Let the equation of the parabola be #y=ax^2bxc# We have to find the values of the parameters #a, b and c# to fix the equation Its slope #(dy/dx)# of the function #y=ax^2bxc# is defined by its first derivative #dy/dx=2axb# Then, at #x=1;Find in the form y= ax^2 bx c, the equation of the quadratic whose graph a) touches the xaxis at 4 and passes through (2,12) b) has vertex (4,1) and passes through (1,11) Answer provided by our tutors y= ax^2 bx c a) touches the xaxis at 4 and passes through (2,12) touches the xaxis at 4 means that passes trough (4,0) and b^2 4*a*c = 0 (the quadratic has 1 solution) passesLearn how to graph a parabola of the form f(x)=ax^2bxc with integer coefficients, and see examples that walk through sample problems stepbystep for
Quadratic Function
