Exponential and Logarithmic Graphs

Exponential graphs are recognised by a horizontal asymptote on one side and a curve that steepens without limit on the other; logarithmic graphs are their mirror images, with a vertical asymptote and slow, unbounded growth. P3 questions ask you to sketch them with their key features, to read off ranges, to find where they meet other curves, and, very often, to sketch two graphs on the same axes to show how many roots an equation has before solving it numerically. Every sketch comes down to three facts: where the asymptote is, where the curve crosses the axes, and which way it goes.

The basic exponential shapes

Start from $y = e^x$: through $(0, 1)$, always positive, increasing, with the $x$-axis as an asymptote on the left. Everything else is a transformation of it.

• $y = e^{-x}$ is the reflection of $y = e^x$ in the $y$-axis: decreasing, asymptote $y = 0$ on the right.
• $y = e^{kx}$ for $k > 0$ is a stretch of $y = e^x$ parallel to the $x$-axis with scale factor $\tfrac{1}{k}$. The larger $k$, the steeper the curve; all pass through $(0, 1)$.
• $y = a^x$ with $a > 1$ has the same shape as $e^x$, because $a^x = e^{x\ln a}$ with $\ln a > 0$. With $0 < a < 1$, $\ln a < 0$, so it has the shape of $e^{-x}$. For example, $\left(\tfrac{1}{2}\right)^x = 2^{-x}$.

Transformations of exponentials

Most exam curves have the form $y = Ae^{kx} + c$, possibly with $x$ replaced by $x - p$. Use the transformations from P1 (transformations of graphs):

Change	Effect on $y = e^x$
$y = e^x + c$	translation $\begin{pmatrix} 0 \\ c \end{pmatrix}$; asymptote becomes $y = c$
$y = e^{x - p}$	translation $\begin{pmatrix} p \\ 0 \end{pmatrix}$; asymptote unchanged
$y = Ae^x$, $A > 0$	stretch parallel to the $y$-axis, factor $A$
$y = -e^x$	reflection in the $x$-axis; curve now below its asymptote
$y = e^{kx}$	stretch parallel to the $x$-axis, factor $\tfrac{1}{k}$
$y = e^{-x}$	reflection in the $y$-axis

Note that $e^{x - p} = e^{-p}e^x$, so a horizontal translation of an exponential is the same as a vertical stretch. Either description is correct.

The curve $y = 3e^{-x} - 6$ with its asymptote $y = -6$. It crosses the axes at $(-\ln 2, 0)$ and $(0, -3)$.

Ranges from the asymptote

Because an exponential never reaches its asymptote, the range of $y = Ae^{kx} + c$ over all real $x$ is $y > c$ (if $A > 0$) or $y < c$ (if $A < 0$). On a restricted domain such as $x \ge 0$, the endpoint is included: for $f(x) = 3 - 2e^{-x}$ with $x \ge 0$, $f(0) = 1$ and $f(x) \to 3$, so the range is $1 \le f(x) < 3$.

Logarithmic graphs

$y = \ln x$ is the reflection of $y = e^x$ in $y = x$. Its features are the exponential's features with $x$ and $y$ swapped.

The three curves are $y = \ln x$ (through $(1, 0)$), $y = \ln(x + 3)$ (asymptote $x = -3$, through $(-2, 0)$) and $y = -\ln x$ (the reflection of $\ln x$ in the $x$-axis).

For a log graph the asymptote is where the argument becomes zero:

• $y = \ln(x - a)$: asymptote $x = a$, domain $x > a$, $x$-intercept at $x = a + 1$.
• $y = \ln(ax + b)$ with $a > 0$: asymptote $x = -\tfrac{b}{a}$, domain $x > -\tfrac{b}{a}$.
• $y = \ln(b - x)$: domain $x < b$, a decreasing curve with asymptote $x = b$.
• $y = \ln(kx) = \ln k + \ln x$: the graph of $\ln x$ translated by $\ln k$ upwards. (It is also a stretch parallel to the $x$-axis with factor $\tfrac{1}{k}$; the two descriptions give the same curve.)
• $y = \ln(x^2) = 2\ln x$ only for $x > 0$. For all $x \ne 0$, $\ln(x^2) = 2\ln|x|$, a curve symmetric in the $y$-axis.

Exponential and log graphs as inverses

If $f$ is built from an exponential, $f^{-1}$ is built from a logarithm, and the graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in $y = x$. Horizontal asymptotes of $f$ become vertical asymptotes of $f^{-1}$, intercepts swap axes, and the domain and range swap.

Graphs that count roots

A very common P3 question has the form: "By sketching a suitable pair of graphs, show that the equation $g(x) = h(x)$ has exactly $n$ real roots." Each root is an $x$-coordinate where the curves $y = g(x)$ and $y = h(x)$ cross. You do not need to find the roots; you need sketches clear enough to show every crossing and to make it obvious there are no others. This is the first step of locating roots before solving numerically.

Worked examples

Practice

Answers

1. Asymptote $y = 1$ (approached as $x \to -\infty$); $y$-intercept $2 + 1 = 3$; increasing, always above $y = 1$, so no $x$-intercept.

2. Asymptote $y = -4$. $y$-intercept $1 - 4 = -3$, so $(0, -3)$. $x$-intercept: $e^{2x} = 4 \Rightarrow x = \tfrac{1}{2}\ln 4 = \ln 2$, so $(\ln 2, 0)$.

3. $2^x = 8 \times 2^{-x} \Rightarrow 2^{2x} = 8 = 2^3 \Rightarrow x = \tfrac{3}{2}$. Then $y = 2^{3/2} = 2\sqrt{2}$. Point $\left(\tfrac{3}{2}, 2\sqrt{2}\right)$.

4. Reflect $y = \ln x$ in the $x$-axis, then translate up by $1$: a decreasing curve with asymptote $x = 0$ ($y \to +\infty$ as $x \to 0^+$). It crosses the axis where $\ln x = 1$, at $(e, 0)$.

5. Domain $x < 3$; asymptote $x = 3$, with $y \to -\infty$ as $x \to 3^-$. $x$-intercept: $3 - x = 1 \Rightarrow (2, 0)$. $y$-intercept: $(0, \ln 3)$. The curve decreases as $x$ increases (it is $y = \ln x$ reflected in the $y$-axis then translated $3$ to the right).

6. $\ln(2x) = \ln 2 + \ln x$, so a translation $\begin{pmatrix} 0 \\ \ln 2 \end{pmatrix}$. Alternatively, replacing $x$ by $2x$ is a stretch parallel to the $x$-axis with scale factor $\tfrac{1}{2}$.

7. $y = e^x$ is increasing everywhere and $y = 3 - x$ is decreasing everywhere, so they can cross at most once; they do cross (the exponential is below the line at $x = 0$ and above it at $x = 1$), so exactly one root. With $f(x) = e^x + x - 3$: $f(0) = -2 < 0$ and $f(1) = e - 2 = 0.718 > 0$. Sign change, $f$ continuous, so the root lies between $0$ and $1$.

8. $e^{2x} - 3e^x - 4 = 0 \Rightarrow (e^x - 4)(e^x + 1) = 0$. $e^x = -1$ is impossible, so there is only one intersection: $e^x = 4$, $x = \ln 4$ ($= 2\ln 2$), and $y = e^{2x} = 16$. Point $(\ln 4, 16)$.

9. (a) $f(0) = 1$ and $f(x) \to 5$ as $x \to \infty$, so $1 \le f(x) < 5$. (b) $e^{-x/2} = \dfrac{5 - y}{4} \Rightarrow -\tfrac{x}{2} = \ln\dfrac{5 - y}{4} \Rightarrow x = 2\ln\dfrac{4}{5 - y}$. So $f^{-1}(x) = 2\ln\left(\dfrac{4}{5 - x}\right)$ with domain $1 \le x < 5$. (c) $5 = 4e^{-x/2} \Rightarrow e^{-x/2} = \tfrac{5}{4} \Rightarrow x = -2\ln\tfrac{5}{4}$ ($\approx -0.446$). This is negative, so it is outside the domain $x \ge 0$ of $f$; indeed the range of $f$ shows $f(x) \ge 1$, so $f$ never takes the value $0$.