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Question 8 Given the function \( P(n)=n(n+1)(n-4) \) its \( P \)-intercept is its \( n \)-intercepts are Question Help: \( \square \) Message instructor Add Work

Ask by May Phillips. in the United States
Mar 10,2025

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Answer

The \( P \)-intercept is \((0,0)\), and the \( n \)-intercepts are \( n=-1 \), \( n=0 \), and \( n=4 \).

Solution

1. To find the \( P \)-intercept, substitute \( n=0 \) into the function: \[ P(0)=0\cdot(0+1)\cdot(0-4)=0. \] Therefore, the \( P \)-intercept is \((0,0)\). 2. The \( n \)-intercepts are the values of \( n \) that make \( P(n)=0 \). Set the function equal to zero: \[ n(n+1)(n-4)=0. \] Solve each factor equal to zero: - \( n=0 \) - \( n+1=0 \) which gives \( n=-1 \) - \( n-4=0 \) which gives \( n=4 \) Thus, the \( n \)-intercepts are \( n=-1 \), \( n=0 \), and \( n=4 \).

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The \( P \)-intercept of the function \( P(n) = n(n+1)(n-4) \) can be found by evaluating \( P(0) \). Plugging in \( n=0 \), we get \( P(0) = 0(0+1)(0-4) = 0 \). Thus, the \( P \)-intercept is at the point (0, 0). For the \( n \)-intercepts, we set \( P(n) = 0 \). This gives us the equation \( n(n+1)(n-4) = 0 \). Solving for \( n \), we find \( n = 0 \), \( n = -1 \), and \( n = 4 \). Hence, the \( n \)-intercepts are at the points (0, 0), (-1, 0), and (4, 0).

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