Question 1001249
A parabola whose axis of symmetry is parallel to the y-axis has an equation of the type
{{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
So, {{{y=a(x-1)^2-3}}}--->{{{y=a(x^2+2x+1)-3}}}--->{{{y=ax^2+2ax+a-3}}}
Making {{{x=0}}} , we find the y-intercept: {{{a-3}}} .
{{{a-3=6}}}--->{{{a=6+3}}}--->{{{a=9}}} .
So, the equation is
{{{y=9(x-1)^2-3}}} or {{{y=9x^2+18x+6}}} .