## Precalculus (6th Edition) Blitzer

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 $\left( -\infty,\infty \right)$.