Chapter 2.7, Problem 30E

To determine

To sketch the function $g(x)=\frac{5(x+4)}{x^2+x-12}$ , also find their intercepts, asymptotes and holes.

Answer to Problem 30E

  $x=3,y=0$

Explanation of Solution

Given:

Function: $g(x)=\frac{5(x+4)}{x^2+x-12}$

Calculation for graph:

Consider $g(x)=\frac{5(x+4)}{x^2+x-12}$

Values of x	Values of g(x)
0	-1.666
1	-2.5
-1	-1.25
2	-5
-2	-1

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

Graph:

  

Calculation:

  $\begin{array}{l}\Rightarrow g(x)=\frac{5(x+4)}{x^2+x-12}\\ \Rightarrow g(x)=\frac{5(x+4)}{x^2+4x-3x-12}\\ \Rightarrow g(x)=\frac{5(x+4)}{x(x+4)-3(x+4)}\\ \Rightarrow g(x)=\frac{5(x+4)}{(x-3)(x+4)}\\ \Rightarrow g(x)=\frac{5}{x-3}\end{array}$

Intercepts:

Let $g(x)=y.$

  $\Rightarrow y=\frac{5}{x-3}$

To find x intercepts, put y = 0,

  $\begin{array}{l}\Rightarrow y=0\\ \Rightarrow \frac{5}{x-3}=0\\ \Rightarrow 5=0\end{array}$

Here, the answer is wrong. It means that, the given function does not have x intercepts.

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{5}{x-3}\\ \Rightarrow y=\frac{5}{0-3}\\ \Rightarrow y=-\frac{5}{3}\\ \Rightarrow y=-1.666\end{array}$

So, y intercept is (0, -1.666).

Asymptotes:

Vertical asymptotes:

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

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

Horizontal asymptotes:

As the degree of numerator is less than the degree of the denominator, the horizontal asymptote is $y=0.$

Holes:

Here, the given function contains on

  

common factor in numerator and denominator i.e.,

  $(x+4).$

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

  $\begin{array}{l}\Rightarrow x+4=0\\ \Rightarrow x=-4\end{array}$

So, the hole of function is $x=-4.$