Question

given a rational function f. the polynomial in the denominator has zeroes at x = 1 and at x = 2. the polynomial in the numerator has a zero at x = 1 but not at x = 2. the multiplicity of the zero x = 1 is greater in the denominator than in the numerator. use the dropdown menus to analyze the zeroes in the denominator and determine if they correspond to holes or vertical asymptotes. answer attempt 1 out of 1 at x = 1, 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 \( x = 1 \)

For a rational function, a zero in the denominator and numerator at the same \( x \)-value creates a hole if the multiplicity in the denominator is greater than or equal to that in the numerator. Here, denominator's zero at \( x = 1 \) has greater multiplicity than numerator's zero at \( x = 1 \), so it's a hole. Thus, at \( x = 1 \), the zero is in the denominator and also in the numerator (since numerator has a zero at \( x = 1 \), but multiplicity is less). Wait, no—wait: the problem says "the polynomial in the numerator has a zero at \( x = 1 \) but not at \( x = 2 \). The polynomial in the denominator has zeros at \( x = 1 \) and \( x = 2 \). The multiplicity of the zero \( x = 1 \) is greater in the denominator than in the numerator." So for \( x = 1 \): denominator has zero, numerator has zero (but lower multiplicity). So the zero in denominator is also in numerator? Wait, no—wait, the numerator has a zero at \( x = 1 \), so \( x = 1 \) is a zero in numerator and denominator. But since denominator's multiplicity is higher, it's a hole. For \( x = 2 \): numerator does NOT have a zero at \( x = 2 \), so denominator's zero at \( x = 2 \) is not in numerator. So the first dropdown (for \( x = 1 \)): "and also" (because numerator has zero at \( x = 1 \)), and for \( x = 2 \): "but not" (because numerator doesn't have zero at \( x = 2 \)). Wait, the question's dropdown is for the first part: "At \( x = 1 \), \( f \) has a zero in the denominator [dropdown: but not / and also] in the numerator. At \( x = 2 \), \( f \) has a zero in the denominator [dropdown: but not / and also] in the numerator." Wait, the first dropdown (for \( x = 1 \)): since numerator has a zero at \( x = 1 \), the zero in denominator is also in numerator. So "and also". Wait, no—wait, the problem says "the polynomial in the numerator has a zero at \( x = 1 \) but not at \( x = 2 \)". So numerator has zero at \( x = 1 \), denominator has zero at \( x = 1 \) (with higher multiplicity). So at \( x = 1 \), the zero is in denominator and also in numerator (because numerator has zero there). At \( x = 2 \), numerator has no zero, so zero in denominator but not in numerator. So the first dropdown (for \( x = 1 \)): "and also", second (for \( x = 2 \)): "but not". Wait, the image shows the first dropdown (for \( x = 1 \)) with options "but not" and "and also", and the blue is on "but not"—wait, maybe I misread. Wait, the problem's first part: "At \( x = 1 \), \( f \) has a zero in the denominator [dropdown] in the numerator." Since numerator has a zero at \( x = 1 \), the zero in denominator is also in numerator, so "and also". But the image's dropdown for \( x = 1 \) has "but not" highlighted? Wait, no—maybe the question is about the first blank (for \( x = 1 \)): let's re-express.

Rational function: \( f(x) = \frac{N(x)}{D(x)} \), where \( N(x) \) has zero at \( x = 1 \) (not at \( x = 2 \)), \( D(x) \) has zeros at \( x = 1 \) (multiplicity \( m \)) and \( x = 2 \) (multiplicity \( n \)), with \( m_{\text{denominator}}(x=1) > m_{\text{numerator}}(x=1) \).

- At \( x = 1 \): \( D(x) \) has zero (so \( x = 1 \) is a zero in denominator), \( N(x) \) has zero (so \( x = 1 \) is a zero in numerator). So the zero in denominator is also in numerator: "and also".
- At \( x = 2 \): \( D(x) \) has zero (so \( x = 2 \) is a zero in denominator), \( N(x) \) does NOT have zero (so \( x = 2 \) is not a zero in numerator): "but not".

Wait, but the image's first dropdown (for \( x = 1 \)) is shown with "but not" highlighted? Maybe I made a mistake. Wait, no—wait, the problem says "the polynomial in the numerator has a zero at \( x = 1 \) but not at \( x = 2 \)". So numerator has zero at \( x = 1 \), so \( x = 1 \) is a zero in numerator and denominator. So the zero in denominator is also in numerator: "and also". For \( x = 2 \), numerator has no zero, so denominator's zero is not in numerator: "but not". So the first dropdown (for \( x = 1 \)) should be "and also", second (for \( x = 2 \)) "but not". But the image's first dropdown is for \( x = 1 \): the options are "but not" and "and also", with "but not" highlighted. Wait, maybe the question is phrased as "At \( x = 1 \), \( f \) has a zero in the denominator [dropdown] in the numerator"—so if numerator has zero at \( x = 1 \), then it's "and also". But maybe the user's dropdown is for the first part ( \( x = 1 \) )—wait, the image shows:

"At \( x = 1 \), \( f \) has a zero in the denominator
[dropdown: but not / and also]
in the numerator.

At \( x = 2 \), \( f \) has a zero in the denominator
[dropdown: but not / and also]
in the numerator."

So for \( x = 1 \): numerator has zero at \( x = 1 \), so denominator's zero is also in numerator: "and also". For \( x = 2 \): numerator has no zero, so denominator's zero is not in numerator: "but not".

But the image's first dropdown (for \( x = 1 \)) is shown with "but not" highlighted—maybe that's a mistake, but according to the problem statement, numerator has zero at \( x = 1 \), so it's "and also". Wait, no—wait, the problem says "the polynomial in the numerator has a zero at \( x = 1 \) but not at \( x = 2 \)". So numerator has zero at \( x = 1 \), so \( x = 1 \) is a zero in numerator and denominator. So the zero in denominator is also in numerator: "and also". So the first dropdown (for \( x = 1 \)) should be "and also", second (for \( x = 2 \)) "but not".

But maybe the user's question is to fill the dropdowns. Let's confirm:

- For \( x = 1 \): denominator has zero, numerator has zero (since numerator's zero is at \( x = 1 \)), so "and also".
- For \( x = 2 \): denominator has zero, numerator does NOT have zero (since numerator's zero is not at \( x = 2 \)), so "but not".

So the first dropdown ( \( x = 1 \) ) is "and also", second ( \( x = 2 \) ) is "but not". But the image's first dropdown is shown with "but not" highlighted—maybe a typo, but based on the problem, it's "and also" for \( x = 1 \), "but not" for \( x = 2 \).

Wait, maybe I misread: the problem says "the polynomial in the numerator has a zero at \( x = 1 \) but not at \( x = 2 \)". So numerator has zero at \( x = 1 \), so \( x = 1 \) is a zero in numerator and denominator. So the zero in denominator is also in numerator: "and also". For \( x = 2 \), numerator has no zero, so denominator's zero is not in numerator: "but not".

Answer:

For \( x = 1 \): and also
For \( x = 2 \): but not

(Assuming the dropdowns are for \( x = 1 \) and \( x = 2 \) respectively. If the first dropdown is for \( x = 1 \), the answer is "and also"; for \( x = 2 \), "but not".)