Chapter 4.3, Problem 40AYU

To determine

To analyze: the graph of given function

Answer to Problem 40AYU

  

Explanation of Solution

Given:

  $f\left(x\right)=2x+\frac{9}{x}$

Calculation:

Here, we have to analyze the graph of the rational function

  $R\left(x\right)=2x+\frac{9}{x}$

Let us consider it as

  $y=2x+\frac{9}{x}$

Too, so that it will be easy to graph it on the Cartesian plane.

This can be done simply in a few steps. Let us do it.

Step1: We will factorize the numerator and denominator and find the domain of the rational function. Therefore,

  $R\left(x\right)=2x+\frac{9}{x}$

  $\begin{array}{l}Fromtheaboveexpression,weconcludethatRisdefinedforallrealexceptwhenx=0.\hfill \\ Domain=x;x\cancel{=}0\hfill \end{array}$

Step2: Now, we will write $R$ in lowest terms. Here, as we see there are no common factors between denominator and numerator, therefore $R$ is in its lowest terms.

Step3: We have to find the intercepts on the graph. The x intercept can be found by determining the real zeroes of the numerator of $R$ written in the lowest terms. Therefore, bysolving

No real solutions.

The y intercept can be found by determining the value of $R$ when $x=0$ and here the value of $R$ is $-5forx=0$ . Hence, y intercept for this is $-5$ .

Step4: Let us now test for the symmetry of the graph. A graph is symmetric about $y-axis$ when $R\left(-x\right)=R\left(x\right)$ and the function is said to be even function. And a graph is symmetric aboutorigin if $R\left(-x\right)=-R\left(x\right)$ and the function is said to be odd function. Let us check the symmetry of the given graph now. Therefore,

  $\begin{array}{l}R\left(-x\right)=2\left(-x\right)+\frac{9}{\left(-x\right)}\hfill \\ =-\left(2x+\frac{9}{x}\right)\hfill \end{array}$

From this, we conclude that there is symmetry with respect to with $y-axisasR\left(-x\right)=-R\left(x\right)$ .

Step5: Now, we will locate the vertical asymptotes. This is done by finding the zeroes of the denomination of the rational function in lowest terms. With R in the lowest terms we find that the graph of $Rhasverticalasymptoteaty=0$ .

Step6: Now let us locate any horizontal or oblique asymptote and determine the point if any which intersect the asymptotes at any point.

There is no horizontal asymptote of the graph. Let us calculate the oblique asymptotes. This can be found by dividing the numerator by denominator and the part except the remainder is the asymptote.

In $R$ when we do this procedure, we find that $R$ can again be written as

  $y=2x+\frac{9}{x}$

Therefore, the polynomial has oblique asymptote as $y=2x$ .

Step7: Now let us graph $R$ using a graphing utility. So, we have the graph as

  

Step8: Now let us use the results we have from step 1 to 7 and the make the final graph by hand. The graph for this can be shown as:

  

Conclusion:

Thus, the graph of $f$ is obtained which illustrates the information obtained from Steps 1 through 6.