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, (y-k)^2 = 4a(x-h) gives a parabola, but it's sideways. We can still derive it, but the only way it's a function is if y is the domain and x 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 a , h , and k , like you would if they were real numbers. Since I don't know exactly what the problem is, I'll derive:
(y-k) = 4a(x-h)^2
the first thing we'll do is expand it out, and solve for y
y-k = 4a(x^2-2hx+h^2)= 4ax^2-8hax+4ah^2
\Rightarrow y = 4ax^2-8hax+4ah^2+k

\frac{dy}{dx} = 8ax-8ha = 8a(x-h)

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 of x when the derivative is zero. That isn't to hard, 8a(x-h) is zero only when x = h , (i'm assuming a is nonzero). The orientation of \frac{dy}{dx} on either side of h depends on the orientation of a . If a is positive, then \frac{dy}{dx} is positive for x >h and negative for x<h . If a is negative, then \frac{dy}{dx} is negative for x >h and positive for x<h .

I hope this helps