I have to create parabola that represents the ark of a projectile, meaning it starts and (0,0) and has a downward ark. My problem is that I only know the vertex (6,3) and I know it ends at 12. Could anyone help me turn that into a quadratic form to graph as a parabola?

Answer 1

Recall that if the vertex is (h,k) , the parabola is
y = a(x-h)^2 + k
so you have (6,3) as your vertex, then
y = a(x-6)^2 + 3
but (0,0) lies on it, then
0 = a(0-6)^ + 3
a = -3/36 = -1/12

y = (-1/12)(x-6)^2 + 3 or y = (-1/12)x^2 + x
---- you will get the same if you sub in (12,0)

or

You know the x-intercepts are 0 and 12
so the equation is
y = a(x-0)(x-12)
but (6,3) lies on it, so
3 = a(6)(-9)
a = 3/-36 = -1/12

y = (-1/12)x(x-12)
y = (-1/12)x^2 + x

Answer 2

Thank you, that was extremely helpful!