Chapter 2.6, Problem 39E

Solution

The asymptotes and intercepts, describe the behaviour of vertical asymptotes using limits, analyse and graph the function.

The vertical asymptote is $x=-3$ and $x=4$ , horizontal asymptotes is $y=0$ with intercept $\left(0,0.083\right)$ .

Given:

The function is $h\left(x\right)=\frac{x-1}{x^2-x-12}$ .

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, $h\left(x\right)=\frac{x-1}{x^2-x-12}$ .

To find vertical asymptotes, find the zeros of denominator, so factorize the denominator $x^2-x-12$ :

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

Thus, the zeros of denominator of $f\left(x\right)$ is $x=-3$ and $x=4$ , Means the vertical asymptotes are $x=-3$ and $x=4$ .

To find the behaviour of vertical asymptotes, compute the limit when x approaches to $x=-3$ and $x=4$ from left and right,

  $\begin{array}{c}\underset{x\to -3^{-}}{\lim }h\left(x\right)=\underset{x\to -3^{-}}{\lim }\frac{x-1}{x^2-x-12}\\ =-\infty \text{ since, }{x}^2-x-12>0,x-1<0\text{ as }x\to -3^{-}\end{array}$

  $\begin{array}{c}\underset{x\to -3^{+}}{\lim }h\left(x\right)=\underset{x\to -3^{+}}{\lim }\frac{x-1}{x^2-x-12}\\ =\infty \text{ since, }{x}^2-x-12<0,x-1<0\text{ as }x\to -3^{+}\end{array}$

  $\begin{array}{c}\underset{x\to 4^{-}}{\lim }h\left(x\right)=\underset{x\to 4^{-}}{\lim }\frac{x-1}{x^2-x-12}\\ =-\infty \text{ since, }{x}^2-x-12<0,x-1>0\text{ as }x\to -1^{-}\end{array}$

  $\begin{array}{c}\underset{x\to 4^{+}}{\lim }h\left(x\right)=\underset{x\to 4^{+}}{\lim }\frac{x-1}{x^2-x-12}\\ =\infty \text{ since, }{x}^2-x-12>0,x-1>0\text{ as }x\to -1^{+}\end{array}$

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $h\left(x\right)=\frac{x-1}{x^2-x-12}$ 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, $y=0$ .

Now find the intercept, the x -intercept is given by zeros of numerator that are not zero of denominator.

Hence, solve $x-1=0$ . That gives $x=1$ . Means the x -intercept of $h\left(x\right)$ is $\left(1,0\right)$ .

And y -intercept is given by $h\left(0\right)$ , so from $h\left(x\right)=\frac{x-1}{x^2-x-12}$ .

  $\begin{array}{c}h\left(0\right)=\frac{0-1}{0^2-0-12}\\ =\frac{1}{12}\end{array}$

Thus y -intercept is $\left(0,\frac{1}{12}\right)\text{ or }\left(0,0.083\right)$ .

Hence, using vertical asymptotes, $x=-3$ and $x=4$ , horizontal asymptotes, $y=0$ with intercept is $\left(1,0\right)$ , $\left(0,0.083\right)$ the graph obtain is as:

  

Interpretations form graph:

1) Domain of $f\left(x\right)$ is $\left(-\infty ,-3\right)\cup \left(-3,4\right)\cup \left(4,\infty \right)$

2) Range is $\left(-\infty ,\infty \right)$ .

3) Continuous everywhere except $x=4,-3$ .

4) Decreasing in $\left(-\infty ,-3\right)\cup \left(-3,4\right)\cup \left(4,\infty \right)$ .

5) No extremes.

6) Not symmetric.

7) Unbounded.

Conclusion:

The vertical asymptote is $x=-3$ and $x=4$ , horizontal asymptote is $y=0$ with intercept $\left(0,0.083\right)$ .