Chapter 2.6, Problem 52E

Solution

The intercepts, analyse and graph the function.

The intercept of $f\left(x\right)$ are $\left(0,0\right)\text{ and }\left(-1/2,0\right)$ .

Given:

The function is $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ .

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, $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ .

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,

  $\begin{array}{c}4x^2+2x=0\\ 2x\left(2x+1\right)=0\end{array}$

Thus, $x=0\text{ or }x=-\frac{1}{2}$ . The x- intercepts are $\left(0,0\right)\text{ and }\left(-1/2,0\right)$ .

And y -intercept is given by $f\left(0\right)$ , so from $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ .

  $\begin{array}{c}f\left(0\right)=\frac{4{\left(0\right)}^2+2\left(0\right)}{{\left(0\right)}^2-4\left(0\right)+8}\\ =0\end{array}$

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

Hence the intercept are $\left(0,0\right)\text{ and }\left(-1/2,0\right)$ .

Find vertical asymptotes, that is given by zeros of denominator $x^2-4x+8$ ,

  $\begin{array}{c}x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\cdot 1\cdot 8}}{2\cdot 1}\\ =\frac{4\pm 4i}{2}\\ =2\pm 2i\end{array}$

Since, the zeros of denominator of $f\left(x\right)$ is not real, Vertical asymptotes is not present.

To find end behaviour asymptotes, find $m\text{ and }n$ by comparing, $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ 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, $y=4$ .

To graph $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ , compute the value of $f\left(x\right)$ for chosen values of x as follows:

  

The graph of $f\left(x\right)=\frac{4x^2+2x}{x^2-4x+8}$ obtain is as:

  

Interpretations form graph:

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

2) Range is $\left[0,9\right]$ .

3) Continuous everywhere.

4) Increasing $\left(-0.23,3.79\right)$ and decreasing in $\left(-\infty ,-0.23\right)\cup \left(3.79,\infty \right)$ .

5) Local maxima $\left(3.79,9.02\right)$ and local minima $\left(-0.23,-0.02\right)$ .

6) Not symmetric.

7) Unbounded.

Conclusion:

The intercept of $f\left(x\right)$ are $\left(0,0\right)\text{ and }\left(-1/2,0\right)$ .