Question 360030
The vertex form of the parabola is {{{y = a(x-h)^2+k}}}, where(h,k) is the vertex of the parabola. By direct substitution, {{{y = a(x--1)^2-1}}}, or
{{{y = a(x+1)^2-1}}}.  We use the coordinates of the origin (0,0) to solve for a:
{{{0 = a(0+1)^2-1}}}, or
{{{0 = a-1}}}, or {{{a = 1}}}.
Therefore the function is {{{f(x) = (x+1)^2-1}}}, or {{{f(x) = x^2+2x}}}.