Question

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

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 \(x
ightarrow\infty\), we have \(u=x + 1
ightarrow\infty\).
Since the function \(y=\log_{4}u\) (\(a = 4>1\)) is an increasing function, when \(u
ightarrow\infty\), \(\log_{4}(u)
ightarrow\infty\).
Then \(y=\log_{4}(x + 1)-3\). Let \(t=x + 1\), as \(x
ightarrow\infty\), \(t
ightarrow\infty\) and \(y=\log_{4}t-3\). Since \(\log_{4}t
ightarrow\infty\) as \(t
ightarrow\infty\), then \(y=\log_{4}(x + 1)-3
ightarrow\infty\) as \(x
ightarrow\infty\).

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 \(x
ightarrow-1^{+}\) (approaching from the right side of \(x =-1\)), \(u=x + 1
ightarrow0^{+}\).
Since \(y=\log_{4}u\) and as \(u
ightarrow0^{+}\), \(\log_{4}u
ightarrow-\infty\). Then \(y=\log_{4}(x + 1)-3
ightarrow-\infty\) as \(x
ightarrow-1^{+}\).

Answer:

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