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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
Answer:
Step-by-step explanation:
is found by first, siplifying the radicals in the parenthesis.
can be simplified as
and
stays the same. Now we have to multiply these two by
which then results in
Assuming the function is
For
to be differentiable, it necessarily has to be continuous. For this condition to be met, we need
For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have
Now we solve for
and 
:
For
For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have
Now we solve for