Finding the Domain and the Intercepts of a Rational Function

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

1. \(R( x) =\dfrac{2x^{2}-4}{x^{2}-4}\)
2. \(R( x) =\dfrac{x}{x^{2}+1}\)
3. \(R(x) =\dfrac{x^{2}-1}{x-1}\)

Solution

1. 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 \( \sqrt{2}\).
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 \(0\).
3. 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 \(-1\).