Where is the removable discontinuity of f (x) = x+5/x^2+3x-10

x = - 5

x = - 2

x = 2

x = 5

Rational functions are discontinuous where the denominator is equal to zero. If f (x) = (x+5) / (x^2 + 3x - 10), we need to find where x^2 + 3x - 10 = 0.

Lets factor the denominator: (x+5) (x-2). So the denominator is 0 in x=-5 and x=2. Nominator is also 0 in x=-5. So we have a hole in x=-5 because the numerator and the denominator are both zero at x=-5 and a vertical asymptote in x=2. The hole is removable discontinuity so the correct answer is x=-5.