Exponential functions and their graphsEdexcel International A Level Maths: Revision notes

An exponential function has the form $y=a^x$, where the base $a$ is a positive constant ($a>0$) with $a\neq1$, and the index $x$ is the variable. This is different from a power function such as $y=x^2$, where the variable is the base. The restrictions are needed. If $a=1$ then $y=1^x=1$ for every $x$, a horizontal line rather than an exponential. If $a<0$ then values such as $(-2)^{\frac12}$ do not exist, so the function could not be defined for all real $x$. With $a>0$ and $a\neq1$, $y=a^x$ is defined for every real $x$ and takes every positive value exactly once: the domain is all real $x$ and the range is $y>0$.

Every graph of $y=a^x$ has the same features:

• It passes through $(0,1)$, because $a^0=1$ for any base.
• It passes through $(1,a)$, because $a^1=a$.
• It lies entirely above the $x$-axis, because $a^x>0$ for all $x$, so there is no $x$-intercept.
• The $x$-axis, $y=0$, is a horizontal asymptote: the curve approaches it but never reaches it. For $y=3^x$: at $x=-1,0,1,2$ the values of $y$ are $\frac13,1,3,9$. Because the function is one-to-one, $a^p=a^q$ implies $p=q$.

The size of the base decides the shape.

• If $a>1$ the function is increasing (exponential growth). As $x\to\infty$, $y\to\infty$; as $x\to-\infty$, $y\to0$.
• If $0<a<1$ the function is decreasing (exponential decay). As $x\to\infty$, $y\to0$; as $x\to-\infty$, $y\to\infty$. All the curves pass through $(0,1)$. For $x>0$ a larger base gives a steeper curve, so $3^x$ lies above $2^x$; for $x<0$ the order reverses and $3^x$ lies below $2^x$.

Since $\left(\frac1a\right)^x=a^{-x}$, the curve $y=\left(\frac1a\right)^x$ is the reflection of $y=a^x$ in the $y$-axis (replace $x$ by $-x$). For example $y=\left(\frac12\right)^x$ is the mirror image of $y=2^x$: the point $(1,2)$ on $y=2^x$ corresponds to $(-1,2)$ on $y=\left(\frac12\right)^x$. A growth curve therefore reflects into a decay curve and vice versa. Both pass through $(0,1)$ and both have the $x$-axis as an asymptote.

To find $a$, substitute a known point and solve. If $y=a^x$ passes through $(2,9)$ then $a^2=9$, so $a=3$ (reject $a=-3$ because $a>0$). If it passes through $\left(-2,\frac1{25}\right)$ then $a^{-2}=\frac1{25}$, so $a^2=25$ and $a=5$. To solve an equation such as $3^x=27$, write both sides as powers of the same base: $3^x=3^3$, so $x=3$. Because the function is one-to-one, the indices must then be equal. With a linear index, $5^{2x-1}=5^6$ gives $2x-1=6$, so $x=\frac72$. Use the index laws for negative and fractional values: $a^{-n}=\frac1{a^n}$, so $2^{-3}=\frac18$.

A full sketch of $y=a^x$ shows: the $y$-intercept $(0,1)$; the correct direction (rising for $a>1$, falling for $0<a<1$); the asymptote $y=0$ with the curve flattening towards it and not touching it; and, if asked, a second point such as $(1,a)$. On a shared set of axes, label which curve is which, for example $y=4^x$ and $y=\left(\frac14\right)^x$, which meet only at $(0,1)$ and are mirror images in the $y$-axis.