At what elevation did julio begin ?
2 answers:
You might be interested in
From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm