Chapter 13, Problem 2GR

Solution

To find: The vertical asymptotes and holes of the graph of the rational function $y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}$ .

The vertical asymptote is at $x=-\frac{1}{2},4$ and there are no holes of the rational function.

Given:

The rational function is $y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}$ .

Concept:

1. Vertical asymptotes of a rational function: The vertical asymptotes of a rational function can be evaluated by reducing the rational function first and equate the denominator equal to zero. If the denominator is already in the reduced form, then simply equate it to zero to find the vertical asymptotes of the function.
2. 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 $y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}$ is in reduced form because it has no common factors in numerator and denominator. Therefore, using the clause $\left(a\right)$ , the vertical asymptote of the function is evaluated as:

  $\begin{array}{c}y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}\mathrm{....}\left(\text{Given}\right)\\ \left(2x+1\right)\left(x-4\right)=0\mathrm{....}\left(\text{Set denominator equal to zero}\right)\\ 2x+1=0\text{or}x-4=0\\ x=-\frac{1}{2}\text{ or}x=4\end{array}$

The vertical asymptote is at $x=-\frac{1}{2},4$ .

Now, the rational function $y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}$ is already in reduced form. Therefore, using the clause $\left(b\right)$ , the rational function $y=\frac{x+2}{\left(2x+1\right)\left(x-4\right)}$ has no holes.

Conclusion:

The vertical asymptote is at $x=-\frac{1}{2},4$ and there are no holes of the rational function.