If P is a second degree polynomial such that... (CALCULUS 1 QUESTION)

P(2) = 5,
P'(2) = 6,
P′′(2) = 4

find the value of P(1).

1.P(1)=1/2
2.P(1)=5/2
3.P(1)=1
4.P(1)=3/2
5.P(1)=2

Since P is a second degree polynomial, we know that for some constants a, b, and c, we have:
P(x) = ax^2 + bx + c
P'(x) = 2ax + b
P"(x) = 2a

Since 4 = P"(2) = 2a, we have a = 2.
Since 6 = P'(2) = 2a(2) + b, we have b = 6 - 4a = 6 - 4(2) = -2.
Since 5 = P(2) = a(2)^2 + b(2) + c, we have c = 5 - 4a - 2b = 5 - 4(2) - 2(-2) = 5 - 8 + 4 = 1

So we have P(x) = 2x^2 - 2x + 1. Thus, we have:
P(1) = 2(1)^2 - 2(1) + 1 = 2 - 2 + 1 = 1.
The answer is 3.