Chapter 2.7, Problem 67E

To determine

To find the asymptotes and holes of the function $f(x)=\frac{2x^3-x^2-2x+1}{x^2+3x+2}$ and verifying them using graph.

Answer to Problem 67E

  $x=-2,y=2x-7$

Explanation of Solution

Given:

Function: $f(x)=\frac{2x^3-x^2-2x+1}{x^2+3x+2}$

Calculation:

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

Intercepts:

Let $f(x)=y.$

  $\Rightarrow y=\frac{(x-1)(2x-1)}{(x+2)}$

To find x intercepts, put y = 0,

  $\begin{array}{l}\Rightarrow y=0\\ \Rightarrow \frac{(x-1)(2x-1)}{(x+2)}=0\\ \Rightarrow (x-1)(2x-1)=0\\ \Rightarrow x=\frac{1}{2},1\\ \Rightarrow x=0.5,1\end{array}$

So, x intercepts are (0.5, 0) and (1, 0)

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{(0-1)(2(0)-1)}{(0+2)}\\ \Rightarrow y=\frac{1}{2}\\ \Rightarrow y=0.5\end{array}$

So, y intercept is (0, 0.5).

Asymptotes:

Vertical asymptotes:

To find vertical asymptotes, put denominator of the given function equal to zero.

  $\begin{array}{l}\Rightarrow x+2=0\\ \Rightarrow x=-2\end{array}$

Horizontal asymptotes:

As the degree of numerator is larger than the degree of the denominator, the horizontal asymptote does not exist for given function.

Slant asymptotes:

To find slant asymptote, divide given function numerator with denominator using long division.

Dividend: $2x^3-x^2-2x+1$

Divisor: $x^2+3x+2$

  $x^2+3x+2\begin{array}{c}\hfill 2x-7\\ \hfill \overline{)\begin{array}{l}2x^3-x^2-2x+1\\ \begin{array}{c}2x^3+6x^2+4x\end{array}\\ -7x^2-6x+2\\ \begin{array}{c}-7x^2-21x-14\end{array}\\ 15x+15\end{array}}\end{array}$  $\begin{array}{l}\frac{2x^3}{x^2}=2x\\ \text{Multiply}2x(x^2+3x+2)\text{and subtract}\text{.}\\ \\ \frac{-7x^2}{x^2}=-7\\ \text{Multiply}-7(x^2+3x+2)\text{and subtract}\text{.}\end{array}$

Here, quotient is $2x-7$ and remainder is 15x + 15.

So, the slant asymptote is $y=2x-7.$

Holes:

Here, the given function contains one common factor in numerator and denominator i.e.., $(x+1).$

To find hole of the function, equate common factor equal to zero.

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

So, the hole of function is $x=-1.$

Calculation for graph:

Consider $f(x)=\frac{2x^3-x^2-2x+1}{x^2+3x+2}$

Values of x	Values of f (x)
0	0.5
1	0
-1	6
2	0.75
-2	Infinity

By taking different values of x, the graph can be plotted.

Graph:

  

Interpretation:

By observing graph, it is clear that

The x intercepts are (0.5, 0) and (1, 0),

The y intercept is (0, 0.5),

The vertical asymptotes are $x=-2$ ,

The slant asymptote is $y=2.$