Question

given a rational function f. 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. 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. (dropdown options: but not, and also)

Explanation:

Step1: Analyze \( x = 8 \)

For a rational function, if a zero \( x = a \) has the same multiplicity in the numerator and denominator, it means the factor \((x - a)\) cancels out, resulting in a hole at \( x = a \). Since \( x = 8 \) has the same multiplicity in numerator and denominator, \( f \) has a zero in the denominator (from the denominator's zero) and also in the numerator (so the factor cancels, leading to a hole, not a vertical asymptote).

Step2: Analyze \( x = -2 \)

The problem states the multiplicity of \( x = -2 \) is the same in numerator and denominator? Wait, no—wait, the denominator has a zero at \( x = -2 \), and the numerator: the problem says "the multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The zero \( x = -2 \) is the same in both the numerator and the denominator?" Wait, re-reading: "The polynomial in the denominator has zeros at \( x = 8 \) and \( x = -2 \). The multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The zero \( x = -2 \) is the same in both the numerator and the denominator." Wait, no, maybe the first part: at \( x = 8 \), since multiplicity is same in numerator and denominator, so denominator has zero (so \( f \) has a zero in denominator) and also in numerator (so the factor \((x - 8)\) is in both, so cancels, hole). At \( x = -2 \): the problem says "the zero \( x = -2 \) is the same in both the numerator and the denominator"—wait, maybe the dropdown is for \( x = -2 \): "At \( x = -2 \), \( f \) has a zero in the denominator but not in the numerator" or "and also in the numerator". Wait, no—wait, the first blank: "At \( x = 8 \), \( f \) has a zero in the denominator \(\boldsymbol{\text{and also}}\) in the numerator" (because multiplicity same, so numerator has the same zero, so factor cancels, hole). For \( x = -2 \): wait, no, maybe I misread. Wait, the problem says "the multiplicity of the zero \( x = 8 \) is the same in both the numerator and the denominator. The zero \( x = -2 \) is the same in both the numerator and the denominator." Wait, no—maybe the first part: \( x = 8 \): denominator has zero, numerator has same multiplicity zero, so "and also" in numerator. For \( x = -2 \): if the zero is only in the denominator (but the problem says "the zero \( x = -2 \) is the same in both"? Wait, no, the original problem's dropdown: "At \( x = 8 \), \( f \) has a zero in the denominator \(\boldsymbol{\text{and also}}\) in the numerator" (because the factor is in both, so cancels, hole). For \( x = -2 \): wait, maybe the second part, but the first dropdown is for \( x = 8 \): the options are "but not" or "and also" in the numerator. Since multiplicity is same, numerator has the zero, so "and also".

Answer:

For \( x = 8 \): and also

(Assuming the dropdown for \( x = 8 \) is to choose between "but not" and "and also" in the numerator. Since multiplicity is same, numerator has the zero, so "and also".)