Chapter 3, Problem 27CLT

To determine

To draw: The graph of the rational function $f\left(x\right)=\frac{x^3-2x^2-9x+18}{x^2+4x+3}$ . Also identify all intercepts and find all asymptotes.

Answer to Problem 27CLT

x − intercept are 2,3 and y − intercept is 6. $x=-1$ is the asymptote.

Explanation of Solution

Given information:

Given rational function is, $f\left(x\right)=\frac{x^3-2x^2-9x+18}{x^2+4x+3}$ .

Consider, the given rational function as $f\left(x\right)=\frac{x^3-2x^2-9x+18}{x^2+4x+3}$ .

Graph of the function is,

  

From the graph it can be observed that the curve cuts the x − axis only at $x=2\text{ and }x=3$ and y − axis at $y=6$

Hence, the x − intercept are 2,3 and y − intercept is 6.

Also, the function $f\left(x\right)=\frac{x^3-2x^2-9x+18}{x^2+4x+3}$ can be written as

  $\begin{array}{c}f\left(x\right)=\frac{x^2\left(x-2\right)-9\left(x+2\right)}{x^2+3x+x+3}\\ =\frac{\left(x-2\right)\left(x^2-9\right)}{\left(x+3\right)\left(x+1\right)}\\ =\frac{\left(x-2\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+1\right)}\\ =\frac{\left(x-2\right)\left(x-3\right)}{\left(x+1\right)}\end{array}$

This implies that the function is not defined at $x=-1$ .

The asymptote of the function is $x=-1$