Chapter 2.6, Problem 51E

Solution

The intercepts, analyse and graph the function.

The intercept of $f\left(x\right)$ is $\left(0,0.8\right)$ .

Given:

The function is $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ .

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, $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ .

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

Since, the numerator is not factorable, use quadratic formula to find the roots:

  $\begin{array}{c}x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\cdot 3\cdot 4}}{2\cdot 3}\\ =\frac{2\pm \sqrt{4-48}}{6}\\ =\frac{2\pm 2i\sqrt{11}}{6}\\ =\frac{1\pm i\sqrt{11}}{3}\end{array}$

The roots are not real so there is no x- intercept.

And y -intercept is given by $f\left(0\right)$ , so from $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ .

  $\begin{array}{c}f\left(0\right)=\frac{3{\left(0\right)}^2-2\left(0\right)+4}{{\left(0\right)}^2-4\left(0\right)+5}\\ =\frac{4}{5}\end{array}$

Thus y -intercept is $\left(0,\frac{4}{5}\right)$ or $\left(0,0.8\right)$ .

Hence the only intercept is $\left(0,0.8\right)$ .

Find vertical asymptotes, so find the zeros of denominator,

  $\begin{array}{c}x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\cdot 1\cdot 5}}{2\cdot 1}\\ =\frac{4\pm 2i}{2}\\ =2\pm i\end{array}$

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, $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ 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, $y=3$ .

To graph $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ , compute the value of $f\left(x\right)$ for chosen values of x as follows:

  

The graph of $f\left(x\right)=\frac{3x^2-2x+4}{x^2-4x+5}$ obtain is as:

  

Interpretations form graph:

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

2) Range is $\left[0.77,14.22\right]$ .

3) Continuous everywhere.

4) Increasing $\left(-0.24,2.44\right)$ and decreasing in $\left(-\infty ,0.24\right)$ .

5) Local maxima $\left(2.44,14.22\right)$ and local minima $\left(-0.24,0.77\right)$ .

6) Not symmetric.

7) Unbounded.

Conclusion:

The intercept of $f\left(x\right)$ is $\left(0,0.8\right)$ .