A-Level Further Maths 9231 / Topic 1.2 / AS & A Level
One curve, four relatives.
Once you can sketch \(y = f(x)\), you can sketch \(y^{2} = f(x)\), \(y = \dfrac{1}{f(x)}\), \(y = |f(x)|\) and \(y = f(|x|)\) without starting again. Each one follows a few fixed rules about zeros, asymptotes, turning points and sign, and the sketches then count solutions and solve inequalities for you.
Start from \(y = f(x)\). For \(y^{2} = f(x)\): keep only where \(f(x) \ge 0\) and draw \(y = \pm\sqrt{f(x)}\), symmetric in the \(x\)-axis. For \(y = \dfrac{1}{f(x)}\): zeros become vertical asymptotes, vertical asymptotes become zeros, the sign stays the same and a maximum becomes a minimum. For \(y = |f(x)|\): reflect the parts below the \(x\)-axis upwards. For \(y = f(|x|)\): keep the part for \(x \ge 0\) and reflect it in the \(y\)-axis.
Fig. 1: y = f(x) with f(x) = x² - 2x - 3, and y = 1/f(x). The zeros x = -1 and x = 3 become vertical asymptotes, the minimum (1, -4) becomes the maximum (1, -1/4), and the four points where f(x) = ±1 lie on both curves
Section 01
y squared: take the root both ways.
If \(y^{2} = f(x)\) then \(y = \pm\sqrt{f(x)}\). A square cannot be negative, so the curve exists only where \(f(x) \ge 0\), and every point \((x, y)\) on it has a partner \((x, -y)\): the curve is symmetric in the \(x\)-axis. Fig. 2 shows \(y^{2} = \dfrac{x - 1}{x + 2}\) on top of \(y = \dfrac{x - 1}{x + 2}\).
Where \(f(x) < 0\): no curve at all. Turning points of \(f\) below the \(x\)-axis disappear.
Where \(f(x) = 0\): the curve crosses the \(x\)-axis with a vertical tangent (the upper and lower halves join smoothly, not in a point).
Where \(f(x) = 1\): the points \((x, \pm 1)\) are on the curve, and \((x, 1)\) is on \(y = f(x)\) too, so it stays fixed.
Where \(0 < f(x) < 1\): \(\sqrt{f(x)} > f(x)\), so the curve moves away from the \(x\)-axis; where \(f(x) > 1\), \(\sqrt{f(x)} < f(x)\) and it moves towards it.
Turning points of \(f\) with \(f > 0\) stay at the same \(x\), with \(y = \pm\sqrt{f}\). A horizontal asymptote \(y = c\) with \(c > 0\) becomes \(y = \pm\sqrt{c}\); a vertical asymptote where \(f \to +\infty\) stays.
Fig. 2: y = (x - 1)/(x + 2) (dashed) and y² = (x - 1)/(x + 2). The new curve exists only for x < -2 or x ≥ 1, crosses the x-axis vertically at (1, 0) and approaches y = 1 and y = -1
Worked example Read the features from f
For the curve \(y^{2} = \dfrac{x - 1}{x + 2}\) (Fig. 2), state the set of values of \(x\) for which it exists, its asymptotes and where it meets the axes.
\(\dfrac{x - 1}{x + 2} \ge 0\) when the top and bottom have the same sign (or the top is zero): \(x \ge 1\) or \(x < -2\). The value \(x = -2\) is excluded because the fraction is undefined there.
As \(x \to \pm\infty\), \(f(x) \to 1\), so \(y \to \pm 1\): asymptotes \(y = 1\) and \(y = -1\).
As \(x \to -2\) from below, \(f(x) \to +\infty\), so \(y \to \pm\infty\): the asymptote \(x = -2\) stays.
\(f(x) = 0\) only at \(x = 1\), so the curve meets the \(x\)-axis at \((1, 0)\) with a vertical tangent; \(x = 0\) is not allowed, so it does not meet the \(y\)-axis.
Answer: Exists for \(x < -2\) or \(x \ge 1\); asymptotes \(x = -2\), \(y = 1\), \(y = -1\); meets the axes only at \((1, 0)\)
Worked example A turning point carried across
The curve \(y = \dfrac{2x}{x^{2} + 1}\) has a maximum at \((1, 1)\), a minimum at \((-1, -1)\) and the asymptote \(y = 0\). Describe the curve \(y^{2} = \dfrac{2x}{x^{2} + 1}\) and state the set of possible values of \(y\).
\(x^{2} + 1 > 0\), so \(f(x) \ge 0\) exactly when \(x \ge 0\). The curve exists only for \(x \ge 0\), and the minimum at \((-1, -1)\) disappears.
At the maximum \(f = 1\), so \(y = \pm 1\): the curve has a maximum at \((1, 1)\) and a minimum at \((1, -1)\).
\(f(0) = 0\): the curve passes through the origin with a vertical tangent.
As \(x \to \infty\), \(f(x) \to 0\), so \(y \to 0\) from above and below: \(y = 0\) is still an asymptote.
The largest value of \(\sqrt{f}\) is \(\sqrt{1} = 1\).
Answer: Exists for \(x \ge 0\) only; symmetric in the \(x\)-axis; vertical tangent at the origin; turning points \((1, 1)\) and \((1, -1)\); asymptote \(y = 0\); \(-1 \le y \le 1\)
Your turn
State the set of values of \(x\) for which the curve \(y^{2} = \dfrac{x - 2}{x + 1}\) exists.
Top and bottom have the same sign, or the top is zero: \(x < -1\) or \(x \ge 2\).
Examiner trap
Do not draw only the top half: \(y^{2} = f(x)\) always gives \(y = +\sqrt{f(x)}\) AND \(y = -\sqrt{f(x)}\). And at a zero of \(f\) the curve is vertical, not a sharp corner or a cusp.
Section 02
Reciprocal: zeros and asymptotes swap.
Each \(y\)-value of \(y = \dfrac{1}{f(x)}\) is the reciprocal of the \(y\)-value of \(y = f(x)\) at the same \(x\). Small values become large, large values become small, and the sign never changes. Fig. 1 shows the effect on \(f(x) = x^{2} - 2x - 3\).
On \(y = f(x)\)
On \(y = \dfrac{1}{f(x)}\)
zero at \(x = a\)
vertical asymptote \(x = a\)
vertical asymptote \(x = a\)
zero at \(x = a\), point \((a, 0)\)
horizontal asymptote \(y = c\), \(c \ne 0\)
horizontal asymptote \(y = \dfrac{1}{c}\)
\(f(x) \to \pm\infty\) as \(x \to \pm\infty\)
asymptote \(y = 0\)
maximum \((p, q)\), \(q \ne 0\)
minimum \(\left(p, \dfrac{1}{q}\right)\)
minimum \((p, q)\), \(q \ne 0\)
maximum \(\left(p, \dfrac{1}{q}\right)\)
\(f(x) = 1\) or \(f(x) = -1\)
same point: fixed
Worked example Reciprocal of a quadratic
Given \(f(x) = x^{2} - 2x - 3\), describe the key features of \(y = \dfrac{1}{f(x)}\) (Fig. 1).
\(f(x) = (x + 1)(x - 3)\) has zeros \(-1\) and \(3\), so \(\dfrac{1}{f}\) has vertical asymptotes \(x = -1\) and \(x = 3\).
\(f(x) \to +\infty\) as \(x \to \pm\infty\), so \(\dfrac{1}{f(x)} \to 0^{+}\): the asymptote is \(y = 0\), approached from above.
\(f\) has a minimum at \((1, -4)\), so \(\dfrac{1}{f}\) has a maximum at \(\left(1, -\dfrac{1}{4}\right)\).
\(f(0) = -3\), so the \(y\)-intercept is \(\left(0, -\dfrac{1}{3}\right)\).
Signs: \(f < 0\) for \(-1 < x < 3\), so the middle branch lies below the \(x\)-axis; the outer branches lie above it.
Answer: Asymptotes \(x = -1\), \(x = 3\), \(y = 0\); maximum \(\left(1, -\frac{1}{4}\right)\); \(y\)-intercept \(-\frac{1}{3}\); no \(x\)-intercepts
Worked example Reciprocal of a rational function
Given \(f(x) = \dfrac{x - 1}{x + 2}\), find the asymptotes and the intercepts of \(y = \dfrac{1}{f(x)}\).
\(\dfrac{1}{f(x)} = \dfrac{x + 2}{x - 1}\).
The zero of \(f\) at \(x = 1\) becomes the vertical asymptote \(x = 1\).
The vertical asymptote of \(f\) at \(x = -2\) becomes a zero: the point \((-2, 0)\).
The horizontal asymptote \(y = 1\) of \(f\) becomes \(y = \dfrac{1}{1} = 1\): unchanged.
\(f(0) = -\dfrac{1}{2}\), so the \(y\)-intercept of \(\dfrac{1}{f}\) is \(-2\).
Answer: Asymptotes \(x = 1\) and \(y = 1\); intercepts \((-2, 0)\) and \((0, -2)\)
Your turn
A curve \(y = f(x)\) has a vertical asymptote \(x = -3\), a horizontal asymptote \(y = 4\), a zero at \(x = 2\) and a maximum at \((5, 2)\). State the corresponding features of \(y = \dfrac{1}{f(x)}\).
Point \((-3, 0)\); horizontal asymptote \(y = \frac{1}{4}\); vertical asymptote \(x = 2\); minimum at \(\left(5, \frac{1}{2}\right)\).
Examiner trap
\(\dfrac{1}{f(x)}\) is the reciprocal of the \(y\)-value, not the inverse function \(f^{-1}(x)\), and not a reflection. The minimum \((1, -4)\) becomes \(\left(1, -\frac{1}{4}\right)\): same \(x\), same sign. It does not become \((1, 4)\) or \(\left(1, \frac{1}{4}\right)\).
Stage 1 · Learn
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The curve \(y^{2} = f(x)\) exists only for the values of \(x\) where
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\(y = f(x)\) has a zero at \(x = 2\). What does \(y = \dfrac{1}{f(x)}\) have at \(x = 2\)?
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\(y = f(x)\) has a minimum at \((1, -4)\). Which point is a turning point of \(y = \dfrac{1}{f(x)}\)?
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\(f(a) = 0.25\). Which points with \(x = a\) lie on the curve \(y^{2} = f(x)\)?
Section 03
Modulus graphs: reflect, then count.
Two different modulus graphs come from \(y = f(x)\). In \(y = |f(x)|\) the modulus acts on the \(y\)-values: any part below the \(x\)-axis is reflected in the \(x\)-axis. In \(y = f(|x|)\) it acts on the \(x\)-values: \(f(|x|) = f(|-x|)\), so the graph is symmetric in the \(y\)-axis; keep the part for \(x \ge 0\) and reflect it. Anything to the left of the \(y\)-axis on \(y = f(x)\), including a vertical asymptote, is lost.
Count solutions of \(|f(x)| = k\) or \(f(|x|) = k\) by drawing the horizontal line \(y = k\) on the sketch and counting meetings.
Solve an inequality such as \(|f(x)| < 1\): find where \(f(x) = 1\) or \(f(x) = -1\) exactly, then read off from the sketch where the curve lies between the lines.
Because both sides are non-negative, \(\left|\dfrac{p(x)}{q(x)}\right| < 1\) can also be squared: \(p(x)^{2} < q(x)^{2}\) with \(q(x) \ne 0\).
Worked example Use the graph of f(|x|)
Given \(f(x) = \dfrac{x - 1}{x + 2}\), describe the graph of \(y = f(|x|)\) and find the set of values of \(k\) for which \(f(|x|) = k\) has exactly two solutions.
For \(x \ge 0\) the graph is \(y = f(x)\): it starts at \(\left(0, -\dfrac{1}{2}\right)\) and increases towards the asymptote \(y = 1\) (it crosses the \(x\)-axis at \(x = 1\)).
Reflect this in the \(y\)-axis. The asymptote \(x = -2\) lies in the discarded part, so \(y = f(|x|)\) has no vertical asymptote: it is a smooth trough with lowest point \(\left(0, -\dfrac{1}{2}\right)\) and \(y = 1\) as its asymptote on both sides.
A line \(y = k\) with \(-\dfrac{1}{2} < k < 1\) meets each half once: two solutions.
\(k = -\dfrac{1}{2}\) gives one solution (\(x = 0\)); \(k \ge 1\) or \(k < -\dfrac{1}{2}\) gives none.
Answer: \(-\frac{1}{2} < k < 1\)
Worked example Solve |f(x)| < 1
Find the set of values of \(x\) for which \(\left|\dfrac{x - 1}{x + 2}\right| < 1\).
Boundary values: \(f(x) = 1\) gives \(x - 1 = x + 2\), impossible (\(y = 1\) is the asymptote). \(f(x) = -1\) gives \(x - 1 = -x - 2\), so \(x = -\dfrac{1}{2}\).
Use Fig. 2 (dashed curve) reflected to \(y = |f(x)|\): for \(x < -2\), \(f(x) > 1\); for \(-2 < x < -\dfrac{1}{2}\), \(f(x) < -1\) so \(|f(x)| > 1\); for \(x > -\dfrac{1}{2}\), \(-1 < f(x) < 1\).
Using Fig. 1, how many real solutions does \(|x^{2} - 2x - 3| = 3\) have?
Four. \(y = |f(x)|\) has the middle part reflected up to a maximum at \((1, 4)\). The line \(y = 3\) meets the outer arms twice (\(x = 1 \pm \sqrt{7}\)) and the reflected arch twice (\(x = 0\) and \(x = 2\)).
Examiner trap
Do not confuse \(|f(x)|\) with \(f(|x|)\). \(|f(x)|\) is never below the \(x\)-axis but can lose its symmetry; \(f(|x|)\) is always symmetric in the \(y\)-axis and CAN go below the \(x\)-axis, as the trough at \(\left(0, -\frac{1}{2}\right)\) shows.
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Given \(f(x) = \dfrac{x + 3}{x - 1}\), which statement about \(y = \dfrac{1}{f(x)}\) is true?
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Find the set of values of \(x\) for which \(\left|\dfrac{x + 1}{x - 2}\right| < 1\).
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Find the set of values of \(x\) for which the curve \(y^{2} = \dfrac{x - 4}{x + 1}\) exists.
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How many real solutions does the equation \(|x^{2} - 4x + 3| = 1\) have?
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