Answer
Domain: All real numbers except $x= \frac{3}{2},\ x= \frac{7}{4}$
Vertical asymptotes: $x= \frac{3}{2},\ x= \frac{7}{4}$
Work Step by Step
Given \begin{equation}
h(x)=\frac{5 x-11}{8 x^2-26 x+21}.
\end{equation} The domain of a rational function consists of all real numbers except numbers that makes the denominator equal to zero. We need to set the expression of the denominator to zero to find the numbers that make the denominator zero and exclude them from the domain of the function. \begin{equation}
\begin{aligned}
8 x^2-26 x+21&=0 \\
x&=\frac{-(-26) \pm \sqrt{(-26)^2-4 \cdot 8 \cdot 21}}{2 \cdot 8}\\
& =\frac{26 \pm 2}{2 \cdot 8}\\
\therefore x&=\frac{26 + 2}{2 \cdot 8}= \frac{7}{4}\\
x&=\frac{26 - 2}{2 \cdot 8}= \frac{3}{2}\\
\end{aligned}
\end{equation} Domain: All real numbers except $x= \frac{3}{2},\ x= \frac{7}{4}$
$x= \frac{3}{2},\ x= \frac{7}{4}$ represent vertical asymptotes because the factors of the denominator do not simplify with the numerator. Use your calculator to check the graph.
