Chapter 8.3, Problem 22PPSE

Solution

To find: The vertical asymptotes and holes of the graph of the rational function $y=\frac{x+5}{x^2+9}$ .

The vertical asymptote is none and there are no holes of the rational function.

Given:

The rational function is $y=\frac{x+5}{x^2+9}$ .

Concept:

  $\left(a\right)$ Vertical asymptotes of a rational function: The rational function $f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ has vertical asymptote at each real zero of $q\left(x\right)$ , if $p\left(x\right)$ and $q\left(x\right)$ have no common zeros.

  $\left(b\right)$ Holes of a rational function: The holes of a rational function can be evaluated by reducing the rational function first and look for common factors in numerator and denominator and equate the common term in numerator and denominator equal to zero to get the holes of the rational function. And, if the rational function is already in reduced form, then there are no holes in the rational function.

Calculation:

The rational function $y=\frac{x+5}{x^2+9}$ is already in reduced form. Therefore, using the clause $\left(a\right)$ , the vertical asymptote of the function is evaluated as:

  $\begin{array}{c}y=\frac{x+5}{x^2+9}\mathrm{....}\left(\text{Given}\right)\\ x^2+9=0\mathrm{....}\left(\text{Set denominator equal to zero}\right)\\ x^2+9=0\mathrm{....}\left(\text{which is impossible}\right)\mathrm{....}\left(1\right)\end{array}$

From $\left(1\right)$ , the vertical asymptote of the rational function is none.

Again, the rational function $y=\frac{x+5}{x^2+9}$ is already in reduced form. Therefore, using the clause $\left(b\right)$ , the rational function $y=\frac{x+5}{x^2+9}$ has no holes.

Conclusion:

The vertical asymptote is none and there are no holes of the rational function.