Chapter 2.6, Problem 54E

Solution

The intercepts, analyse and graph the function.

The intercept are $\left(1.26,0\right)$ and $\left(0,-1\right)$ .

Given:

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

Concept Used:

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

And,

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.

Calculation:

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

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

So, find the zeros of the numerator as,

  $\begin{array}{c}{x}^3-2=0\\ x^3=2\\ x=1.26\end{array}$

Thus, $x=1.26$ . The x- intercepts is $\left(1.26,0\right)$ .

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

  $\begin{array}{c}f\left(x\right)=\frac{0^3-2}{0+2}\\ =-1\end{array}$

Thus y -intercept is $\left(0,-1\right)$ .

Hence the intercept are $\left(1.26,0\right)$ and $\left(0,-1\right)$ .

Find vertical asymptotes, that is given by zeros of denominator $x+2$ ,

Since, the zeros of denominator of $f\left(x\right)$ is $x=-2$ , Vertical asymptotes is line $x=-2$ .

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{x^3-2}{x+2}$ 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 the quotient of $f\left(x\right)=\frac{x^3-2}{x+2}$ . Rewrite $f\left(x\right)$ using division as follows,

  

Thus, write $f\left(x\right)$ as,

  $\begin{array}{c}f\left(x\right)=\frac{x^3-2}{x+2}\\ =x^2-2x+4-\frac{10}{x+2}\end{array}$

Hence, end behaviour asymptotes is $y=x^2-2x+4$ .

To graph $f\left(x\right)=\frac{x^3-2}{x+2}$ , compute the value of $f\left(x\right)$ for chosen values of x as follows:

  

The graph of $f\left(x\right)=\frac{x^3-2}{x+2}$ obtain is as:

  

Interpretations form graph:

1) Domain of $f\left(x\right)$ is $\left(-\infty ,-2\right)\cup \left(-2,\infty \right)$

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

3) Continuous everywhere except $x=-2$

4) Increasing in $\left(-3.10,-2\right)\cup \left(-2,\infty \right)$ and decreasing in $\left(-\infty ,-3.10\right)$ .

5) No Local maxima and local minima at $\left(-3.10,28.90\right)$ .

6) Not symmetric.

7) Unbounded.

Conclusion:

The intercept are $\left(1.26,0\right)$ and $\left(0,-1\right)$ ..