Chapter 8, Problem 16MCQ

Solution

Toidentify:The any hole and vertical or horizontal asymptotes for the graph of rational function is $y=\frac{5x-2}{x+2}$

The vertical asymptotes is $x=\frac{2}{5}$

The horizontal asymptote is $y=5$

Given information:

The given rational function is $y=\frac{5x-2}{x+2}$ .

Explanation:

Consider the given rational function.

  $y=\frac{5x-2}{x+2}$

Determine the zero of the denominator.

  $\begin{array}{c}x+2=0\\ x=-2\end{array}$

Therefore zero of a denominator is $x=-2$

Determine the zeros of numerator

Numerator is zero for.

  $\begin{array}{c}5x-2=0\\ x=\frac{2}{5}\end{array}$

Therefore numerator is zero at $x=\frac{2}{5}$

Determine the vertical asymptotes and holes.

Clearly zero of denominator is not a zero of numerator.

Hence the vertical asymptotes are $x=-2$

Thereis no holes

Determine the horizontal asymptotes

Here degree of the numerator is equal to the degree of the denominator.

  $y=\frac{\text{leading}\text{coefficient}\text{of}\text{numerator}}{\text{leading}\text{coefficient}\text{of}\text{denminator}}$

  $\begin{array}{c}y=\frac{\text{leading}\text{coefficient}\text{of}\text{numerator}}{\text{leading}\text{coefficient}\text{of}\text{denminator}}\\ y=\frac{5}{1}\\ y=5\end{array}$

Hence the horizontal asymptote is $y=5$