Question 135108
Using {{{f(x)=a(x-h)^2+k}}} means the vertex is at (h,k).  Since you are given the vertex of (0,3), that means the equation must be {{{f(x)=a(x-0)^2+3}}}, or more simply:  {{{f(x)=ax^2+3}}}.  For the equation to pass through the point (2,4), {{{f(2)=4}}}, so {{{a(2)^2+3=4}}} => {{{4a=4-3}}} => {{{a=1/4}}}, and the equation must be:


{{{f(x)=x^2/4+3}}}


{{{drawing(600,600,-5,5,-1,9,
grid(1),
graph(600,600,-5,5,-1,9,x^2/4+3),
circle(0,3,.07),
locate(.2,2.8,V(0,3)),
circle(2,4,.07),
locate(2.2,3.8,P(2,4))
)}}}


You can see that "a" in the general equation cannot be negative in this case because that would mean the parabola would open downward.  A downward opening parabola with a vertex at (0,3) cannot possibly contain the point (2,4).