Derivative of high school mathematics known function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e is even function The tangent equation passing through point a (0. - 1) and at x = 1 is 2x + Y-2 = 0. Find the analytic formula of y = f (x)

Derivative of high school mathematics known function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e is even function The tangent equation passing through point a (0. - 1) and at x = 1 is 2x + Y-2 = 0. Find the analytic formula of y = f (x)

Substituting point a (0. - 1) gives e = - 1
From the tangent equation at x = 1 is 2x + Y-2 = 0, we know that f (x) passes through the point (1,0)
If f (x) is an even function, then f (x) passes through the point (- 1,0) and the tangent equation at this point is - 2x + Y-2 = 0
Substituting into the equation, a + B + C + D + e = 0; A-B + C-D + e = 0
Then a + C = 1, B + D = 0
F '(x) = 4x ^ 3 + 3x ^ 2 + 2x ^ + D
Then f '(1) = 4A + 3B + 2C + D = - 2; f' (- 1) = - 4A + 3b-2c + D = 2
A + B = - 2, C + D = 3; - A + B = 2
To sum up: a = - 2, B = 0, C = 3, d = - 3, e = - 1
Substituting it into the equation, it is obtained that:
f(x)= -2x^4+3x^2-3x-1
Note: I haven't touched these things for a long time, so you'd better just take a close look