Question 434359
{{{y = 3x^2}}}

To show this, Let {{{y = a(x-h)^2 + k}}}
==> {{{3 = a(1-h)^2 + k}}}
{{{12 = a(2-h)^2 + k}}}, and
{{{48 = a(4-h)^2 + k}}}

Subtracting the 1st from the 2nd, we get {{{9 = a(2-h)^2 - a(1-h)^2}}}
<==> 9 = 3a - 2ah 
Subtracting the 2nd from the 3rd, we get  {{{36= a(4-h)^2 - a(2-h)^2}}}
<==> 36 = 12a - 4ah, or 9 = 3a - ah.

Now we have the equations 9 = 3a - 2ah, and 9 = 3a - ah.
subtracting the first from the second, we get ah = 0.
Since {{{a<>0}}}, we have h = 0.
Substituting h = 0 into {{{3 = a(1-h)^2 + k}}} and {{{12 = a(2-h)^2 + k}}}, we get
3 = a + k and 12 = 4a + k.
From this we get a = 3.
Finally putting h = 0 and a = 3 into {{{3 = a(1-h)^2 + k}}}, we get k = 0.