Chapter 3.7, Problem 90E

To determine

To calculate : The function, $r\left(x\right)=\frac{x^6+10}{x^4+8x^2+15}$ has no x -intercept, and no horizontal, vertical or slant asymptotes. Also, explain its end behavior.

Answer to Problem 90E

The rational function, r(x) also approaches to $\infty$ when x approaches to $\infty$ and $-\infty$ .

Explanation of Solution

Given information : The function is $r\left(x\right)=\frac{x^6+10}{x^4+8x^2+15}$ .

Concept used : The function has x -intercept for the value of x at which the function is 0.

The horizontal asymptotes exists when the degree of numerator is less than or equal to the degree of denominator.

The vertical asymptotes exists where the denominator of the rational function is 0.

The slant asymptotes exists when the degree of numerator is 1 more than the degree of denominator.

The end behavior is the value of function when x approaches to $\infty$ and $-\infty$ .

Calculation : For x -intercepts, equate the function to 0.

  $\begin{array}{c}\frac{x^6+10}{x^4+8x^2+15}=0\\ x^6+10=0\\ x^6=-10\end{array}$

As the exponent of x is an even number, it will always give the positive value. So, the equation will have no solution.

Thus, the function has no x -intercepts.

The degree of the numerator is 6 and the degree of the denominator is 4. As the degree of numerator is greater than the degree of denominator.

Thus, the function has no horizontal asymptotes.

For vertical asymptotes, equate the denominator of the function to 0.

  $\begin{array}{c}{x}^4+8x^2+15=0\\ x^4+3x^2+5x^2+15=0\\ x^2\left(x^2+3\right)+5\left(x^2+3\right)=0\\ \left(x^2+5\right)\left(x^2+3\right)=0\\ x^2=-5,-3\end{array}$

The value of square of a function cannot be negative.

Thus, the rational function, $r\left(x\right)=\frac{x^6+10}{x^4+8x^2+15}$ has no vertical asymptotes.

The degree of the numerator is 6 and the degree of the denominator is 4. As the degree of numerator is 2 more than the degree of denominator.

Thus, the function has no slant asymptotes.

Divide the numerator and denominator of the rational function by $x^4$ .

  $\begin{array}{c}r\left(x\right)=\frac{\frac{x^6+10}{x^4}}{\frac{x^4+8x^2+15}{x^4}}\\ =\frac{x^2+\frac{10}{x^4}}{1+\frac{8}{x^2}+\frac{15}{x^4}}\end{array}$

Find the value of the function when x approaches to $\infty$ .

  $\begin{array}{c}\underset{x\to \infty }{\lim }r\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^2+\frac{10}{x^4}}{1+\frac{8}{x^2}+\frac{15}{x^4}}\\ =\frac{{\infty }^2+\frac{10}{{\infty }^4}}{1+\frac{8}{{\infty }^2}+\frac{15}{{\infty }^4}}\\ =\frac{\infty +0}{1+0+0}\\ =\infty \end{array}$

Thus, when x approaches to $\infty$ the rational function, r(x) also approaches to $\infty$ .

Find the value of the function when x approaches to $-\infty$ .

  $\begin{array}{c}\underset{x\to -\infty }{\lim }r\left(x\right)=\underset{x\to -\infty }{\lim }\frac{x^2+\frac{10}{x^4}}{1+\frac{8}{x^2}+\frac{15}{x^4}}\\ =\frac{{\left(-\infty \right)}^2+\frac{10}{{\left(-\infty \right)}^4}}{1+\frac{8}{{\left(-\infty \right)}^2}+\frac{15}{{\left(-\infty \right)}^4}}\\ =\frac{\infty +0}{1+0+0}\\ =\infty \end{array}$

Thus, when x approaches to $-\infty$ the rational function, r(x) also approaches to $\infty$ .