Discontinuity for functions

Question

Describe the discontinuity for the function 

Options:

1. There is a hole at x = -9.
2. There is a vertical asymptote at x = 3.
3. There is a removable discontinuity at x = 3.
4. There is no discontinuity at x = 3.

Answer 1

If you graph this function, you can easily see that there is a vertical asymptote at x = 3 but how would you know if you didn't graph it?

Discontinuities in rational functions (when the numerator and denominator are polynomials) are found when the denominator is zero (because dividing by zero is undefined). So you would find the discontinuity by making the denominator equal to zero and x - 3 = 0 when x = 3 so this automatically eliminates answers 1 and 4.. To have a removable discontinuity (or hole) you would be able to factor and cancel to eliminate the issue. If x² + 9 could be factored into (x - 3) times something then you could cancel an "x - 3" from the top and bottom and eliminate the discontinuity but this is NOT the case. x² + 9 does not factor. So the answer is not 3.