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????
To find the equation of a parabola given its vertex and one point on the curve, you can follow these steps:
1. Start with the general equation for a parabola using the vertex form: (y - k)^2 = 4a(x - h), where (h, k) represents the vertex of the parabola.
2. Substitute the given vertex coordinates into the equation: (y - 2)^2 = 4a(x - 0).
3. Use the given point on the curve (1, 0) to solve for 'a'. Plug in the coordinates into the equation and solve for 'a':
(0 - 2)^2 = 4a(1 - 0)
(-2)^2 = 4a
4 = 4a
a = 1
4. Substitute the value of 'a' back into the equation: (y - 2)^2 = 4(x - 0).
5. Simplify if possible: (y - 2)^2 = 4x.
So, the equation of the parabola is (y - 2)^2 = 4x.
To find the focus of the parabola, you can use the formula: (h + a, k), where (h, k) are the coordinates of the vertex and 'a' is the distance between the vertex and the focus.
In this case, the vertex is (0, 2) and 'a' is 1. Thus, the focus coordinates would be (0 + 1, 2) which simplifies to (1, 2).
Therefore, the equation of the parabola is (y - 2)^2 = 4x and the focus is located at (1, 2).