Logarithmic function graph and asymptote (III)

Statement

Draw the following function after finding its vertical asymptote: $y=-\dfrac{3}{2}-\dfrac{5}{2}\log_{1/2}(4-x)$

Solution

Domain and vertical asymptote. The argument must be positive: $4-x>0 \iff x<4$. The vertical asymptote is the line $x=4$.

Rewriting. Since $\log_{1/2}(u)=-\log_2(u)$: $y=-\tfrac32+\tfrac52\log_2(4-x).$ As $x$ increases the argument $4-x$ decreases, so the function is decreasing; as $x\to4^{-}$ we have $y\to-\infty$.

Useful points:

• $4-x=1 \Rightarrow x=3,\ y=-\tfrac32$;
• $4-x=2 \Rightarrow x=2,\ y=-\tfrac32+\tfrac52=1$;
• $4-x=4 \Rightarrow x=0,\ y=-\tfrac32+5=\tfrac72$.

A decreasing curve that from $x=0$ falls toward the asymptote $x=4$ (where $y\to-\infty$) through $(0,\tfrac72)$, $(2,1)$, $(3,-\tfrac32)$.

$\boxed{\text{vertical asymptote }x=4}$