Question

indicate which of the following four graphs is the correct graph of this function.
q(x) = \log_{6}(x) + 2

Explanation:

Step1: Analyze the parent function

The parent function is \( y = \log_6(x) \), which has a vertical asymptote at \( x = 0 \) (the y - axis) and passes through the point \( (1,0) \) since \( \log_6(1)=0 \). The function \( y=\log_6(x) \) is defined for \( x>0 \) and is an increasing function.

Step2: Analyze the transformation

The given function is \( q(x)=\log_6(x)+2 \). This is a vertical shift of the parent function \( y = \log_6(x) \) upward by 2 units. So, the vertical asymptote remains \( x = 0 \), and the point \( (1,0) \) on the parent function will shift to \( (1,0 + 2)=(1,2) \). Also, the function is still defined for \( x>0 \) and is increasing.

Step3: Analyze the graphs

• The first graph: Has a vertical asymptote at \( x = 4 \), which is incorrect as our function should have a vertical asymptote at \( x = 0 \).
• The second graph: The function is decreasing and defined for \( x<0 \) (left of the y - axis) in parts, which is incorrect as \( \log_6(x) \) is only defined for \( x>0 \) and is increasing.
• The third graph: The function is defined for \( x>0 \) but has a y - intercept (starts near \( x = 0 \) with \( y\approx - 4\)) and is increasing but does not pass through \( (1,2) \) (when \( x = 1 \), \( y=-4 + 0=-4

eq2\)).

• The fourth graph: Has a vertical asymptote at \( x = 0 \) (y - axis), is defined for \( x>0 \), is increasing, and when \( x = 1 \), \( y=\log_6(1)+2=0 + 2 = 2\), which matches our transformed function.

Answer:

The fourth graph (the bottom - right graph)