Chapter 13, Problem 25CCSR

Solution

To find: All the asymptotes of the rational function $y=\frac{3x^2-5x-2}{x^2+2x-8}$ .

The vertical asymptote is at $x=-4$ and the horizontal asymptote is at $y=3$ .

Given:

The rational function is $y=\frac{3x^2-5x-2}{x^2+2x-8}$ .

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. Horizontal asymptote of a rational function: The horizontal asymptotes of a rational function can be evaluated by reducing the rational function first and then dividing the leading coefficient term of denominator by leading coefficient term of numerator, if the leading coefficients in numerator and denominator has the same degree. And is written as $y=\frac{\text{coeeficient of leading term in denominator}}{\text{coeeficient of leading term in numerator}}$ .

Calculation:

The rational function $y=\frac{2}{x-3}$ in reduced form is written as:

  $\begin{array}{c}y=\frac{3x^2-5x-2}{x^2+2x-8}\mathrm{....}\left(\text{Given}\right)\\ y=\frac{3x^2-6x+x-2}{x^2+4x-2x-8}\\ y=\frac{3x\left(x-2\right)+1\left(x-2\right)}{x\left(x+4\right)-2\left(x+4\right)}\\ y=\frac{\left(x-2\right)\left(3x+1\right)}{\left(x+4\right)\left(x-2\right)}\\ y=\frac{3x+1}{x+4}\mathrm{....}\left(\text{Reduced form}\right)\mathrm{....}\left(1\right)\end{array}$

Therefore, using the clause $\left(a\right)$ , the vertical asymptote of the function from the reduced form $\left(1\right)$ is evaluated as:

  $\begin{array}{c}y=\frac{3x+1}{x+4}\mathrm{....}\left(\text{Reduced form}\right)\\ x+4=0\mathrm{....}\left(\text{Set denominator equal to zero}\right)\\ x=-4\end{array}$

The vertical asymptote is at $x=-4$ .

Now, the rational function $y=\frac{3x^2-5x-2}{x^2+2x-8}$ is has same degrees in leading coefficient terms of numerator and denominator. Therefore, using the clause $\left(b\right)$ , the horizontal asymptote of the rational function $y=\frac{3x^2-5x-2}{x^2+2x-8}$ is written as:

  $\begin{array}{c}y=\frac{\text{coeeficient of leading term in denominator}}{\text{coeeficient of leading term in numerator}}\\ =\frac{3}{1}\\ =3\end{array}$

The horizontal asymptote is at $y=3$ .

Conclusion:

The vertical asymptote is at $x=-4$ and the horizontal asymptote is at $y=3$ .