Question

identify the end behavior of the given functions. select the correct answer from each drop - down menu. the function ( m(x)=2^{x}+1 ) toward as ( x ) increases. the function ( n(x)=10^{x - 2}+5 ) toward as ( x ) decreases. the function ( p(x)=-e^{x}-2 ) toward as ( x ) decreases.

Explanation:

Step1: Analyze the function \(m(x)=2^{x}+1\)

As \(x\) increases, \(2^{x}\) increases (since the base \(2>1\) for an exponential function \(y = a^{x},a>1\)). So \(m(x)=2^{x}+1\) increases. As \(x\) decreases, \(2^{x}\) approaches \(0\) (\(\lim_{x
ightarrow-\infty}2^{x}=0\)), so \(m(x)=2^{x}+1\) approaches \(1\).

Step2: Analyze the function \(n(x)=10^{x - 2}+5\)

As \(x\) increases, \(10^{x-2}\) increases (since the base \(10 > 1\) for an exponential function \(y=a^{x},a > 1\)). So \(n(x)=10^{x - 2}+5\) increases. As \(x\) decreases, \(10^{x-2}\) approaches \(0\) (\(\lim_{x
ightarrow-\infty}10^{x-2}=0\)), so \(n(x)=10^{x - 2}+5\) approaches \(5\).

Step3: Analyze the function \(p(x)=-e^{x}-2\)

As \(x\) increases, \(e^{x}\) increases (since the base \(e\approx2.718>1\) for an exponential function \(y = a^{x},a>1\)), and \(p(x)=-e^{x}-2\) decreases. As \(x\) decreases, \(e^{x}\) approaches \(0\) (\(\lim_{x
ightarrow-\infty}e^{x}=0\)), so \(p(x)=-e^{x}-2\) approaches \(- 2\).

Answer:

For \(m(x)=2^{x}+1\): toward \(+\infty\) as \(x\) increases, toward \(1\) as \(x\) decreases.
For \(n(x)=10^{x - 2}+5\): toward \(+\infty\) as \(x\) increases, toward \(5\) as \(x\) decreases.
For \(p(x)=-e^{x}-2\): toward \(-\infty\) as \(x\) increases, toward \(-2\) as \(x\) decreases.