Question

begin by graphing ( f(x)=log _{4} x ). then use transformations of this graph to graph the given function. what is the vertical asymptote? use the graphs to determine the given functions domain and range.
( g(x)=log _{4}(x + 2) )
graph the function ( g(x)=log _{4}(x + 2) ). graph the asymptote of ( g(x) ) as a dashed line. use the graphing tool to graph the function.
click to enlarge graph

Explanation:

Step1: Analyze the transformation

The function \(g(x)=\log_{4}(x + 2)\) is a horizontal shift of \(f(x)=\log_{4}x\). The transformation rule for \(y = f(x + h)\) is a shift to the left by \(h\) units. Here \(h = 2\), so the graph of \(y=\log_{4}x\) is shifted 2 units to the left.

Step2: Find the vertical asymptote

For \(y=\log_{4}x\), the vertical asymptote is \(x = 0\). After the shift \(x+2=0\) (since for \(y=\log_{4}(u)\) with \(u=x + 2\), the vertical asymptote occurs when \(u = 0\)). Solving \(x+2=0\) gives \(x=-2\).

Step3: Determine the domain

For \(y=\log_{4}(x + 2)\), the argument of the logarithm \(x+2>0\). Solving \(x+2>0\) gives \(x>-2\). So the domain is \((-2,\infty)\).

Step4: Determine the range

The range of a logarithmic function \(y = \log_{a}(u)\) (where \(a>0,a
eq1\)) is \((-\infty,\infty)\). Since the transformation is a horizontal shift (which does not affect the range), the range of \(y=\log_{4}(x + 2)\) is \((-\infty,\infty)\).

Answer:

• Vertical asymptote: \(x=-2\)
• Domain: \((-2,\infty)\)
• Range: \((-\infty,\infty)\)