Question

13
drag the tiles to the correct boxes to complete the pairs. not all tiles will be used
consider the graph of the function ( f(x)=ln x )
match each transformation of function ( f ) with a feature of the transformed function
( g(x)=-\frac{1}{2} f(x - 2) )
( j(x)=f(x)-\frac{1}{2} )
( h(x)=fleft(x-\frac{1}{2}
ight) )

Explanation:

Step1: Analyze \(g(x)=-\frac{1}{2}f(x - 2)\)

The original function \(y = f(x)=\ln x\) has a vertical asymptote \(x = 0\). For \(y=-\frac{1}{2}\ln(x - 2)\), the vertical asymptote is \(x=2\). The negative sign reflects the function over the \(x\) - axis. When \(y = 0\), \(0=-\frac{1}{2}\ln(x - 2)\), then \(\ln(x - 2)=0\), so \(x-2 = 1\), \(x=3\). When \(x = 2+\frac{1}{e}\), \(y=-\frac{1}{2}\ln(2+\frac{1}{e}-2)=-\frac{1}{2}\ln(\frac{1}{e})=\frac{1}{2}\). The function \(y = g(x)\) is decreasing as \(x\) increases (because of the negative coefficient \(-\frac{1}{2}\)).

Step2: Analyze \(j(x)=f(x)-\frac{1}{2}\)

The function \(j(x)=\ln x-\frac{1}{2}\). The vertical asymptote remains \(x = 0\) (since the transformation is a vertical shift). When \(x = 1\), \(y=\ln(1)-\frac{1}{2}=-\frac{1}{2}\). So the \(y\) - intercept: When \(x = 1\) (but for \(y\) - intercept \(x = 0\) is not in the domain of \(\ln x\), we consider the limit as \(x\to1\) in terms of the transformation. The original \(y=\ln x\) has no \(y\) - intercept. But for \(j(x)\), when \(x = 1\), \(j(1)=\ln(1)-\frac{1}{2}=-\frac{1}{2}\).

Step3: Analyze \(h(x)=f(x-\frac{1}{2})\)

The function \(h(x)=\ln(x-\frac{1}{2})\). When \(y = 0\), \(\ln(x-\frac{1}{2})=0\), then \(x-\frac{1}{2}=1\), \(x = 1.5\). The vertical asymptote is \(x=\frac{1}{2}\).

Answer:

\(g(x)=-\frac{1}{2}f(x - 2)\) \(\to\) function decreases as \(x\) increases; \(j(x)=f(x)-\frac{1}{2}\) \(\to\) \(y\) - intercept at \((0,-\frac{1}{2})\) (Note: technically for \(y = \ln x-\frac{1}{2}\), the \(y\) - intercept is approached as \(x\to1\) in the context of the transformation); \(h(x)=f(x-\frac{1}{2})\) \(\to\) \(x\) - intercept at \((1.5,0)\)