[B]Gratia Mindaro - Mathematics[/B]

"[color=66cd00]I let your [i]daddy[/i] do the caring for me.[/color]"

Gratia leaned forwards once more, already moving to open up the heavy textbook lying on the desk before her. Her fingers rapidly flipped through the pages, the flitter-flutter of paper distracting her from the rest of the room. She really didn't have the fucking patience to care right now, no matter if it was some ego-filled comment from ... the cheese-eating member of the team sitting beside her, or if anybody else wished to receive her attention. There was no need for her to deign herself to socialise in a fucking maths class - all that was necessary was to take down any notes and complete any assigned work.

So there was the calculus section of the book.

Material she was already relatively familiar with.

Beacon's textbooks, however, seemed to prefer functional notation over the Leibniz that was far more common back in her homeland, but personally she actually considered the former to be far more streamlined and simpler to use. One could only go so far in defining [i]u[/i] and [i]v[/i] in order to master the chain, product and quotient rules before it became all too boring to undertake. Mathematics was far more tolerable if one cut down on the amount they needed to write.

>The graph of f(x)=x^3 - 4x^2 is as shown.
>>Find f'(x)

That was simple. F'(x)=3x^2 - 8x.

>>Find the gradient of the tangent of the curve at the point (2,-8).

A tangent's gradient would be the rate of change of the graph at that point. The Mistralese huntress would simply need to calculate the graph's gradient at x=2 in order to obtain her answer.

F'(2)	=3(2)^2 - 8(2)
	=12 - 16
	=-4
∴tangm	=-4

>>Find the equation of the tangent at the point (2,-8).

A follow-up. This was ... excessively simple. But then again, it was admittedly the second week of the semester (and her first as an exchange).

y-y1	=m(x-x1)
y-(-8)	=-4(x-2)
y+8	=-4x+8
y	=-4x

>>Find the coordinates of the point Q where the tangent crosses the curve again.

y=-4x and y=F(x)=x^3 - 4x^2
∴-4x	=x^3 - 4x^2
  0	=x^3 - 4x^2 +4x
	=x(x^2 -4x +4)
	=x(x-2)^2
∴x=0, x=2
∴F(0)	=(0)^3 - 4(0)^2
	=0

∴Q is (0,0).