Question

Which answers describe the end behaviors of the function modeled by the graph?

Answer 1

Answer:

Options 1 and 2 are correct.

Step-by-step explanation:

The graph represents the function

$f(x)=(\frac{1}{2})^{x+1}-2$

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

$lim_{x\rightarrow \infty}f(x)=lim_{x\rightarrow \infty}(\frac{1}{2})^{x+1}-2$

Apply limit.

$lim_{x\rightarrow \infty}f(x)=(\frac{1}{2})^{\infty+1}-2=0-2=-2$

As x decreases without bound

$lim_{x\rightarrow -\infty}f(x)=lim_{x\rightarrow -\infty}(\frac{1}{2})^{x+1}-2$

Apply limit.

$lim_{x\rightarrow -\infty}f(x)=(\frac{1}{2})^{-\infty+1}-2=-2=\infty-2=\infty$

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.

Answer 2

For this case, we observe that the graph approaches the line y = -2 as x increases. It is observed that the behavior of the graph for x = inf is y = -2.
 answer
 As x increases without bound, f (x) approaches the line y = -2y = -2.