Figure

Find the domain and the intercepts (if any) of each rational function:

Solution

The domain of $R ( x) =\dfrac{2x^{2}-4}{ x^{2}-4}$ is $\{ x|x\neq -2;x\neq 2\}$ . Since $0$ is in the domain of $R$ and $R ( 0) =1$ , the $y$ -intercept is $1$ . The zeros of $R$ are solutions of the equation $2x^{2}-4=0$ or $x^{2}=2.$ Since $- \sqrt{2}$ and $\sqrt{2}$ are in the domain of $R,$ the $x$ -intercepts are $- \sqrt{2}$ and $\sqrt{2}$ .
The domain of $R ( x) =\dfrac{x}{x^{2}+1}$ is the set of all real numbers. Since $0$ is in the domain of $R$ and $R ( 0) =0,$ the $y$ -intercept is $0$ , and the $x$ -intercept is also $0$ .
The domain of $R ( x) =\dfrac{x^{2}-1}{x-1}$ is $\{ x|x\neq 1\}$ . Since $0$ is in the domain of $R$ and $R ( 0) =1$ , the $y$ -intercept is $1$ . The $x$ -intercept(s), if any, satisfy the equation $\begin{eqnarray*} x^{2}-1 &=&0 \\[0pt] x^{2} &=&1 \\[0pt] x &=&-1\quad\hbox{or}\quad x=1 \end{eqnarray*}$ Since $1$ is not in the domain of $R,$ the only $x$ -intercept is $-1$ .