Question

1. state the points of discontinuity and classify the type of each2. state the vertical and horizontal asymptotesSHOW ALL WORK

Answer 1

Remember that the points of discontinuity are found where the denominator and numerators are zero is zero.

$x^2-9\ne0$

Let's solve it x

$\begin{gathered} x^2\ne9 \\ x\ne\pm\sqrt[]{9} \\ x\ne\pm3 \end{gathered}$

Let's evaluate the function when x = 3.

$f(3)=\frac{(3)^2+2(3)-15}{(3)^2-9}=\frac{9+6-15}{9-9}=\frac{15-15}{0}=\frac{0}{0}$

This means x = 3 is a discontinuity value.

Let's evaluate the function when x = -3.

$f(-3)=\frac{(-3)^2+2(-3)-15}{(-3)^2-9}=\frac{9-6-15}{9-9}=\frac{3-15}{0}=\frac{-12}{0}=-\infty$

This means x = -3 is a discontinuity value.

However, if we factor the numerator and denominator, we get the following

$f(x)=\frac{(x+5)(x-3)}{(x+3)(x-3)}=\frac{x+5}{x+3}$

After rewriting the function, the only value that makes the function determined is x = -3.

The first discontinuity is a Removable Discontinuity because we were able to remove it using the factorization process.

The second discontinuity is an essential discontinuity because it can't be avoided.

Hence, the point of discontinuity is all the set of points on the line x = -3.

The vertical asymptote is x = -3.

The horizontal asymptote y = 1.