Chapter 11.7, Problem 4LC

To determine

Find the excluded value and the vertical and horizontal asymptote of the given function.

Answer to Problem 4LC

Excluded value $x=5$

Vertical asymptote $x=5$ , Horizontal asymptote $y=1$

Explanation of Solution

Given function:

  $y=\frac{7}{x-5}+1$

Concept used:

• Excluded value are the values that makes the denominator of a fraction equal to zero and are not part of the domain.
• The vertical asymptotes of a function occur at the points where the function is undefined. Vertical asymptote is the vertical line that the graph of rational function follows till infinity but never intersects.

The horizontal asymptote of a rational function is given by:

• If the degree of the polynomial in numerator is less than the degree of polynomial in denominator, then the horizontal asymptote is the horizontal axis or y=0.
• If the degree of the polynomial in numerator is equal to the degree of polynomial in denominator, then the horizontal asymptote is the ratio of the leading coefficients of numerator and denominator.
• If the degree of the polynomial in numerator is greater than the degree of polynomial in denominator, then there is no horizontal asymptote. In this case there occurs a slant asymptote.
• We know that the vertical asymptotes occur at the values of x for which the denominator is zero.

Calculation:

Excluded value is

  $\begin{array}{c}y=\frac{7}{x-5}+1\\ x-5=0\\ x=5\end{array}$

So the excluded value $x=5$

Vertical asymptote at $x=5$

Horizontal asymptote at $y=1$