well, horizontal tangent,y'=0
y'=0=3ax^2+bx+c
so you solve that quadratic equation
x=(-b+-sqrt(b^2-4ac))/2a
Find a cubic function
y = ax3 + bx2 + cx + d
whose graph has horizontal tangents at the points
1 answer
y = ax3 + bx2 + cx + d
whose graph has horizontal tangents at the points
1 answer