Derive the Standard Equation of Parabola with vertex at (h,k):
(y-k)^2 = 4a(x-h)
Please derive it with diagram.
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Derive the Standard Equation of Parabola with vertex at (h,k):
(y-k)^2 = 4a(x-h)
Please derive it with diagram.
First of all,gives a parabola, but it's sideways. We can still derive it, but the only way it's a function is if
is the domain and
is the range. It's a little weird, but it's more likely you copied the problem wrong. Deriving it isn't to hard, you just treat
,
, and
, like you would if they were real numbers. Since I don't know exactly what the problem is, I'll derive:
the first thing we'll do is expand it out, and solve for y
I'm not sure what you mean by a diagram, but I think it's the chart that shows the inflection points, and the orientation of the derivative. The inflection points are the values ofwhen the derivative is zero. That isn't to hard,
is zero only when
, (i'm assuming
is nonzero). The orientation of
on either side of
depends on the orientation of
. If
is positive, then
is positive for
and negative for
. If
is negative, then
is negative for
and positive for
.
I hope this helps
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