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?
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
Thank you, that was extremely helpful!
Sure! To create a parabola given the vertex and one other point, you can use the vertex form of a quadratic equation: y = a(x - h)^2 + k, where (h, k) represents the vertex.
In your case, the vertex is given as (6,3), so h = 6 and k = 3. We can plug these values into the equation to get: y = a(x - 6)^2 + 3.
Now, we need to find the value of 'a', which determines the shape and direction of the parabola. Since you mentioned that the parabola ends at x = 12 and starts at (0,0), we can substitute these values into the equation to find 'a'.
At the start of the parabola, x = 0 and y = 0, so we have: 0 = a(0 - 6)^2 + 3.
At the end of the parabola, x = 12 and y = 0, so we have: 0 = a(12 - 6)^2 + 3.
Simplifying these equations, we get: 0 = 36a + 3 and 0 = 36a + 3.
Now we have a system of equations:
36a + 3 = 0,
36a + 3 = 0.
Solving the system, we find that a = -1/12.
Substituting this value back into the equation, our final equation for the parabola is: y = (-1/12)(x - 6)^2 + 3.
Now you can plot this equation on a graph, and it will represent a parabola that starts at (0,0), has the vertex at (6,3), and ends at x = 12.