86.6k views
2 votes
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
by
7.3k points

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
by
9.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories