Chapter 1, Problem 21RE

Solution

(a).

Find vertical asymptotes of the graph of the function. Be sure to state your answer as equation of line.

  $y=\frac{5}{x^2-5x}$

  $x=0,5$

Given:

The given function, $y=\frac{5}{x^2-5x}$

Concept Used:

A vertical asymptote is a vertical line that has the property that either: 1. 2. That is, as approaches from either the positive or negative side, the function approaches infinity. Vertical asymptotes occur at the values where a rational function has a denominator of 0.

Calculation:

The given function, $y=\frac{5}{x^2-5x}$

  $\begin{array}{c}y=\frac{5}{x^2-5x}\\ x^2-5x=0\\ x(x-5)=0\\ x=0\\ x-5=0\\ x=5\end{array}$

The vertical asymptotes of the function $y=\frac{5}{x^2-5x}$ is $x=0,5$ .

(b).

Find horizontal asymptotes of the graph of the function. Be sure to state your answer as equation of line.

  $y=\frac{5}{x^2-5x}$

  $y=0$

Given:

The given function, $y=\frac{5}{x^2-5x}$

Concept Used:

• A horizontal asymptote is a horizontal line that tells you how the function will behave at the very edges of a graph.
• If degree of numerator N < degree of denominator D, then the horizontal asymptote is y = 0.
• If degree of numerator N =degree of denominator D, then the horizontal asymptote is y = ratio of the leading coefficients.
• If degree of numerator N > degree of denominator D, then there is no horizontal asymptote.

Calculation:

The given function, $y=\frac{5}{x^2-5x}$

Here degree of numerator (N) = 0

And If degree of denominator (D) = 2

If degree of numerator N < degree of denominator D, then the horizontal asymptote is y = 0.

So N < D

Then the horizontal asymptote is y = 0.