The vertex form of a quadratic is `y=a(x-h)^2+k` with the vertex at `(h,k)` , and `a` determines if the parabola opens up or down and how wide.

Given the vertex at (1,2) and the point (0,4) we have:

`y=a(x-h)^2+k`

`y=a(x-1)^2+2`

For the point (0,4) we have:

`4=a(0-1)^2+2`

`4=a+2`

`a=2`

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Thus the equation is `y=2(x-1)^2+2`

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Expanding you get `y=2(x^2-2x+1)+2` or `y=2x^2-4x+4`

Let the parabola be :

f(x)= ax^2 + bx + c

Given the y-intercept is 4

==> Then, the point (0,4) is on the graph.

==> f(0)= 4 ==> c = 4

==> f(x)= ax^2 + bx + 4

Also, given that the vertex is (1,2) ==> f(1) = 2

Also, the vertex is the extreme point of the function.

==> f'(1)= 0

==> f'(x)= 2ax + b

==> f'(1)= 2a + b= 0 ==> b= -2a .....(1)

Also, f(1)= 2

==> a+ b+ 4= 2 .........(2)

==> a+ (-2a) = -2

==> -a = -2 ==> a = 2

==> b= -4

`==gt f(x)= 2x^2 -4x + 4 `