Q:

Which answers describe the end behaviors of the function modeled by the graph? f(x)=(12)x+1−2f(x)=(12)x+1−2Select each correct answer.As x decreases without bound, f(x) increases without bound.As x increases without bound, f(x) approaches the line y=−2y=−2 .As x decreases without bound, f(x) approaches the line y=−2y=−2 .As x decreases without bound, f(x) decreases without bound.

Accepted Solution

A:
Answer:Options 1 and 2 are correct.Step-by-step explanation:The graph represents the function[tex]f(x)=(\frac{1}{2})^{x+1}-2[/tex]From the given graph it is noticed that the value of f(x) approaches to infinity as x approaches to negative infinity and the value of f(x) approaches to -2 as x approaches to positive infinity.It can also represented by limits.As x increases without bound[tex]lim_{x\rightarrow \infty}f(x)=lim_{x\rightarrow \infty}(\frac{1}{2})^{x+1}-2[/tex]Apply limit.[tex]lim_{x\rightarrow \infty}f(x)=(\frac{1}{2})^{\infty+1}-2=0-2=-2[/tex]As x decreases without bound[tex]lim_{x\rightarrow -\infty}f(x)=lim_{x\rightarrow -\infty}(\frac{1}{2})^{x+1}-2[/tex]Apply limit.[tex]lim_{x\rightarrow -\infty}f(x)=(\frac{1}{2})^{-\infty+1}-2=-2=\infty-2=\infty[/tex]Therefore as x decreases without bound, f(x) increases without bound and as x increases without bound, f(x) approaches the line y=−2. Options 1 and 2 are correct.