Chapter 4, Problem 5CT

To determine

The domain and any horizontal, vertical or oblique asymptotes of the rational function $g\left(x\right)=\frac{2x^2-14x+24}{x^2+6x-40}$ .

Answer to Problem 5CT

Solution:

The domain of the rational function is $\left(-\infty ,-10\right)\cup \left(-10,4\right)\cup \left(4,\infty \right)$ and the function has no oblique asymptote while the horizontal asymptote is $y=2$ and vertical asymptote is $x=-10$ .

Explanation of Solution

Given information:

The rational function, $g\left(x\right)=\frac{2x^2-14x+24}{x^2+6x-40}$

The domain of a rational function is set of all real numbers except those for which the denominator of the function is $0$ .

For the rational function $g\left(x\right)=\frac{2x^2-14x+24}{x^2+6x-40}$ , the denominator is $q\left(x\right)=x^2+6x-40$ .

Therefore the values of $x$ , for which the denominator of the rational function is $0$ are the zeros of the function $q\left(x\right)$ .

Set $q\left(x\right)=0$ .

  $\Rightarrow x^2+6x-40=0$

By factoring,

  $\Rightarrow \left(x-4\right)\left(x+10\right)=10$

  $\Rightarrow x=4$ or $x=-10$

Hence the denominator of the rational function is $0$ , when the value of $x$ is $4$ or $-10$ .

Therefore, the domain of the function is the set of all real numbers except $x=4,-10$ .

Thus, the domain of the rational function is $\left\{x\in ℝ|x\ne 4,-10\right\}$ .

In interval notation domain is $\left(-\infty ,-10\right)\cup \left(-10,4\right)\cup \left(4,\infty \right)$ .

Next, find the asymptotes of the rational function $g\left(x\right)$ .

Here the rational function $g\left(x\right)=\frac{2x^2-14x+24}{x^2+6x-40}$ is not in the lowest term.

Firstly, convert the rational function in the lowest term.

The numerator $2x^2-14x+24$ can factor as $2\left(x-3\right)\left(x-4\right)$ and the denominator $x^2+6x-40$ can be factor as $\left(x-4\right)\left(x+10\right)$ .

  $\Rightarrow g\left(x\right)=\frac{2x^2-14x+24}{x^2+6x-40}$

  $=\frac{2\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+10\right)}$

  $=\frac{2\left(x-3\right)}{\left(x+10\right)}$

Thus, the rational function becomes $g\left(x\right)=\frac{2\left(x-3\right)}{\left(x+10\right)}$ .

If $x-r$ is a factor of the denominator of $g\left(x\right)$ , in lowest terms, then $g\left(x\right)$ will have the vertical asymptote $x=r$ .

Here, $x+10$ is a factor of the denominator of $g\left(x\right)$ , then $x=-10$ is a vertical asymptote.

Here the numerator is $2\left(x-3\right)$ and the denominator is $x+10$ , that means the degree of the numerator is equal to the degree of the denominator.

Thus, the rational function $g\left(x\right)$ has a horizontal asymptote equal to the ratio of the leading coefficients; $y=\frac{2}{1}=2$ .

Hence the line $y=2$ is a horizontal asymptote.

As there is horizontal asymptote, oblique asymptote does not exist.