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 ) increases 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 \( m(x)=2^{x}+1 \)

As \( x\) increases, \( 2^{x}\) increases (since the base \( a = 2>1\) in the exponential function \( y = a^{x}\)). So \( m(x)=2^{x}+1\) increases toward positive infinity as \( x\) increases. As \( x\to-\infty\), \( 2^{x}\to0\), so \( m(x)=2^{x}+1\) approaches the line \( y = 1\) (horizontal - asymptote).

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

As \( x\to-\infty\), \( 10^{x-2}=\frac{10^{x}}{10^{2}}\to0\) (since the base \( a = 10>1\) in the exponential function \( y=a^{x}\)), so \( n(x)=10^{x - 2}+5\) approaches the line \( y = 5\). As \( x\) increases, \( 10^{x-2}\) increases, and \( n(x)=10^{x - 2}+5\) increases toward positive infinity.

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

As \( x\) increases, \( e^{x}\) increases (base \( a = e\approx2.718>1\) in the exponential function \( y = a^{x}\)), and since there is a negative sign in front (\( y=-e^{x}-2\)), \( p(x)=-e^{x}-2\) decreases toward \(-\infty\). As \( x\to-\infty\), \( e^{x}\to0\), so \( p(x)=-e^{x}-2\) approaches the line \( y=-2\).

Answer:

For \( m(x)=2^{x}+1\): increases toward positive infinity as \( x\) increases and toward the line \( y = 1\) as \( x\) decreases.
For \( n(x)=10^{x - 2}+5\): increases toward positive infinity as \( x\) increases and toward the line \( y = 5\) as \( x\) decreases.
For \( p(x)=-e^{x}-2\): decreases (since it's \(-e^{x}-2\)) toward \(-\infty\) as \( x\) increases and toward the line \( y=-2\) as \( x\) decreases.