Question

given a rational function f. the polynomial in the denominator has zeroes at x = 8 and at x = -2. the polynomial in the numerator also has zeroes at x = 8 and x = -2. the multiplicity of the zero x = 8 is the same in both the numerator and the denominator. the multiplicity of the zero x = -2 is the same in both the numerator and the denominator. use the dropdown menus to analyze the zeroes in the denominator and determine if they correspond to holes or vertical asymptotes. answer: at x = 8, f has a zero in the denominator → in the numerator. at x = -2, f has a zero in the denominator → in the numerator.

Explanation:

Step1: Recall Hole and Asymptote Rules

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), a zero in both numerator and denominator (same multiplicity) creates a hole. A zero only in the denominator creates a vertical asymptote. A zero only in the numerator is an \( x \)-intercept.

Step2: Analyze \( x = 8 \)

At \( x = 8 \), \( f \) has a zero in the denominator and the same multiplicity in the numerator (given "the multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator"). So this is a hole. Thus, "At \( x = 8 \), \( f \) has a zero in the denominator" corresponds to "in the numerator" (since same multiplicity, so zero in numerator too, leading to a hole).

Step3: Analyze \( x = -2 \)

At \( x = -2 \), the multiplicity in the numerator and denominator is the same? Wait, the problem says "the multiplicity of the zero \( x = -2 \) is the same in both the numerator and the denominator"? Wait, no, re-reading: "The polynomial in the denominator has zeroes at \( x = 8 \) and \( x = -2 \). The multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The multiplicity of the zero \( x = -2 \) is the same in both the numerator and the denominator?" Wait, the original says: "The polynomial in the denominator has zeroes at \( x = 8 \) and at \( x = -2 \). The multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The multiplicity of the zero \( x = -2 \) is the same in both the numerator and the denominator?" Wait, the user's text: "the multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The multiplicity of the zero \( x = -2 \) is the same in both the numerator and the denominator." Wait, then for \( x = -2 \), since zero in denominator and same multiplicity in numerator, so zero in numerator too? But the dropdown for \( x = -2 \): "At \( x = -2 \), \( f \) has a zero in the denominator" and the option is "in the numerator" (since same multiplicity, so zero in numerator, leading to a hole? Wait, no—wait, maybe I misread. Wait, the problem is to match:

For \( x = 8 \): "At \( x = 8 \), \( f \) has a zero in the denominator" → match to "in the numerator" (because same multiplicity, so zero in numerator, so hole).

For \( x = -2 \): "At \( x = -2 \), \( f \) has a zero in the denominator" → match to "in the numerator" (same multiplicity, so zero in numerator, hole) or is there a mistake? Wait, maybe the problem is that for \( x = 8 \), zero in denominator and same in numerator (so hole), so "At \( x = 8 \), \( f \) has a zero in the denominator" → "in the numerator" (because numerator also has zero, same multiplicity). For \( x = -2 \), same: "At \( x = -2 \), \( f \) has a zero in the denominator" → "in the numerator" (since same multiplicity, so numerator has zero too, hole). But let's check the dropdowns. The first dropdown for \( x = 8 \): "At \( x = 8 \), \( f \) has a zero in the denominator" → options: "in the numerator" (because same multiplicity, so numerator has zero, so hole). The second dropdown for \( x = -2 \): "At \( x = -2 \), \( f \) has a zero in the denominator" → "in the numerator" (same reason).

Wait, but maybe the problem is that for \( x = 8 \), since same multiplicity, zero in numerator and denominator → hole. For \( x = -2 \), same → hole. So the matches are:

At \( x = 8 \), \( f \) has a zero in the denominator → in the numerator (because numerator has zero, same multiplicity).

At \( x = -2 \), \( f \) has a zero in the denominator → in the numerator (because numerator has zero, same multiplicity).

But let's confirm the rule: A hole occurs when a factor \( (x - a) \) appears in both numerator and denominator with the same multiplicity. A vertical asymptote occurs when \( (x - a) \) appears in the denominator but not the numerator (or higher multiplicity in denominator). An \( x \)-intercept occurs when \( (x - a) \) appears in the numerator but not the denominator (or higher multiplicity in numerator).

So for \( x = 8 \): denominator has \( (x - 8)^m \), numerator has \( (x - 8)^m \) (same multiplicity \( m \)) → hole. So "zero in denominator" and "zero in numerator" (same multiplicity).

For \( x = -2 \): denominator has \( (x + 2)^n \), numerator has \( (x + 2)^n \) (same multiplicity \( n \)) → hole. So "zero in denominator" and "zero in numerator" (same multiplicity).

Thus, the matches are:

At \( x = 8 \), \( f \) has a zero in the denominator → in the numerator.

At \( x = -2 \), \( f \) has a zero in the denominator → in the numerator.

Answer:

For \( x = 8 \): in the numerator
For \( x = -2 \): in the numerator