Chapter 4, Problem 30RE

To determine

To discuss:the given rational function.

Answer to Problem 30RE

  

Explanation of Solution

Given:

  $F\left(x\right)=\frac{3x^2}{{\left(x-1\right)}^2}$

Calculation:

The domain of $R$ is $\left\{x|x\ne 1\right\}$

Since $x$ cannot be zero, there is no $y-\mathrm{int}ercept$ .

The function $F$ is in lowest terms. The real zero of the numerator is $0$ . The only $x-\mathrm{int}ercept$ is $0$ .

The real zeroes of the denominator are the real solutions of the equation.

  ${\left(x-1\right)}^2=0$ , that is, $1$ . The graph of $F$ has one vertical asymptote: $X=1$ . Graph the asymptote using dashed line.

The degree of the numerator is greater than the degree of the denominator. The function $F$ is improper. Use the method of long division to find the horizontal asymptote.

  $\begin{array}{l}{x}^2-2x+1\begin{array}{c}\hfill 3x-6\\ \hfill \overline{)3x^3}\end{array}\\ \underset{\_}{3x^3-6x^2+3x}\\ \underset{\_}{\begin{array}{l}-6x^2+3x\\ -6x^2+12x-6\end{array}}\\ -9x-6\end{array}$

The quotient is $3x-6$ . The line $y=3x-6$ is an oblique asymptote of the graph. Indicate the line $y=3x-6$ on the graph using a dashed line.

The zero of the numerator, $0$ , and the zero of the denominator, $1$ , divide the $X-axis$ into three intervals.

  $\left(-\infty ,0\right),\left(0,1\right),and\left(1,\infty \right)$

Interval	$\left(-\infty ,0\right)$	$\left(0,1\right)$	$\left(1,\infty \right)$
Number chosen	$-1$	$0.5$	$1.25$
Value of $F$	$-\frac{3}{4}$	$1.5$	$17.6$
Location of graph	Below $x-axis$	Above $x-axis$	Above $x-axis$
Point on graph	$\left(-1,-\frac{3}{4}\right)$	$\left(0.5,1.5\right)$	$\left(1.25,17.6\right)$

Plot $y-\mathrm{int}ercept$ and the points obtained from the table on a coordinate plane. Draw the vertical and oblique asymptotes using dashed lines.

The points are not labeled for convenience.

  

Now, determine the behavior of the graph near the asymptotes.

Since $3x-6$ is an oblique asymptote and the graph lies below $X-axis$ for $X<0$ , sketch a portion of the graph by placing a small arrow to the far left and under the $X-axis$ . The line $X=1$ is a vertical asymptote and the graph lies above the $X-axis$ for

  $X>0$ . Continue placing an arrow well above the $X-axis$ and approaching the line $X=1$ on the left.

  

Complete the graph as shown.

  

Conclusion:

Therefore, the given rational function is analyzed