Chapter 2.7, Problem 32E

To determine

To sketch the function $f(x)=\frac{x^2-16}{x-4}$ , also find their intercepts, asymptotes and holes.

Answer to Problem 32E

No asymptotes

Explanation of Solution

Given:

Function: $f(x)=\frac{x^2-16}{x-4}$

Calculation for graph:

Consider $f(x)=\frac{x^2-16}{x-4}$

Values of x	Values of f (x)
0	4
1	5
-1	3
2	6
-2	2

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

Graph:

  

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=\frac{x^2-16}{x-4}\\ \Rightarrow f(x)=\frac{x^2-4^2}{x-4}\\ \Rightarrow f(x)=\frac{(x+4)(x-4)}{x-4}\\ \Rightarrow f(x)=\frac{x+4}{1}\end{array}$

Intercepts:

Let $f(x)=y.$

  $\Rightarrow y=\frac{x+4}{1}$

To find x intercepts, put y = 0,

  $\begin{array}{l}\Rightarrow y=0\\ \Rightarrow \frac{x+4}{1}=0\\ \Rightarrow x+4=0\\ \Rightarrow x=-4\end{array}$

So, x intercept is (-4, 0).

To find y intercepts, put x = 0,

  $\begin{array}{l}\Rightarrow y=\frac{0+4}{1}\\ \Rightarrow y=\frac{4}{1}\\ \Rightarrow y=4\end{array}$

So, y intercept is (0, 4).

Asymptotes:

Vertical asymptotes:

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

  $\Rightarrow 1=0$

Here, the answer is wrong. It means that, the given function does not have vertical asymptotes.

Horizontal asymptotes:

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

Holes:

Here, the given function contains one 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 $x=4.$