Question

x(x+2) / x(x-3)

would the holes be x=-2 & x=3
or just x=3?

Answer 1

hint: solve x(x-3) = 0

Answer 2

well both -2 and 3 are equals to 0

Answer 3

oh wait, the 'x' terms will cancel

Answer 4

x(x+2) / x(x-3)

would turn into

(x+2)/(x-3)

Answer 5

So this means that there is a vertical asymptote at x = 3

and there's a hole at x = 0 because this value must be excluded as well

Answer 6

Waittt, but the (x+2) won't have anything to do with it because that's used for determining the horizontal only right?

Answer 7

the x+2 in the numerator plays no role in figuring out vertical asymptotes or holes (unless it canceled with a x+2 in the denominator)

Answer 8

the leading coefficients determine the horizontal asymptote

Answer 9

thanks!

Answer 10

you're welcome