At what x value does the function given below have a hole?

f(x)=x+3/x2−9

hole at x=-3

Step-by-step explanation:

The hole is the discontinuity that exists after the fraction reduces. (Still doesn't exist for original of course)

The discontinuities for this expression is when the bottom is 0. x^2-9=0 when x=3 or x=-3 since squaring either and then subtracting 9 would lead to 0.

So anyways we have (x+3)/(x^2-9)

= (x+3)/((x-3)(x+3))

Now this equals 1/(x-3) with a hole at x=-3 since the x+3 factor was "cancelled" from the denominator.