Construct a Rational Function that has a vertical asymptote at X = 6, and a removable
discontinuity at x = -2. Write your function, and explain in detail how you determined your answer.

(x+2)/(x-6)(x+2)

Step-by-step explanation:

We know that vertical asymptotes are found in the denominator, so we can start with 1/(x-6)

x-6 will equal zero at 6, so that is where the VA will be.

For the removable discontinuity (hole), it needs to cancel out in the numerator and denominator. (x+2)/(x-6)(x+2)

when x equals -2, there is an undefined y value.