Sketching Graphs of the Form <em>y = ab^x + q</em>

Sketch the graph of \(g(x)=3\times {2}^{x}+2\). Mark the intercept and the asymptote.

Examine the standard form of the equation

From the equation we see that \(a>1\), therefore the graph curves upwards. \(q>0\) therefore the graph is shifted vertically upwards by \(\text{2}\) units.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = 3\times {2}^{x}+2 \\ & = 3\times {2}^{0}+2 \\ & = 3+2 \\ & = 5 \end{align*}

This gives the point \((0;5)\).

For the \(x\)-intercept, let \(y=0\):

\begin{align*} y& = 3\times {2}^{x}+2 \\ 0& = 3\times {2}^{x}+2 \\ -2& = 3\times {2}^{x} \\ {2}^{x}& = -\cfrac{2}{3} \end{align*}

There is no real solution, therefore there is no \(x\)-intercept.

Determine the asymptote

The horizontal asymptote is the line \(y=2\).

Plot the points and sketch the graph

Domain: \(\{x:x\in \mathbb{R}\}\)

Range: \(\{g(x):g(x)>2\}\)

Note that there is no axis of symmetry for exponential functions.

Sketch the graph of \(y=-2\times {3}^{x}+6\)

Examine the standard form of the equation

From the equation we see that \(a<0\) therefore the graph curves downwards. \(q>0\) therefore the graph is shifted vertically upwards by \(\text{6}\) units.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = -2\times {3}^{x}+6 \\ & = -2\times {3}^{0}+6 \\ & = 4 \end{align*}

This gives the point \((0;4)\).

For the \(x\)-intercept, let \(y=0\):

\begin{align*} y& = -2\times {3}^{x}+6 \\ 0& = -2\times {3}^{x}+6 \\ -6& = -2\times {3}^{x} \\ {3}^{1}& = {3}^{x} \\ \therefore x& = 1 \end{align*}

This gives the point \((1;0)\).

Determine the asymptote

The horizontal asymptote is the line \(y=6\).

Plot the points and sketch the graph

Domain: \(\{x:x\in \mathbb{R}\}\)

Range: \(\{g(x):g(x)<6\}\)