SOLUTION
Write out the function given.
[tex]f(x)=1+\log _2x[/tex]
To plot two point in the function, let obtain the point in from the function.
[tex]\begin{gathered} \text{let x=4,} \\ f(x)=1+\log _24 \\ f(x)=1+\log _22^2=1+2\log _22 \\ f(x)=1+2=3 \\ \end{gathered}[/tex]
Then, the first point is
[tex](4,3)[/tex]
Similarly, let x=2
[tex]\begin{gathered} \text{if x=2} \\ f(x)=1+\log _22=1+1=2 \\ \text{Then} \\ (2,2) \end{gathered}[/tex]
Then the point is
[tex](2,2)[/tex]
The two point to use are
[tex](4,3)\text{ and (2,2)}[/tex]
The point are (2,2) and (4,3)
Then the vertical asymptotes is obtain by equating the logatithm expression to zero
Hence
[tex]x=0[/tex]
Vertical asymptotes is x=0
Hence
The image of the graph is given below
