Your welcome. What year are you in?

]]>will do nicely.

(But using the 'a' and 'b' from the new quadratic, of course.)

Bob

]]>Yes, there is a way using just algebra.

]]>

If i'm asking for help too much lately please let me know. I have exams in 3 weeks so trying to get everything cleared up before the exam.

Thanks]]>

d(BA)/dx is a special notation used in the differential calculus.

It gives you a way of finding the maximum value of an expression (amongst many, many uses!).

It's too big a topic to start in answer to the question, if you haven't met it before.

But don't worry. As the expression

is a quadratic there's another way to get the maximum value.

I've put the graph below. As you can see it does have a maximum value. Would you be able to work out the x, at this point?

Bob

]]>Thanks]]>

I got a = -2 b = -4 and c = 0

Then for the points:

So

So you need

Can you take over from here?

Bob

]]>Have you got a, b and c yet?

I'll assume yes.

That fixes the parabola (ie. there's only one answer) and the line is obviously unique.

But B can move about on the parabola and so that means M moves too.

I'd call M (x,0) and write the coordinates of B and A in terms of this.

Then you write an expression for BA in terms of x, differentiate, and hence get the maximum.

I'd better go and get a piece of paper and try it out.

Bob

]]>Thanks a lot in advance]]>