Antiderivative problem

The lines x + y = 0 and x + y = 4/3 are tangent to the graph of y = f(x), where F is an antiderivative of -x^2. Find F

please help me figure it out

f ' (x) = -x^2

integrating both sides

F(x) = -(1/3)x^3 + C -------------eqn(1)

eqn of tangent is given by

y = F(a) + f ' (a)(x - a) -----------------eqn (2)

where [a, f(a))]are x and y-coordinates of point of tangency and f '(a) is slope of tangent

since x + y = 0 and x + y = 4/3 are tangents==> y = -x and y = -x + 4/3.

==> slope of tangent, f ' (a) = -1

since f'(x) = -x^2 ==>f ' (a) = -a^2 ==> -a^2 = -1 and a = -1 and 1

From eqn (1), F(a) = (1/3)a^3 + C ==> F(a) = -(1/3) + C and (1/3) + C

from eqn (2)

y = -(1/3) + C - 1(x - 1) and

y = -(1/3)(-1) + C - 1(x + 1)

=> y = -x + C + 2/3 and

y = -x + C - 2/3

x + y = C + 2/3

x + y = C - 2/3

comparing with x + y = 0 and x + y = 4/3

==> C = 2/3

so F(x) = -(1/3)x^3 + 2/3

= 1/3[- x^3 + 2 ]