Exponential graphs

AS · P1 · 12 min

An exponential function has the variable in the power, as in $y = 2^x$ . Exponentials and logarithms are examined in Paper 3 (and Paper 2), not in Paper 1. This note is here as a bridge: it practises the Paper 1 skills of transformations, ranges, one-one functions and disguised quadratics on a new family of curves, and it prepares you for Exponential graphs in P3. Paper 1 transformation questions may use "other graphs with given features", so being comfortable with an unfamiliar curve and its asymptote is useful.

Extension

Nothing in this note will be examined on Paper 1 as exponential content. Treat it as practice of P1 methods and a preview of P3. Equations here are solved only by matching powers (IGCSE index laws); logarithms belong to P3.

The basic curve

Consider $y = 2^x$ . A few values:

$x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$2^x$	$\tfrac{1}{4}$	$\tfrac{1}{2}$	$1$	$2$	$4$	$8$

Each step of $1$ to the right doubles $y$ . Going left, $y$ halves again and again, getting closer to $0$ but never reaching it.

Key result

For $y = a^x$ with $a > 1$ :

the curve passes through $(0, 1)$ , since $a^0 = 1$ ;
$y > 0$ for every $x$ ; the $x$ -axis ( $y = 0$ ) is a horizontal asymptote;
the function is increasing, so it is one-one;
domain $\mathbb{R}$ , range $y > 0$ .

For $0 < a < 1$ the curve decreases instead. Since $\left(\tfrac{1}{2}\right)^x = 2^{-x}$ , the graph of $y = \left(\tfrac{1}{2}\right)^x$ is the reflection of $y = 2^x$ in the $y$ -axis.

y = 2^x y = (1/2)^x y = 3^x

Larger bases rise more steeply for $x > 0$ : $3^x$ is above $2^x$ to the right of the $y$ -axis and below it to the left. All three curves pass through $(0, 1)$ .

Transformations of exponential graphs

The transformation rules are exactly those from Transformations of graphs. The only new feature is the asymptote, which moves with vertical translations and stretches.

Equation	Transformation of $y = 2^x$	Asymptote	$y$ -intercept
$y = 2^x + q$	translation $\begin{pmatrix} 0 \\ q \end{pmatrix}$	$y = q$	$1 + q$
$y = 2^{x - p}$	translation $\begin{pmatrix} p \\ 0 \end{pmatrix}$	$y = 0$	$2^{-p}$
$y = k \cdot 2^x$	stretch parallel to the $y$ -axis, factor $k$	$y = 0$	$k$
$y = -2^x$	reflection in the $x$ -axis	$y = 0$	$-1$
$y = 2^{-x}$	reflection in the $y$ -axis	$y = 0$	$1$

Sketching a transformed exponential

Identify the transformations and find the new horizontal asymptote; draw it dashed.
Find the $y$ -intercept by putting $x = 0$ .
If the curve crosses the $x$ -axis, find the $x$ -intercept by solving $y = 0$ .
Decide whether the curve increases or decreases, and on which side of the asymptote it lies.
Sketch a smooth curve approaching the asymptote at one end.

A translated exponential

Sketch $y = 2^{x + 2} - 1$ , showing the asymptote and the intercepts.

Solution

This is $y = 2^x$ translated by $\begin{pmatrix} -2 \\ -1 \end{pmatrix}$ . The asymptote moves to $y = -1$ .

$y$ -intercept: $2^2 - 1 = 3$ , so $(0, 3)$ .

$x$ -intercept: $2^{x + 2} = 1 = 2^0$ , so $x + 2 = 0$ and $x = -2$ , giving $(-2, 0)$ .

y = 2^(x + 2) - 1 y = -1 (-2, 0) (0, 3)

The curve increases, lies above $y = -1$ , and approaches it as $x \to -\infty$ .

A reflected and translated exponential

Describe the transformations mapping $y = 2^x$ onto $y = 4 - 2^{-x}$ , and state the range of $f(x) = 4 - 2^{-x}$ for $x \in \mathbb{R}$ .

Solution

$2^x \to 2^{-x} \to -2^{-x} \to 4 - 2^{-x}$ :

reflection in the $y$ -axis;
reflection in the $x$ -axis;
translation by $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ .

Range: $2^{-x}$ takes every positive value, so $-2^{-x} < 0$ and $4 - 2^{-x} < 4$ . Range $f(x) < 4$ .

y = 4 - 2^(-x) y = 4 (0, 3) (-2, 0)

( $y = 0$ when $2^{-x} = 4 = 2^2$ , so $x = -2$ .)

Ranges and inverses

Because $a^x$ is increasing (for $a > 1$ ), every exponential function of the form $k \cdot a^x + q$ with $k \neq 0$ is one-one and has an inverse. Writing that inverse needs logarithms, which are P3.

Range and one-one

The function $f$ is defined by $f(x) = 3 \times 2^x + 5$ for $x \in \mathbb{R}$ . State the range of $f$ and explain why $f$ has an inverse.

Solution

$2^x > 0$ for all $x$ , so $3 \times 2^x > 0$ and $f(x) > 5$ . As $2^x$ takes every positive value, the range is $f(x) > 5$ .

$2^x$ is increasing, so $f$ is increasing for all $x$ ; hence it is one-one, and a one-one function has an inverse.

Even without a formula, you can sketch the inverse: reflect the curve in the line $y = x$ . The image of $y = 2^x$ passes through $(1, 0)$ instead of $(0, 1)$ , and its asymptote is the $y$ -axis ( $x = 0$ ) instead of the $x$ -axis. Its domain is $x > 0$ , the range of $2^x$ , exactly as the rule "domains and ranges swap" from One-one and inverse functions predicts.

y = 2^x y = x 2^y = x (0, 1) (1, 0)

Composite functions with an exponential

The functions $f$ and $g$ are defined for all real $x$ by $f(x) = 2^x$ and $g(x) = x^2 - 1$ .

(a) Find expressions for $fg(x)$ and $gf(x)$ .

(b) State the range of $fg$ and the range of $gf$ .

Solution

(a) $fg(x) = f\left(x^2 - 1\right) = 2^{x^2 - 1}$ . $gf(x) = g\left(2^x\right) = \left(2^x\right)^2 - 1 = 2^{2x} - 1 = 4^x - 1$ .

(b) For $fg$ : $x^2 - 1 \ge -1$ for all $x$ , with equality at $x = 0$ , and $2^t$ is increasing, so $fg(x) \ge 2^{-1} = \tfrac{1}{2}$ . Range $fg(x) \ge \tfrac{1}{2}$ .

For $gf$ : $4^x$ takes every positive value, so $4^x - 1 > -1$ . Range $gf(x) > -1$ .

The order matters: $fg$ means "do $g$ first". See Composite functions.

Solving equations by matching powers

If both sides can be written as powers of the same base, the powers must be equal (because $a^x$ is one-one).

Matching bases

Solve $4^x = 8^{x - 1}$ .

Solution

Write both sides as powers of $2$ : $4^x = \left(2^2\right)^x = 2^{2x}$ and $8^{x - 1} = 2^{3(x - 1)}$ .

2^{2x} = 2^{3x - 3} \quad\Rightarrow\quad 2x = 3x - 3 \quad\Rightarrow\quad x = 3

A disguised quadratic in an exponential

Solve $2^{2x} - 5 \times 2^x + 4 = 0$ .

Solution

$2^{2x} = \left(2^x\right)^2$ , so let $u = 2^x$ (see Equations that are quadratic in a function of x):

u^2 - 5u + 4 = 0 \quad\Rightarrow\quad (u - 1)(u - 4) = 0

$2^x = 1$ gives $x = 0$ ; $2^x = 4$ gives $x = 2$ . (A value $u \le 0$ would have to be rejected, since $2^x > 0$ .)

Counting solutions with a sketch

Some equations, such as $2^x = x + 2$ , mix an exponential with a polynomial and cannot be solved exactly with Paper 1 methods. You can still say how many solutions there are, by sketching both sides on the same axes and counting the intersections, which is the same graph-and-equation idea used in Intersections of lines and curves.

Number of roots from a sketch

(a) By sketching $y = 2^x$ and $y = x + 2$ on the same axes, show that the equation $2^x = x + 2$ has exactly two real roots.

(b) Verify that one root is $x = 2$ , and show that the other lies between $-2$ and $-1$ .

Solution

(a)

y = 2^x y = x + 2 y = 3 - x

The line $y = x + 2$ crosses the $x$ -axis at $-2$ , where the curve is just above the axis ( $2^{-2} = 0.25 > 0$ ), so the line starts below the curve on the left. At $x = 0$ the line is above the curve ( $2 > 1$ ). Far to the right the exponential grows faster than any line, so the curve is above again. The line therefore crosses the curve exactly twice.

(b) $2^2 = 4 = 2 + 2$ , so $x = 2$ is a root. For the other, compare $2^x - (x + 2)$ at the ends of the interval: at $x = -2$ it is $0.25 - 0 = 0.25 > 0$ ; at $x = -1$ it is $0.5 - 1 = -0.5 < 0$ . The sign changes, so a root lies between $-2$ and $-1$ (it is about $-1.69$ ).

(c) $y = 2^x$ is increasing and $y = 3 - x$ is decreasing, so they can cross at most once; and they do cross, since at $x = 0$ the line is above the curve and at $x = 2$ it is below. So there is exactly one root. Trying $x = 1$ : $2^1 = 2 = 3 - 1$ . The root is $x = 1$ .

Watch out

Drawing the curve crossing its asymptote. $y = 2^x + q$ never reaches $y = q$ .

Thinking $2^x$ can be zero or negative. It is always positive, which is why $u = 2^x \le 0$ must be rejected.

Confusing $2^{x + 2}$ with $2^x + 2$ . The first is a horizontal translation (left $2$ ); the second is a vertical one (up $2$ ).

Exam tip

Not examined as content. Paper 1 will not ask you to solve exponential equations with logarithms. It can, however, give an unfamiliar function and ask about its transformations, range, one-one property or composites, and the skills here are exactly those.
Sketches. Draw the asymptote as a dashed line and label it with its equation. Mark the intercepts with coordinates. The curve must approach the asymptote without touching it.
Ranges. Use strict inequalities: the range of $3 \times 2^x + 5$ is $f(x) > 5$ , not $f(x) \ge 5$ , because $2^x$ never reaches $0$ .
Order of transformations. For a combination such as $4 - 2^{-x}$ , list the steps in an order that works and check by tracking one point, such as the $y$ -intercept.

Summary

Exponentials are P2/P3 content; on P1 this note is practice of transformations and functions.
$y = a^x$ ( $a > 1$ ) passes through $(0, 1)$ , is increasing, has range $y > 0$ and asymptote $y = 0$ .
Vertical translations and stretches move or keep the asymptote; horizontal changes do not move it.
Exponential functions are one-one, so they have inverses (logarithms, in P3).
Solve $a^{f(x)} = a^{g(x)}$ by setting $f(x) = g(x)$ ; spot quadratics in $a^x$ .

Practice questions

Question

Sketch $y = 3^x - 9$ , giving the asymptote and the intercepts.
Describe the transformation mapping $y = 2^x$ onto $y = 2^{x - 3}$ , and show that it is also a stretch parallel to the $y$ -axis.
Solve $9^x = 27^{2 - x}$ .
Solve $3^{2x} - 10 \times 3^x + 9 = 0$ .
State the range of $f(x) = 5 - 2^x$ for $x \in \mathbb{R}$ .
Describe a sequence of transformations mapping $y = 2^x$ onto $y = 2^{-x} + 1$ , and state the asymptote and $y$ -intercept.
The curve $y = a \times 2^x + b$ passes through $(0, 5)$ and $(2, 11)$ . Find $a$ and $b$ , and state the equation of the asymptote.
The functions $f$ and $g$ are defined for all real $x$ by $f(x) = 2^x$ and $g(x) = 3x + 1$ . Find $fg(x)$ and solve $fg(x) = 16$ . State the range of $gf$ .
Show that the equation $3^x = 5 - x$ has exactly one real root, and that it lies between $1$ and $2$ .
The function $h$ is defined by $h(x) = 2^{x} - 4$ for $x \in \mathbb{R}$ . Sketch $y = h(x)$ and $y = h^{-1}(x)$ on the same axes, stating the asymptote and the intercepts of each.

Answers

Asymptote $y = -9$ . $y$ -intercept $1 - 9 = -8$ , so $(0, -8)$ . $x$ -intercept: $3^x = 9 = 3^2$ , so $(2, 0)$ .
Translation by $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ . Also $2^{x - 3} = 2^{-3} \times 2^x = \tfrac{1}{8} \times 2^x$ , a stretch parallel to the $y$ -axis with scale factor $\tfrac{1}{8}$ .
$3^{2x} = 3^{3(2 - x)}$ , so $2x = 6 - 3x$ and $x = \tfrac{6}{5}$ .
Let $u = 3^x$ : $u^2 - 10u + 9 = 0$ , $(u - 1)(u - 9) = 0$ . $3^x = 1$ gives $x = 0$ ; $3^x = 9$ gives $x = 2$ .
$2^x > 0$ , so $5 - 2^x < 5$ . Range $f(x) < 5$ .
Reflection in the $y$ -axis, then translation by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ . Asymptote $y = 1$ ; $y$ -intercept $2$ , so $(0, 2)$ .
At $(0, 5)$ : $a + b = 5$ . At $(2, 11)$ : $4a + b = 11$ . Subtracting, $3a = 6$ , so $a = 2$ and $b = 3$ . Asymptote $y = 3$ .
$fg(x) = 2^{3x + 1}$ . $2^{3x + 1} = 16 = 2^4$ , so $3x + 1 = 4$ and $x = 1$ . $gf(x) = 3 \times 2^x + 1$ ; since $2^x > 0$ , the range is $gf(x) > 1$ .
$y = 3^x$ is increasing and $y = 5 - x$ is decreasing, so they meet at most once. At $x = 1$ : $3^1 = 3 < 4 = 5 - 1$ . At $x = 2$ : $3^2 = 9 > 3 = 5 - 2$ . The difference $3^x - (5 - x)$ changes sign between $1$ and $2$ , so there is exactly one root, in that interval (it is about $1.21$ ).
$y = h(x)$ : asymptote $y = -4$ , $y$ -intercept $(0, -3)$ , $x$ -intercept where $2^x = 4$ , i.e. $(2, 0)$ . The inverse is the reflection in $y = x$ : asymptote $x = -4$ , $x$ -intercept $(-3, 0)$ , $y$ -intercept $(0, 2)$ . Domain of $h^{-1}$ : $x > -4$ .

The basic curve

Transformations of exponential graphs

Ranges and inverses

Solving equations by matching powers

Counting solutions with a sketch

Practice questions

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