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Given the function P(r) = 3r⁴ + 21r³ +30r², which of the following points is a P-intercept of the function?

A) (0, 0)
B) (-2, 0)
C) (-5, 0)
D) None of the given points is a P-intercept.

User Kal Zekdor
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1 Answer

4 votes

Final answer:

The P-intercepts of the function P(r) = 3r⁴ + 21r³ +30r² are (0, 0), (-2, 0), and (-5, 0).

Step-by-step explanation:

The P-intercepts of a function are the points at which the function crosses or intersects the x-axis. To find the P-intercepts, we set P(r) equal to zero and solve for r.

Given the function P(r) = 3r⁴ + 21r³ +30r², we can set it equal to zero: 3r⁴ + 21r³ +30r² = 0.

Now we can factor out the common term r² to simplify the equation: r²(3r² + 21r + 30) = 0.

The equation is equal to zero when either r² = 0 or 3r² + 21r + 30 = 0.

From r² = 0, we find r = 0. Therefore, the point (0, 0) is a P-intercept of the function.

From 3r² + 21r + 30 = 0, we find r = -2 and r = -5. Therefore, the points (-2, 0) and (-5, 0) are also P-intercepts of the function.

Therefore, the answer is: A) (0, 0), B) (-2, 0), and C) (-5, 0) are all P-intercepts of the function.

User Ashok Rathod
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