Posted by **bobpursley** on Thursday, November 2, 2006 at 1:41pm.

At the x axis, y is zero. So if the x intercept is (1,0) then

(y-a)^2=x + b

(0-a)^2=1 + b

Now the vertex.

(y-a)^2=x +b

(2-a)^2= 0 + b

or these two equations are..

a^2= 1+b

or b= a^2 - 1 putting this into the second equation.

4-4a + a^2=a^2-1

solve for a, then go back and solve for b

none of those links help for the parabola helped.

it has a vertex of (0,2)

it intercepts the x axis at (1,0)

what is the equation and find the focus???????

Is it (y-2)^2 = x????

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