posted by bobpursley .
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.
(2-a)^2= 0 + b
or these two equations are..
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????