MATHS ASAP!!!!!
posted by bobpursley .
At the x axis, y is zero. So if the x intercept is (1,0) then
(ya)^2=x + b
(0a)^2=1 + b
Now the vertex.
(ya)^2=x +b
(2a)^2= 0 + b
or these two equations are..
a^2= 1+b
or b= a^2  1 putting this into the second equation.
44a + a^2=a^21
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 (y2)^2 = x????
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