select the correct end behavior for the function $f(x)=\\log_4(x + 1)-3$\nright: as $x\\to\\infty,y\\to\\inft…

Explanation:

Step1: Analyze the right - hand end behavior

For the function (y = \log_{4}(x + 1)-3), we know that the domain of (y=\log_{a}u) ((a>1)) is (u>0). Here (u=x + 1). As (x\rightarrow\infty), we have (u=x + 1\rightarrow\infty). Since the function (y=\log_{4}u) ((a = 4>1)) is an increasing function, when (u\rightarrow\infty), (\log_{4}(u)\rightarrow\infty). Then (y=\log_{4}(x + 1)-3). Let (t=x + 1), as (x\rightarrow\infty), (t\rightarrow\infty) and (y=\log_{4}t-3). Since (\log_{4}t\rightarrow\infty) as (t\rightarrow\infty), then (y=\log_{4}(x + 1)-3\rightarrow\infty) as (x\rightarrow\infty).

Step2: Analyze the left - hand end behavior

The vertical asymptote of (y = \log_{4}(x + 1)-3) is found by setting (x+1 = 0), so (x=-1). As (x\rightarrow-1^{+}) (approaching from the right side of (x =-1)), (u=x + 1\rightarrow0^{+}). Since (y=\log_{4}u) and as (u\rightarrow0^{+}), (\log_{4}u\rightarrow-\infty). Then (y=\log_{4}(x + 1)-3\rightarrow-\infty) as (x\rightarrow-1^{+}).

Answer:

Right: as (x\rightarrow\infty,y\rightarrow\infty)