Chapter 2.7, Problem 29E

To determine

To sketch the function $f(x)=\frac{x^2+3x}{x^2+x-6}$ , also find their intercepts, asymptotes and holes.

Answer to Problem 29E

  $x=2,y=1$

Explanation of Solution

Given:

Function: $f(x)=\frac{x^2+3x}{x^2+x-6}$

Calculation for graph:

Consider $f(x)=\frac{x^2+3x}{x^2+x-6}$

Values of x	Values of f (x)
0	0
1	-1
-1	0.333
2	Infinity
-2	0.5

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

Graph:

  

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=\frac{x^2+3x}{x^2+x-6}\\ \Rightarrow f(x)=\frac{x(x+3)}{x^2+x-6}\\ \Rightarrow f(x)=\frac{x(x+3)}{x^2+3x-2x-6}\\ \Rightarrow f(x)=\frac{x(x+3)}{x(x+3)-2(x+3)}\\ \Rightarrow f(x)=\frac{x(x+3)}{(x-2)(x+3)}\\ \Rightarrow f(x)=\frac{x}{x-2}\end{array}$

Intercepts:

Let $f(x)=y.$

  $\Rightarrow y=\frac{x}{x-2}$

To find x intercepts, put y = 0,

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

So, x intercept is (0, 0).

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{0}{0-2}\\ \Rightarrow y=\frac{0}{-2}\\ \Rightarrow y=0\end{array}$

So, y intercept is (0, 0).

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 equal to the degree of the denominator, the horizontal asymptote is

  $\begin{array}{l}\Rightarrow y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}\\ \Rightarrow y=\frac{1}{1}\\ \Rightarrow y=1\end{array}$

Holes:

Here, the given function contains one common factor in numerator and denominator i.e.., $(x+3).$

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

  $\begin{array}{l}\Rightarrow x+3=0\\ \Rightarrow x=-3\end{array}$

So, the hole of function is $x=-3.$