Chapter 2, Problem 134CR

To determine

To graph: the function $f(x)$ and check for intercept and various asymptotes.

Answer to Problem 134CR

y-intercept is $y=3$

x-intercept is $x=-3,-\frac{7}{2}$

Vertical asymptote is at $x=-1$

No horizontal asymptote

No hole

Slant asymptote is $2x+5$

Explanation of Solution

Given information:

  $f(x)=\frac{2x^2+7x+3}{x+1}$

Graph: Assuming the value of x to find $f(x)$ and plot the graph

  

  

Interpretation :

To determine y-intercept put $x=0$ and solve for $y$ or $f(x)$

  $f(x)=\frac{2x^2+7x+3}{x+1}=\frac{0+0+3}{0+1}=3$

To determine x-intercept put $y=0$ and solve for $x$

  $\begin{array}{l}f(x)=\frac{2x^2+7x+3}{x+1}\\ 0=\frac{2x^2+7x+3}{x+1}\\ 2x^2+7x+3=0\\ 2x^2+6x+x+3=0\\ \left(2x+7\right)\left(x+3\right)=0\\ x=-3,-\frac{7}{2}\end{array}$

To find the vertical asymptotes we have to solve the denominator by equating it equal to zero:

  $x+1=0\Rightarrow x=-1$

A horizontal asymptotes is defined when the degree of the denominator is greater than or equal to degree of the numerator.

Here the degree of denominator is less than numerator therefore we cannot calculate horizontal asymptotes

Slant asymptote occur when the degree of denominator is lower than that of the numerator.

And here degree of denominator is less than that of the numerator and horizontal asymptote is also absent, therefore there slant asymptote is given by solving the function-

  $f\left(x\right)=2x+5-\frac{2}{x+1}$

Slant asymptote is $2x+5$

Now, since the degree of the denominator is less than the degree of the numerator therefore there will be no hole possible in the graph.

On solving the above equation we find that

  $f(x)=\frac{2x^2+7x+3}{x+1}=\frac{\left(x+3\right)\left(2x+7\right)}{\left(x+1\right)}$

There is no common factor on both numerator and denominator, hence no hole.