Hi, I was working on my review sheet for my midterm on friday, and I came across a problem that my professor never explained how to solve. Anyone know how to do this? And please give detailed steps as to how you solved it, not just the answer.
Find a second-degree polynomial P such that P(3) = 10, P'(3) = 8, and P''(3) = 4.
Find a second-degree polynomial P such that P(3) = 10, P'(3) = 8, and P''(3) = 4.
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Answer: P(x) = 2x² - 4x + 4
You are told you need to find a 2nd degree polynomial. That will be in the form:
P(x) = ax² + bx + c
To find a, b, and c, you have to use the information you are given. So, you will have to find the first and second derivatives of P(x).
P'(x) = 2ax + b
P"(x) = 2a
It is easiest to now work "backwards", starting with the second derivative to solve for "a".
P"(x) = 2a
P"(3) = 2a ... and ... P"(3) = 4
2a = 4
a = 2
sub your value of "a" in P'(x) and solve for "b".
P'(x) = 2ax + b
P'(3) = 2(2)(3) + b = 12 + b ... and ... P'(3) = 8
12 + b = 8
b = -4
sub "a" and "b" in P(x) and solve for "c".
P(x) = 2x² - 4x + 4
P(3) = (2)(3)² + (-4)(3) + c = 6 + c ... and ... P(3) = 10
6 + c = 10
c = 4
So your polynomial is:
P(x) = 2x² - 4x + 4
Hope that helps!
You are told you need to find a 2nd degree polynomial. That will be in the form:
P(x) = ax² + bx + c
To find a, b, and c, you have to use the information you are given. So, you will have to find the first and second derivatives of P(x).
P'(x) = 2ax + b
P"(x) = 2a
It is easiest to now work "backwards", starting with the second derivative to solve for "a".
P"(x) = 2a
P"(3) = 2a ... and ... P"(3) = 4
2a = 4
a = 2
sub your value of "a" in P'(x) and solve for "b".
P'(x) = 2ax + b
P'(3) = 2(2)(3) + b = 12 + b ... and ... P'(3) = 8
12 + b = 8
b = -4
sub "a" and "b" in P(x) and solve for "c".
P(x) = 2x² - 4x + 4
P(3) = (2)(3)² + (-4)(3) + c = 6 + c ... and ... P(3) = 10
6 + c = 10
c = 4
So your polynomial is:
P(x) = 2x² - 4x + 4
Hope that helps!