Chapter 1, Problem 23RE

Solution

(a).

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

  $y=\frac{7x}{\sqrt{x^2+10}}$

None.

Given:

The given function, $y=\frac{7x}{\sqrt{x^2+10}}$

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, $y=\frac{7x}{\sqrt{x^2+10}}$

  $\begin{array}{l}y=\frac{7x}{\sqrt{x^2+10}}\\ \sqrt{x^2+10}=0\\ onsquaringbothside\\ x^2+10=0\\ x^2=-10\\ x=\sqrt{-10}\\ x=i\sqrt{10}\end{array}$

The vertical asymptotes of the function $y=\frac{7x}{\sqrt{x^2+10}}$ is None.

(b).

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

  $y=\frac{7x}{\sqrt{x^2+10}}$

  $y=+7,-7$

Given:

The given function, $y=\frac{7x}{\sqrt{x^2+10}}$

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, $y=\frac{7x}{\sqrt{x^2+10}}$

Here degree of numerator (N) = 1

And If degree of denominator (D) = 1

If degree of numerator N =degree of denominator D, then the horizontal asymptote is y = ratio of the leading coefficients.

So N = D

Horizontal asymptote is −

  $\begin{array}{l}y=\frac{7x}{\sqrt{x^2}}\\ y=\frac{7x}{\pm x}\\ y=\frac{7x}{x},\frac{7x}{-x}\\ y=7,-7\end{array}$

Thus the horizontal asymptote of the function $y=\frac{7x}{\sqrt{x^2+10}}$ is $y=+7,-7$ .