Question

at x = 8, f has a zero in the denominator and also in the numerator. there is a hole at x = 8 because the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator. at x = -2, f has a zero in the denominator but not in the numerator.

Explanation:

Step1: Recall Hole Condition

A hole in a rational function occurs when a root is present in both the numerator and the denominator, and the multiplicity of the root in the numerator is greater than or equal to that in the denominator. For \( x = 8 \), \( f \) has a zero in the denominator and also in the numerator. So, we check the condition for a hole: multiplicity of the root in the numerator (let's say \( m \)) and in the denominator (let's say \( n \)). A hole exists when \( m \geq n \). So the first blank (after "also"): when a root is in both numerator and denominator with numerator's multiplicity \( \geq \) denominator's, there is a hole. So the word here is "and also" (wait, no, the first part: "At \( x = 8 \), \( f \) has a zero in the denominator and also \(\underline{}\) in the numerator. There is a hole at \( x = 8 \) because the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator." Wait, the first blank: "has a zero in the denominator and also \(\underline{}\) in the numerator" – so it should be "has a zero in the denominator and also \(\boldsymbol{has\ a\ zero}\) in the numerator"? Wait, no, the options? Wait, the image has a dropdown? Wait, the user's image: "At \( x = 8 \), \( f \) has a zero in the denominator and also \(\underline{}\) in the numerator. There is a hole at \( x = 8 \) because the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator." Wait, maybe the first blank is "has a zero" (but the options? Wait, the second part: "At \( x = -2 \), \( f \) has a zero in the denominator \(\underline{}\) in the numerator." The options are "but not" or "and also". So for \( x = -2 \), if there's a zero in the denominator but not in the numerator, that's a vertical asymptote. If it's also in the numerator, maybe a hole (but multiplicity? Wait, let's analyze:

For a rational function \( f(x)=\frac{N(x)}{D(x)} \):

- **Hole**: When a root \( c \) is in both \( N(x) \) and \( D(x) \), and \( \text{multiplicity of } c \text{ in } N(x) \geq \text{multiplicity in } D(x) \) (or vice versa? Wait, actually, if a root \( c \) is common to both numerator and denominator, we can cancel the common factor. If the multiplicity in numerator is equal to that in denominator, it's a hole. If numerator's multiplicity is less, then after canceling, denominator still has the root, so vertical asymptote? Wait, no: Let's take \( f(x)=\frac{(x - 8)(x + 2)}{(x - 8)(x + 2)^2} \). Here, \( x = 8 \): numerator and denominator have multiplicity 1. So cancel \( (x - 8) \), we get \( \frac{1}{(x + 2)^2} \), so hole at \( x = 8 \). For \( x = -2 \): numerator has multiplicity 1, denominator has multiplicity 2. After canceling, denominator still has \( (x + 2) \), so vertical asymptote at \( x = -2 \). So for \( x = 8 \): zero in denominator **and also** zero in numerator (so common root), hence hole. For \( x = -2 \): zero in denominator **but not** zero in numerator (after canceling, numerator has no \( (x + 2) \) left), so vertical asymptote.

So first blank (at \( x = 8 \)): "and also" (has zero in denominator and also in numerator). Second blank (at \( x = -2 \)): "but not" (has zero in denominator but not in numerator).

Wait, the first part: "At \( x = 8 \), \( f \) has a zero in the denominator and also \(\underline{}\) in the numerator. There is a hole at \( x = 8 \) because the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator." So the first blank: "has a zero" (but the options? Wait, the user's image: the first dropdown is after "and also" – maybe the first sentence is "At \( x = 8 \), \( f \) has a zero in the denominator and also \(\boldsymbol{has\ a\ zero}\) in the numerator..." Then the second part: "At \( x = -2 \), \( f \) has a zero in the denominator \(\underline{}\) in the numerator." The options are "but not" or "and also". So:

1. At \( x = 8 \): zero in denominator **and also** in numerator (so hole, because multiplicity in numerator \( \geq \) denominator? Wait, no, if multiplicity in numerator is equal to denominator, it's a hole. If numerator's multiplicity is less, then after canceling, denominator still has the root, so vertical asymptote. Wait, the problem says "the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator" for the hole. So that means numerator has at least as many of that root as denominator. So for \( x = 8 \), numerator has the root (so "and also" in numerator), and multiplicity in numerator \( \geq \) denominator, hence hole. For \( x = -2 \), denominator has the root, but numerator does not (so "but not" in numerator), so vertical asymptote.

So filling the blanks:

- First blank (after "and also"): Wait, the first sentence: "At \( x = 8 \), \( f \) has a zero in the denominator and also \(\underline{}\) in the numerator. There is a hole at \( x = 8 \) because the multiplicity of the root in the numerator is greater than or equal to the multiplicity of the root in the denominator." So the blank is "has a zero" (but the options? Wait, the user's image shows a dropdown for the first part? Maybe the first part is "in the numerator" – so "and also in the numerator". Then the second part: "At \( x = -2 \), \( f \) has a zero in the denominator \(\underline{}\) in the numerator." The options are "but not" or "and also". So:

- At \( x = 8 \): zero in denominator **and also** in numerator (so common root, hole because numerator's multiplicity \( \geq \) denominator's).

- At \( x = -2 \): zero in denominator **but not** in numerator (so vertical asymptote, since denominator has the root but numerator doesn't, so can't cancel, hence vertical asymptote).

So the answers are:

First blank (at \( x = 8 \)): "and also" (in the numerator)

Second blank (at \( x = -2 \)): "but not" (in the numerator)

Answer:

At \( x = 8 \): "and also" (in the numerator)
At \( x = -2 \): "but not" (in the numerator)