What is the vertical asymptote and the hole in this equation?

Question

4x/x³-9x

Answer 1

The first thing we would do with this function is factor the denominator and we are left with:

4x / x(x² - 9)

The x in the numerator cancels with the x in the denominator which creates a hole at x = 0 and we are left with:

4 / x² - 9

To find the vertical asymptote, we set the denominator equal to zero and solve for x:

x² - 9 = 0

x² = 9

x = 3 and x = -3

So there is a vertical asymptote at x = 3 and x = -3