Chapter 4, Problem 37RE

To determine

To analyze: The graph of Rational function $R(x)=\frac{2x^4}{{(x-1)}^2}$ .

Answer to Problem 37RE

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

The vertical asymptote is $x=1$ .

The oblique asymptote is $y=2x^2$.

Explanation of Solution

Given information:

The rationalfunction$R(x)=\frac{2x^4}{{(x-1)}^2}$.

Formula used:

Vertical Asymptote: A rational function $R(x)=\frac{p(x)}{q(x)}$ , in lowest terms, will have a

asymptote $x=r$if $r$ is a real zero of the denominator q.

Horizontal Asymptote: If a rational function is proper , the line $y=0$ is a asymptote of the graph.

Oblique Asymptote : it occur when the degree of the denominator of a rational

function is one less than the degree of the numerator. The asymptote will be along line

$y=mx+b$where $m\ne 0$

Calculation:

Consider ,

  $R(x)=\frac{2x^4}{{(x-1)}^2}$

Numerator:-

  $=x^3$

Denominator:-

  $\begin{array}{l}={(}^x\\ =(x-1)(x-1)\end{array}$

Therefore,

  $R(x)=\frac{x^3}{(x-1)(x-1)}$

Step 1.

The domain of R is the set of all real numbers except $x=1$ .

Step 2.

Now, R isin the lowest terms.

Step 3.

Now, the intercepts of the graph R are as follows:

x-intercept of the graph R

  $\begin{array}{l}=2x^4=0\\ =x=0\end{array}$

The x-intercept is $x=0$ .

y-intercept:-Since 0 is included in domain.

  $\begin{array}{l}R(x)=\frac{2x^4}{{(x-1)}^2}\\ R(0)=\frac{0}{{(0-1)}^2}\end{array}$

  $\begin{array}{l}=\frac{0}{1}\\ =0\end{array}$

They-intercept is $y=0$ .

Step 4.

Denominator:-

  $=(x-1)(x-1)=0$

  $\begin{array}{l}=x-1=0\\ =x=1\end{array}$ or $\begin{array}{l}=x-1=0\\ =x=1\end{array}$

The real zeros of the denominator R is $x=1,1$ . So, the graph of the R has one vertical asymptote$x=1$.

Step 5.

Since the degree of the numerator ,4,is greater than the degree of denominator 2.The rational function is improper.

Using long division method we get,

  $2x^2$

  $x^2-2x+1\overline{)\begin{array}{l}2x^4\\ 2x^4-4x^3+2x^2\end{array}}$

  $x^2-2x+1\overline{)\begin{array}{l}2x^4\\ -2x^4-4x^3+2x^2\end{array}}$

  $x^2-2x+1\overline{)-4x^3+2x^2}$

The quotient is $2x^2$ , so the line $y=2x^2$ is an oblique asymptote of the graph.

Now, solving the equation we get,

  $R(x)=\frac{2x^4}{{(x-1)}^2}$

  $\begin{array}{l}=\frac{2x^4}{{(x-1)}^2}=2x^2\\ =2x^4=2x^2{(}^x\\ =2x^4=2x^2(x^2-2x+1)\\ =2x^4=2x^4-4x^3+2x^2\\ =2x^4-2x^4+4x^3-2x^2=0\\ =4x^3-2x^2=0\\ =2x^2(2x-1)=0\end{array}$

  $\begin{array}{l}=2x^2=0\\ =x=0\end{array}$or$\begin{array}{l}=2x-1=0\\ =2x=1\\ =x=\frac{1}{2}\end{array}$

This implies the graph intersect the line $y=2x^2$at $x=0,\frac{1}{2}$ is a oblique asymptote.

Hence, $\left(0,2x^2\right),\left(\frac{1}{2},2x^2\right)$is a point on the graph of R.

Step 6.

  

Figure 1.

Step 7.

Using the information gathered from Step 1 − Step 5 the graph of R has been drawn. The green color shows the graph R(x). The graph has the vertical asymptote at $x=1$and oblique asymptote at $y=2x^2$represented by black color. The graph intersects$y=2x^2$ atpoint $\left(0,0\right)$ and $\left(0.5,0.5\right)$ .