), you may multiply straight through.
Check the answer against the sketch, and exclude any \(x\) where the function is undefined.
Worked example A line in the gap
Solve \(\dfrac{x(x + 3)}{x - 1} < 2\).
From Fig. 1, \(y = 2\) lies in the gap \(1 < y < 9\): the left branch is always below it and the right branch always above it. Expect \(x < 1\).
\((x - 1)(x - 2)(x - 5) > 0\): the critical values are 1, 2 and 5.
The cubic is positive for \(1 < x < 2\) and for \(x > 5\). The sketch agrees: the right branch dips below \(y = 10\) only between \(x = 2\) and \(x = 5\), round its minimum \((3, 9)\).
Multiplying \(\dfrac{x(x + 3)}{x - 1} < 2\) by \(x - 1\) gives \(x^{2} + x + 2 < 0\), which has no solutions, yet every \(x < 1\) works. For \(x < 1\) the factor \(x - 1\) is negative and the inequality sign should have reversed. Use \((x - 1)^{2}\).
Stage 2 · Exam
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Find the set of values of \(y = \dfrac{x^{2} + 5}{x - 2}\) for real \(x \ne 2\).
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Find the turning points of \(y = \dfrac{x^{2} + 5}{x - 2}\) and their nature.
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Solve \(\dfrac{x^{2} + 5}{x - 2} > 14\).
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For \(y = \dfrac{x^{2} - 3}{x - 2}\), between which two values can \(y\) not lie?
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