Find the cubic function given the y-int, tangent line, slope of the tangent line, and the horizontal tangent:

Question

Find the cubic function in the form of f(x) = ax^3+b×^2+cx+d in which the y-intercept is (0,2), the tangent line at x=0 has a slope of 3, and at which the function has a horizontal tangent line at the point (1, 3).

I feel like I am slow for not understanding how to do this.

Answer 1

f(0) = d = 2; f'(x) = 3ax² + 2bx + c; f'(0) = c = 3.

f'(1) = 3a + 2b + c = 0 and f(1) = a + b + c + d = 3.

3a + 2b = - 3; a + b = - 2;

a = 1 and b = - 3.

So, f(x) = x³ - 3x² + 3x + 2