f(x)=-1/3x^2-5x+2
Roots are:x = -15/2-sqrt(249)/2 and x = sqrt(249)/2-15/2
The vertex is where df(x)/dx = 0
df(x)/dx = -(2/3)x-5
This is zero when x = -15/2, at which point f(x) = -(1/3)(15/2)^2 - 5*(-15/2) + 2 = 83/4
so the vertex is at (-15/2, 83/4)
Hallo guest!
