Question

sketch the graph of the function.
$f(x)=2(4)^{x}$

Explanation:

Step1: Find the y - intercept

When \(x = 0\), \(f(0)=2(4)^{0}\).
Since \(a^{0}=1\) (\(a
eq0\)), then \(f(0)=2\times1 = 2\). So the point \((0,2)\) is on the graph.

Step2: Find another point

When \(x = 1\), \(f(1)=2(4)^{1}=2\times4 = 8\). So the point \((1,8)\) is on the graph.

Step3: Analyze the behavior as \(x\to-\infty\)

As \(x\to-\infty\), \(y = 2(4)^{x}\to0\) (because \(4^{x}=\frac{1}{4^{-x}}\) and as \(x\to-\infty\), \(4^{-x}\to+\infty\)). The horizontal asymptote is \(y = 0\).

Step4: Analyze the behavior as \(x\to+\infty\)

As \(x\to+\infty\), \(y = 2(4)^{x}\to+\infty\) since the base \(a = 4>1\) and the function \(y = a^{x}\) is an exponential growth function.

Plot the points \((0,2)\) and \((1,8)\), draw a smooth curve approaching the horizontal asymptote \(y = 0\) as \(x\to-\infty\) and increasing without bound as \(x\to+\infty\).

Answer:

The graph of \(y = 2(4)^{x}\) has a \(y\) - intercept at \((0,2)\), passes through \((1,8)\), has a horizontal asymptote \(y = 0\), and is an increasing exponential curve.