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 = -a ). 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 = -b ).

Explanation:

First, let's simplify the rational function \( f(x)=\frac{(x + a)(x + b)}{x^{2}+ax} \). Notice that the denominator can be factored: \( x^{2}+ax=x(x + a) \). So we can rewrite the function as:

Step 1: Factor the denominator

The denominator \( x^{2}+ax \) factors to \( x(x + a) \), so \( f(x)=\frac{(x + a)(x + b)}{x(x + a)} \).

Step 2: Simplify the function (considering domain restrictions)

We can cancel out the common factor of \( (x + a) \) from the numerator and the denominator, but we have to note that \( x
eq - a \) (because that would make the original denominator zero before cancellation) and \( x
eq0 \) (also makes the original denominator zero). After canceling \( (x + a) \) (for \( x
eq - a \)), the function simplifies to \( f(x)=\frac{x + b}{x} \), which is equivalent to \( f(x)=1+\frac{b}{x} \) (for \( x
eq - a,x
eq0 \)).

Now let's analyze each statement:

1. "There is a removable discontinuity at \( x=-a \)."

A removable discontinuity occurs when a factor cancels out from the numerator and the denominator. We canceled \( (x + a) \), so at \( x=-a \), the original function is undefined (since plugging \( x = - a \) into the original denominator \( x(x + a) \) gives \( (-a)(0)=0 \)), but after canceling, we can define the function at \( x=-a \) (by plugging into the simplified function \( \frac{x + b}{x} \), we get \( \frac{-a + b}{-a} \)). So this is a removable discontinuity. This statement is true.

1. "There are no removable discontinuities."

From the first analysis, we saw there is a removable discontinuity at \( x=-a \), so this is false.

1. "There is a vertical asymptote at \( x = 0 \)."

For the simplified function \( f(x)=\frac{x + b}{x}=1+\frac{b}{x} \), when \( x\) approaches \( 0 \), \( \frac{b}{x} \) approaches \( \pm\infty \) (depending on the sign of \( b \) and the side of approach). Also, the original function has \( x = 0 \) in the denominator and we can't cancel \( x \) (since the numerator after factoring was \( (x + a)(x + b) \), and \( x \) isn't a factor of the numerator unless \( b = 0 \), but generally, \( x \) isn't canceled). So \( x = 0 \) is a vertical asymptote. This statement is true.

1. "There is a vertical asymptote at \( x=-a \)."

We canceled \( (x + a) \), so \( x=-a \) is a removable discontinuity, not a vertical asymptote (vertical asymptotes occur when a factor remains in the denominator after canceling common factors). So this is false.

1. "There is a removable discontinuity at \( x=-b \)."

Let's check the original function at \( x=-b \). The numerator becomes \( (-b + a)(-b + b)=(-b + a)(0)=0 \), and the denominator becomes \( (-b)^{2}+a(-b)=b^{2}-ab \). For there to be a removable discontinuity, the denominator should also be zero at \( x=-b \) (so that we can cancel a common factor). Let's see if \( b^{2}-ab = 0 \) when \( x=-b \). \( b^{2}-ab=b(b - a) \). This is zero only if \( b = 0 \) or \( b = a \), which isn't a general case. So in general, at \( x=-b \), the denominator isn't zero (unless specific conditions on \( a \) and \( b \) hold), so there's no discontinuity (removable or otherwise) here. So this statement is false.

Answer:

The two correct statements are:

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