Let f be a function such that f "(x) = 6x + 8.
find f(x) if the graph of f is tangent to the line
3x - y = 2 at the point (0,-2).

F''(x) = 6x + 8
∫f''(x) = f'(x) = 3x^2 + 8x + c1
∫f'(x) = f(x) = x^3 + 4x^2 + c1x + c2

At point (0, -2),
c2 = -2

3x - y = 2
y = 3x - 3

f'(0) = 3
c1 = 3

Therefore, the required function is
f(x) = x^3 + 4x^2 + 3x - 2

Don’t worry about it