Logarithmic function graph and asymptote (II)

Statement

Draw the following function after finding its vertical asymptote: $y=1+\log_2(x+1)$

Solution

Domain and vertical asymptote. The argument must be positive: $x+1>0 \iff x>-1$. The vertical asymptote is the line $x=-1$.

Behaviour. The coefficient $+1$ of the logarithm makes the function increasing; as $x\to-1^{+}$ we have $y\to-\infty$.

Useful points:

• $x+1=\tfrac12 \Rightarrow x=-\tfrac12,\ y=1-1=0$;
• $x+1=1 \Rightarrow x=0,\ y=1$;
• $x+1=2 \Rightarrow x=1,\ y=1+1=2$.

An increasing curve rising from $-\infty$ (approaching the asymptote $x=-1$) through $(-\tfrac12,0)$, $(0,1)$, $(1,2)$.

$\boxed{\text{vertical asymptote }x=-1}$