Answer
See the graph below:
Work Step by Step
Let us draw the curve of $ f\left( x \right)={{\log }_{2}}x $ as follows:
Construct the table of coordinates for $ f\left( x \right)={{\log }_{2}}x $
Choose the appropriate values for $ x $ and calculate the corresponding $ y\text{-values}$:
Now, plot every point given in the above table in the graph and join them using a smooth curve, and the $ y\text{-axis}$ or $ x=0$ as the vertical asymptote.
Now, the curve of $ r\left( x \right)={{\log }_{2}}\left( -x \right)$ is the same as the graph of $ f\left( x \right)={{\log }_{2}}x $ reflected about the $ y\text{-axis}$ (by multiply each $ x\text{-coordinate}$ by −1).
As observed in the graph, the vertical asymptote is $ x=0$.
Observe the graph of $ r\left( x \right)={{\log }_{2}}\left( -x \right)$: the $ x\text{-intercept}$ of the function $ r\left( x \right)={{\log }_{2}}\left( -x \right)$ is $\left( -1,0 \right)$, the vertical asymptote is $ x=0$, and the domain of the provided function is $\left( -\infty,0 \right)$, and the range is $\left( -\infty,\infty \right)$.
Thus, the $ x\text{-intercept}$ of the function $ r\left( x \right)={{\log }_{2}}\left( -x \right)$ is $\left( -1,0 \right)$, the vertical asymptote is $ x=0$, and the domain of the provided function is $\left( -\infty,0 \right)$, and range is $\left( -\infty,\infty \right)$.
