Question

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

Explanation:

Step1: Recall function - discontinuity concepts

A removable discontinuity occurs when the numerator and denominator have a common factor. A vertical asymptote occurs when the denominator is zero and the numerator is non - zero at that point.

Step2: Analyze the rational function \(f(x)=\frac{x^{2}+ax}{(x + a)(x + b)}\)

Factor the numerator: \(x^{2}+ax=x(x + a)\). So, \(f(x)=\frac{x(x + a)}{(x + a)(x + b)}\), \(x
eq - a,-b\).

Step3: Check for removable discontinuities

When \(x=-a\), as long as \(x
eq - b\), we can cancel out the \((x + a)\) terms. So, there is a removable discontinuity at \(x=-a\).

Step4: Check for vertical asymptotes

The denominator is zero when \(x=-a\) or \(x=-b\). After canceling out the \((x + a)\) terms (for \(x
eq - a\)), the function has a vertical asymptote at \(x=-b\) since substituting \(x =-b\) into the simplified function \(\frac{x}{x + b}\) makes the denominator zero and the numerator non - zero (\(x
eq0\) when \(x=-b\)). And \(x = 0\) is not a point of vertical asymptote.

Answer:

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