Question

which of the following statements are true of this rational function? choose two correct answers.
$f(x) = \frac{(x + a)(x + b)}{x^2 + ax}$

• there is a removable discontinuity at $x = -b$.
• there are no removable discontinuities.
• there is a removable discontinuity at $x = -a$.
• there is a vertical asymptote at $x = -a$.
• there is a vertical asymptote at $x = 0$.

Explanation:

Step1: Simplify the Rational Function

First, factor the denominator: \(x^2 + ax = x(x + a)\). So the function becomes \(f(x)=\frac{(x + a)(x + b)}{x(x + a)}\). We can cancel the common factor \((x + a)\) (for \(x
eq -a\)), so the simplified function is \(f(x)=\frac{x + b}{x}\) (with a restriction \(x
eq -a\) and \(x
eq0\) from the original denominator).

Step2: Analyze Discontinuities

• Removable Discontinuity: A removable discontinuity occurs where a factor cancels out. Here, \((x + a)\) cancels, so at \(x=-a\), there is a removable discontinuity (since the factor cancels, it's a hole, not a vertical asymptote).
• Vertical Asymptote: After canceling, the denominator is \(x\), so vertical asymptote is at \(x = 0\) (since \(x = 0\) makes the simplified denominator zero and wasn't canceled).
• For \(x=-b\): There's no factor canceling or making the original/ simplified denominator zero (unless \(b=a\), but generally, no), so no removable discontinuity at \(x=-b\). And "There are no removable discontinuities" is false because we have a removable discontinuity at \(x=-a\).

Answer:

• There is a removable discontinuity at \(x = -a\).
• There is a vertical asymptote at \(x = 0\).