Question

1. use transformations of the graph ( f(x)=log _{2} x ) to graph the function ( h(x)=log _{2}(x - 2) )

Explanation:

Step1: Analyze the transformation from \(y = \log_{2}x\) to \(y=\log_{2}(x - 2)\)

The general form of a horizontal transformation of a function \(y = f(x)\) is \(y=f(x - h)\), where \(h>0\) shifts the graph of \(y = f(x)\) to the right by \(h\) units. For the function \(h(x)=\log_{2}(x - 2)\) compared to \(f(x)=\log_{2}x\), we have \(h = 2\).

Step2: Apply the horizontal - shift rule

Start with the key points of \(y=\log_{2}x\). The function \(y = \log_{2}x\) has a \(x\) - intercept at \((1,0)\) (since \(\log_{2}1=0\)) and passes through the point \((2,1)\) (since \(\log_{2}2 = 1\)).
For the function \(y=\log_{2}(x - 2)\), we use the transformation rule \(x'=x + 2\) (or \(x=x'-2\)).
For the point \((x,y)\) on \(y = \log_{2}x\), the corresponding point \((x',y)\) on \(y=\log_{2}(x - 2)\) is given by \(x'=x + 2\).
When \(x = 1\) (for \(y = 0\) in \(y=\log_{2}x\)), \(x'=1 + 2=3\). When \(x = 2\) (for \(y = 1\) in \(y=\log_{2}x\)), \(x'=2+2 = 4\).
The vertical asymptote of \(y=\log_{2}x\) is \(x = 0\). For \(y=\log_{2}(x - 2)\), the vertical asymptote is \(x=2\) (since \(x-2=0\) gives \(x = 2\)).

Step3: Sketch the graph

Plot the vertical asymptote \(x = 2\). Then plot the points \((3,0)\) and \((4,1)\). The graph of \(y=\log_{2}(x - 2)\) has the same shape as the graph of \(y=\log_{2}x\) (since the base of the logarithm is the same), but it is shifted 2 units to the right.

Answer:

Shift the graph of \(y = \log_{2}x\) 2 units to the right. The vertical asymptote is \(x = 2\), and key points \((3,0)\) and \((4,1)\) can be used for sketching.