Chapter 8.3, Problem 18PPSE

Solution

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

The vertical asymptote is none and the hole is $x=-5$ of the rational function.

Given:

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

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:

Change the rational function $y=\frac{x+5}{x+5}$ to reduced form and set denominator of the rational function equal to zero. Therefore, using the clause $\left(a\right)$ , the vertical asymptote of the rational function is evaluated as:

  $\begin{array}{c}y=\frac{x+5}{x+5}\mathrm{....}\left(\text{Given}\right)\mathrm{....}\left(1\right)\\ y=1\mathrm{....}\left(\text{Cancel the common factors}\right)\\ 1=0\mathrm{....}\left(\begin{array}{l}\text{Set denominator equal to}\\ \text{zero, which is impossible}\end{array}\right)\\ 1\ne 0\mathrm{....}\left(\text{So, no vertical asymptote}\right)\mathrm{....}\left(2\right)\end{array}$

From $\left(2\right)$ , the vertical asymptote of the rational function is none.

From $\left(1\right)$ , the rational function $y=\frac{x+5}{x+5}$ has common factor $\left(x+5\right)$ in numerator and denominator. Therefore, using the clause $\left(b\right)$ , equate the common factor $\left(x+5\right)$ equal to zero to obtain holes of the rational function.

  $\begin{array}{c}y=\frac{\left(x+5\right)}{\left(x+5\right)}\mathrm{....}\left(\text{From }\left(1\right)\right)\\ x+5=0\mathrm{....}\left(\begin{array}{l}\text{Equate the common}\\ \text{ factor }\left(x+5\right)=0\end{array}\right)\\ x=-5\end{array}$

Conclusion:

The vertical asymptote is none and the hole is $x=-5$ of the rational function.