EL1 - Exponential functions and their graphs

Explore curves whose input appears in a power. You will calculate and plot exponential functions, recognise growth and decay from their graphs, and find the constants that control their height and rate of change.

A constant multiplier changes the shape

A curve passes through $(0, 3)$ and $(1, 6)$ . At $x=2$ , should its height be $9$ or $12$ ? Adding $3$ gives a straight line; multiplying by $2$ gives an exponential curve. Two points alone cannot tell you which rule was intended.

You already know

Substitute an input to calculate a graph's height. A plot uses accurate points on uniform scales; a sketch shows the main shape and labelled features. From GCSE indices, $a^0=1$ and $a^{-n}=1/a^n$ for $a>0$ .

For the exponential rule, the heights are $3, 6, 12, 24$ at inputs $0, 1, 2, 3$ . The differences change, but the multiplier remains $2$ . That distinction has extraordinary consequences: $2^{30}=1\,073\,741\,824$ , so just thirty doublings take $1$ beyond a billion.

Exponential function

An exponential function has the form $y=ka^x$ , with constant $k$ and positive constant base $a$ ; the variable $x$ is in the exponent. For example, $3\times2^x$ is exponential. The function $3x^2$ is a power of the variable, not an exponential function.

The positive-base condition makes the rule meaningful for every real input: for example, $4^{1/2}=2$ . A negative base would fail at inputs such as $1/2$ . Base $0$ is excluded too. For $a>0$ , the value $a^x$ is always positive, so the sign of $y$ is the sign of $k$ .

When $k>0$ , a base $a>1$ gives a rising curve, while $0<a<1$ gives a falling curve. Both have vertical intercept $(0, k)$ . For $a\ne1$ and $k\ne0$ , the horizontal axis is an asymptote: the curve approaches it at one end without reaching it. There is no horizontal intercept. These graph properties are not on the formulae sheet: learn them.

Negative $k$ reflects the corresponding positive curve in the horizontal axis. Thus $-2\times3^x$ falls as $x$ increases, even though its base exceeds $1$ . Two boundary cases deserve their own thought: $a=1$ gives the constant function $y=k$ , and $k=0$ gives $y=0$ for any positive base.

Key idea

For a non-zero exponential, equal steps in the input give equal multipliers in the output. Read both $a$ and the sign of $k$ before describing whether the graph rises or falls.

Test that sign decision on a negative curve.

Plotting growth and decay curves

A table fixes the plotted points; the multiplier and asymptote tell you how to join them.

Choose inputs on both sides of $0$ , calculate negative powers as reciprocals, and include the vertical intercept. Draw a smooth curve, without corners or a final segment lying along the horizontal axis. Half-unit inputs are useful if the curve bends sharply between whole-number inputs.

Worked example 1: a rising exponential

Question. Plot $y=3\times2^x$ for $-2\le x\le2$ , stating its vertical intercept and horizontal asymptote.

Plan. Calculate five heights, plot them on uniform scales, and join with the exponential shape.

\begin{aligned} y(-2)&=3\times\frac14=0.75 && \text{negative power is reciprocal}\\ y(-1)&=3\times\frac12=1.5 && \text{double the previous height}\\ y(0)&=3\times1=3 && \text{vertical intercept}\\ y(1)&=3\times2=6 && \text{double again}\\ y(2)&=3\times4=12 && \text{double again} \end{aligned}

Plot these points using uniform scales. The left panel below shows the smooth curve. It approaches the horizontal axis to the left when extended.

Why retain $0.75$ and $1.5$ ? Rounding them to whole numbers would flatten the small-height portion and spoil an accurate plot.

Answer. The rising curve shown, with intercept $(0, 3)$ and horizontal asymptote $y=0$ .

Check. The extra point $(0.5, 3\sqrt2)$ , approximately $(0.5, 4.24)$ , lies between the heights $3$ and $6$ .

Change only the base to its reciprocal. The same heights appear in the reverse order, giving a reflection in the vertical axis.

Worked example 2: a falling exponential

Question. Plot $y=3(\frac12)^x$ for $-2\le x\le2$ , stating its vertical intercept and horizontal asymptote.

Plan. Calculate the same inputs and halve the height at every one-unit step to the right.

\begin{aligned} y(-2)&=3\times4=12 && \text{reciprocal of }(1/2)^2\\ y(-1)&=3\times2=6 && \text{halve the previous height}\\ y(0)&=3 && \text{vertical intercept}\\ y(1)&=1.5 && \text{halve again}\\ y(2)&=0.75 && \text{halve again} \end{aligned}

Plot the five points on the same scales. Join smoothly, approaching the horizontal axis to the right when extended, as in the right panel.

Why is $(\frac12)^{-2}=4$ ? A negative exponent takes the reciprocal: $1/(\frac12)^2=1/(\frac14)=4$ .

Answer. The falling curve shown, with intercept $(0, 3)$ and horizontal asymptote $y=0$ .

Check. Replacing $x$ by $-x$ in the first equation gives $3\times2^{-x}=3(\frac12)^x$ , confirming the reflection.

The matching scales let you compare the two curves directly.

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EL1 - Exponential functions and their graphs

A constant multiplier changes the shape

Plotting growth and decay curves

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