From calculus we know that the vertex of a quadratic function is its one and only critical point.
f(x) = 3x^2 + bx + c
df/dx = 6x+b is not undefined anywhere so,
6x+b = 0
x = -b/6 (Which you might recall from algebra as x = -b/2a) is the x coordinate. So we assume that equal the x coordinate of our vertex : -b/6 = 4/3 ===(Multiplying both sides by -6)> b = -8
So our quadratic function takes the form,
f(x) = 3x^2 - 8x + c
To find c just plug in 4/3 for x and assume it is -10/3:
3(4/3) ^ 2 - 8*(4/3) + c = -10/3
Solving for c we get: c = 2 [If I did my arithmetic correctly that is]
f(x) = 3x^2 - 8x +2
Comment if you have any questions or you think I made a mistake somewhere.
Hope this helps :)
