Answer
See the graph below:
Work Step by Step
We use the rule of logarithm ${{\log }_{b}}{{b}^{z}}=z $.
Construct the table of coordinates for $ f\left( x \right)={{\log }_{2}}x $ and choose the appropriate values for $ x $ and calculate the corresponding $ y\text{-values}$.
Now, plot every point given in the above table and join them using a smooth curve, and the $ y\text{-axis}$ or $ x=0$ is the vertical asymptote.
Now, the curve of $ h\left( x \right)=-1+{{\log }_{2}}x $ is similar to the curve of $ f\left( x \right)={{\log }_{2}}x $ translated horizontally downward by 1 unit in the graph.
Then, shift the graph of $ f\left( x \right)={{\log }_{2}}x $ downward by 1 unit.
In the graph, that vertical asymptote remains the same, that is, $ x=0$.
Now, observe that in the graph of $ h\left( x \right)=-1+{{\log }_{2}}x $, the x-axis includes all real values greater than 0, and the range includes all real numbers.
So, the domain of the stated function is $\left( 0,\infty \right)$ and the range is $\left( -\infty,\infty \right)$.
The $ x\text{-intercept}$ of the function $ h\left( x \right)=-1+{{\log }_{2}}x $ is $\left( 2,0 \right)$, the vertical asymptote is $ x=0$, the domain of the provided function is $\left( 0,\infty \right)$, and the range is $\left( -\infty,\infty \right)$.
