Question

h(x)= (x^2 -36)/x +6 explain why the following function is not continuous at x=-6

Answer 1

You can solve this problem through factoring.

First, you have the equation,

$h(x) = \frac{x^2-36}{x-6}$

Then, you can factor the numerator.

$h(x) = \frac{(x+6)(x-6)}{x-6}$

You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.

You are left with $h(x) = x+6$

The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.