IGCSE · Practice questions · Graphs of exponential and logarithmic functions

Asymptote, intercepts, exact.

Six original Cambridge-style questions on exponential and logarithmic graphs: exact values of logarithms, sketching \(y = ke^{nx} + a\) and \(y = k\ln(ax + b)\) with their asymptotes, domains, ranges and intercepts, an exponential and its inverse, and finding constants from a graph.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Asymptote, side, intercepts, then shape.

Definition: \(\log_{a} x = y\) means \(a^{y} = x\); \(\ln 1 = 0\), \(\ln e = 1\), \(\ln e^{x} = x\), \(e^{\ln x} = x\).
Exponential \(y = ke^{nx} + a\): asymptote \(y = a\); above it if \(k > 0\), below if \(k < 0\); \(y\)-intercept \(k + a\).
Logarithm \(y = k\ln(ax + b)\): domain \(ax + b > 0\); asymptote \(x = -\frac{b}{a}\); \(x\)-intercept from \(ax + b = 1\).
Exact: on Paper 1 leave intercepts as \(\ln 2\), \(2\ln 3\), \(\frac{1}{2}\ln 5\).
Inverse: reflect in \(y = x\); asymptotes, intercepts, domain and range all swap.

Recall

[3 marks]

Find the exact value of

(a) \(\log_{3} 81\), (b) \(\log_{5} \frac{1}{25}\), (c) \(\ln \frac{1}{e^{3}}\).

(a) \(3^{4} = 81\), so \(\log_{3} 81 = 4\). ✓
(b) \(5^{-2} = \frac{1}{25}\), so \(\log_{5} \frac{1}{25} = -2\). ✓
(c) \(\frac{1}{e^{3}} = e^{-3}\), so \(\ln \frac{1}{e^{3}} = -3\). ✓

Calculation

[4 marks]

Sketch the curve \(y = 3e^{2x} - 6\). State the equation of the asymptote, the range, and the exact coordinates of the points where the curve meets the axes.

As \(x \to -\infty\), \(3e^{2x} \to 0\): asymptote \(y = -6\). \(k = 3 > 0\), so the curve is above it and rises steeply to the right; range \(y > -6\). ✓
\(y\)-intercept: \(3 - 6 = -3\), so \((0, -3)\). ✓
\(x\)-intercept: \(3e^{2x} = 6\), \(e^{2x} = 2\), \(2x = \ln 2\), so \(\left(\frac{1}{2}\ln 2, 0\right)\). ✓
Sketch: the dashed line \(y = -6\), the curve coming in from the left just above it, through \((0, -3)\) and \(\left(\frac{1}{2}\ln 2, 0\right)\), then rising more and more steeply. ✓

Calculation

[4 marks]

Sketch the curve \(y = 3\ln(2x + 1)\), stating its domain, the equation of its asymptote and the coordinates of any points where it meets the axes.

\(2x + 1 > 0\) gives the domain \(x > -\frac{1}{2}\); the asymptote is \(x = -\frac{1}{2}\). ✓
\(x\)-intercept: \(2x + 1 = 1\) gives \(x = 0\); \(y\)-intercept: \(3\ln 1 = 0\). So the curve passes through the origin, and that is its only point on either axis. ✓
\(k = 3 > 0\) and \(a = 2 > 0\): the curve rises from \(-\infty\) beside the asymptote, through \((0, 0)\), and keeps rising more and more slowly. ✓
Range: all real \(y\). ✓

Analysis

[5 marks]

The curve \(C\) has equation \(y = -2\ln(4 - x)\).

(a) State the domain of the function and the equation of the asymptote of \(C\).

(b) Find the exact coordinates of the points where \(C\) meets the axes.

(a) \(4 - x > 0\) gives \(x < 4\); the asymptote is \(x = 4\). The curve lies to the LEFT of it. ✓
(b) \(y = 0\): \(4 - x = 1\), so \(x = 3\): \((3, 0)\). ✓
\(x = 0\): \(y = -2\ln 4\): \((0, -2\ln 4)\), about \((0, -2.8)\). ✓
(c) As \(x \to 4\) from the left, \(\ln(4 - x) \to -\infty\), so \(y = -2\ln(4 - x) \to +\infty\). As \(x \to -\infty\), \(\ln(4 - x) \to +\infty\), so \(y \to -\infty\). ✓
So the curve rises slowly from bottom left, through \((0, -2\ln 4)\) and \((3, 0)\), then shoots up beside the asymptote \(x = 4\). ✓

Analysis

[5 marks]

\(f(x) = e^{2x} - 4\) for \(x \in \mathbb{R}\).

(a) State the range of \(f\).

(b) Find \(f^{-1}(x)\) and state its domain.

(c) On one diagram, sketch \(y = f(x)\) and \(y = f^{-1}(x)\), giving the equations of the asymptotes and the exact coordinates of the points where each graph meets the axes.

(a) \(e^{2x} > 0\), so \(f(x) > -4\). ✓
(b) \(y = e^{2x} - 4\): \(e^{2x} = y + 4\), \(x = \frac{1}{2}\ln(y + 4)\). So \(f^{-1}(x) = \frac{1}{2}\ln(x + 4)\), with domain \(x > -4\) (the range of \(f\)). ✓
(c) \(y = f(x)\): asymptote \(y = -4\); \(y\)-intercept \(1 - 4 = -3\), so \((0, -3)\); \(x\)-intercept \(e^{2x} = 4\), \(2x = \ln 4\), \(x = \frac{1}{2}\ln 4 = \ln 2\), so \((\ln 2, 0)\). ✓
\(y = f^{-1}(x)\): asymptote \(x = -4\); the points reflect in \(y = x\): \((-3, 0)\) and \((0, \ln 2)\). Check: \(\frac{1}{2}\ln(-3 + 4) = 0\) and \(\frac{1}{2}\ln 4 = \ln 2\). ✓
The two curves are reflections of each other in \(y = x\) (draw it dashed); they meet on that line. ✓

Analysis

[6 marks]

The curve \(y = a + ke^{-x}\), where \(a\) and \(k\) are integers, has asymptote \(y = 2\) and passes through the point \((0, 7)\).

(a) Find \(a\) and \(k\).

(b) State the range of values of \(y\) and explain why the curve does not meet the \(x\)-axis.

(a) As \(x \to +\infty\), \(e^{-x} \to 0\), so the asymptote is \(y = a\): \(a = 2\). At \(x = 0\): \(2 + k = 7\), so \(k = 5\) and \(y = 2 + 5e^{-x}\). ✓
(b) \(5e^{-x} > 0\) for every \(x\), so \(y > 2\). Every point of the curve is above \(y = 2\), so it can never reach \(y = 0\). ✓
(c) \(2 + 5e^{-x} = 12\) gives \(e^{-x} = 2\), so \(-x = \ln 2\) and \(x = -\ln 2\). ✓
Check: \(e^{\ln 2} = 2\), so \(y = 2 + 5(2) = 12\). The point is \((-\ln 2, 12)\), to the left of the \(y\)-axis, where the decreasing curve is higher than at \((0, 7)\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.