Question

which of the following statements are true of this rational function? check all of the boxes that apply. $f(x)=\frac{(x + a)(x + b)}{x^{2}+ax}$
there is a removable discontinuity at $x=-b$.
there is a vertical asymptote at $x = 0$.
there are no removable discontinuities.
there is a vertical asymptote at $x=-a$.
there is a removable discontinuity at $x=-a$
done

Explanation:

Step1: Factor the denominator

Factor $x^{2}+ax$ as $x(x + a)$. So $f(x)=\frac{(x + a)(x + b)}{x(x + a)}$.

Step2: Analyze removable discontinuities

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x + a)$ cancels out (assuming $x
eq - a$). So there is a removable discontinuity at $x=-a$.

Step3: Analyze vertical asymptotes

Vertical asymptotes occur at the values of $x$ that make the denominator zero after canceling common factors. The denominator $x(x + a)$ after canceling $(x + a)$ is $x$. When $x = 0$, the function is undefined, so there is a vertical asymptote at $x = 0$.

Answer:

There is a vertical asymptote at $x = 0$.
There is a removable discontinuity at $x=-a$.