Ask Me Help Desk - Question About Calculus..

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