Question 156738
Since the vertex is at {{{x=0}}} and {{{f(0)=8}}} (which means that when {{{x=0}}}, then {{{y=0}}}), this means that the vertex is at the point (0,8). Now remember, the quadratic {{{y=a(x-h)^2+k}}} has the vertex (h,k). So this means that in this case {{{h=0}}} and {{{k=8}}}



{{{y=a(x-h)^2+k}}} Start with the general quadratic in vertex form



{{{y=a(x-0)^2+8}}} Plug in {{{h=0}}} and {{{k=8}}}



{{{y=ax^2+8}}} Subtract



So the quadratic equation which has the vertex at (0,8) is {{{y=ax^2+8}}} where "a" can be any number. For example, {{{y=2x^2+8}}} has the vertex at (0,8)