Chapter 2.6, Problem 28E

Solution

The asymptotes and intercepts of function and graph the function.

The vertical asymptotes is $x=-2$ and $x=2$ , horizontal asymptotes is $y=-3$ with intercepts $\left(2.17,0\right)$ , $\left(-1.84,0\right)$ and $\left(0,-3\right)$ .

Given:

The function is $f\left(x\right)=\frac{-3x^2+x+12}{x^2-4}$ .

Concept Used:

If a polynomial function in the form $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ , then, vertical asymptotes is given by real zeros of denominator provided it should not be zero of numerator.

And,

The end behaviour asymptote given by $\underset{x\to {\infty }^{-}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}\text{ or }\underset{x\to {\infty }^{+}}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ is horizontal asymptotes.

The condition can be concluded as,

1) If $n<m$ , the end behaviour is horizontal asymptotes that is $y=0$ ,

2) If $n=m$ the end behaviour is horizontal asymptotes is horizontal asymptotes that is $y=\frac{a_n}{b_n}$

3) If $n>m$ ,the quotient $q\left(x\right)$ of $f\left(x\right)/g\left(x\right)$ is the end behaviour asymptotes, where $q\left(x\right)$ is given by $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ with $r\left(x\right)$ is remainder of $f\left(x\right)/g\left(x\right)$ . Hence, no horizontal asymptotes.

And,

The x -intercept is given by zeros of numerator that are not zero of denominator. And y -intercept is given by $f\left(0\right)$ .

From using asymptotes and intercepts the graph can be drawn.

Calculation:

Consider the function, $f\left(x\right)=\frac{-3x^2+x+12}{x^2-4}$ .

To find vertical asymptotes, find the zeros of denominator, so factorize the denominator $x^2-4$ :

  $x^2-4=\left(x+2\right)\left(x-2\right)$

Thus, the zeros of denominator of $f\left(x\right)$ is $x=-2$ and $x=2$ , Means the vertical asymptotes are $x=-2$ and $x=2$ .

To find end behaviour asymptotes, find $m\text{ and }n$ compare, $f\left(x\right)=\frac{-3x^2+x+12}{x^2-4}$ with $y=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\left(a_n{x}^n+\mathrm{....}\right)}{\left(b_m{x}^m+\mathrm{....}\right)}$ .

Since, $n=m$ end behaviour asymptotes is horizontal asymptotes is,

  $\begin{array}{c}y=\frac{-3}{1}\\ =-3\end{array}$

Now find the intercept, the x -intercept is given by zeros of numerator that are not zero of denominator. Hence solve $-3x^2+x+12=0$ for x .

  $x=\frac{1\pm \sqrt{145}}{6}$

The x -intercept are $\left(\frac{1+\sqrt{145}}{6},0\right)$ and $\left(\frac{1-\sqrt{145}}{6},0\right)$ or $\left(2.17,0\right)$ and $\left(-1.84,0\right)$

And y -intercept is given by $f\left(0\right)$ , so from $f\left(x\right)=\frac{-3x^2+x+12}{x^2-4}$ .

  $\begin{array}{c}f\left(0\right)=\frac{-3{\left(0\right)}^2+0+12}{0^2-4}\\ =\frac{12}{-4}\\ =-3\end{array}$

Thus y -intercept is $\left(0,-3\right)$ .

Hence, using vertical asymptotes $x=-2$ and $x=2$ , horizontal asymptotes, $y=-3$ with intercepts are $\left(2.17,0\right)$ , $\left(-1.84,0\right)$ and $\left(0,-3\right)$

the graph obtain is as:

  

Conclusion:

The vertical asymptotes is $x=-2$ and $x=2$ , horizontal asymptotes is $y=-3$ with intercepts $\left(2.17,0\right)$ , $\left(-1.84,0\right)$ and $\left(0,-3\right)$ .