Which of these functions represents a vertical asymptote at x=−3x = -3x=−3?
f(x)=x+3x2+9f(x) = \frac{x+3}{x^2+9}f(x)=x2+9x+3
f(x)=1x+3f(x) = \frac{1}{x+3}f(x)=x+31
f(x)=1x−3f(x) = \frac{1}{x-3}f(x)=x−31
f(x)=x+3f(x) = x+3f(x)=x+3