Question

describe the end behavior of the graph of the function.

$f(x)=4(4)^{-x}+9$

as $x\to -\infty,f(x)\to$ as $x\to \infty,f(x)\to$

Explanation:

Step1: Analyze \(x\to-\infty\)

When \(x\to-\infty\), \(-x\to\infty\). For the exponential function \(y = a^{u}\) (\(a = 4>1\), \(u=-x\)), as \(u\to\infty\), \(4^{-x}=4^{u}\to\infty\). Then \(f(x)=4(4)^{-x}+9\). Multiply by \(4\) and add \(9\), so \(f(x)=4\times\infty + 9=\infty\)

Step2: Analyze \(x\to\infty\)

When \(x\to\infty\), \(-x\to-\infty\). For the exponential function \(y = a^{u}\) (\(a = 4>1\), \(u =-x\)), as \(u\to-\infty\), \(4^{-x}=\frac{1}{4^{x}}\to0\). Then \(f(x)=4\times0 + 9=9\)

Answer:

As \(x\to-\infty\), \(f(x)\to\infty\); As \(x\to\infty\), \(f(x)\to9\)