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 $x = -a$

Explanation:

Step1: Factor denominator

Factor the quadratic in the denominator:
$x^2 + ax = x(x + a)$

Step2: Simplify the function

Rewrite the function with factored denominator, then cancel common terms (assuming $x
eq -a$):
$f(x)=\frac{(x+a)(x+b)}{x(x+a)}=\frac{x+b}{x}$

Step3: Identify discontinuities

• Removable discontinuity: Occurs at $x=-a$ (the canceled common factor makes this a hole, not an asymptote).
• Vertical asymptote: Occurs at $x=0$ (this value makes the simplified denominator 0 and is not canceled).
• $x=-b$ is not a discontinuity of the original function (it does not make the original denominator 0).

Answer:

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