Gradient Chord

What does the answer to question 3ii) mean?
How do you answer questions like these?

Question: http://mei.org.uk/files/papers/2011_june_c2.pdf


Answer:http://www.ocr.org.uk/Images/62146-mark-scheme-unit-4752-concepts-for-advanced-mathematics-june.pdf

Well it's asking you for the gradient of a chord between two points on the curve. One of which has gradient $x=4$ and another $x=4.1$. Knowing the eq. of the curve you can calculate the corresponding y coordinates hence you have the two points and working out the gradient between them is elementary.

For part (ii) you need to do the same thing, but you need to pick $x=4$ and a different point that is closer to $x=4$ than $x=4.1$ is. So really, you can pick $x=4.05$ for example and just do the same thing as in part (i)

The idea here is that if you keep dragging you second point closer to $x=4$, the gradient of the chord is going to approach the gradient of the tangent at $x=4$. This is where one of the definitions for the derivative comes from.

Thank you.

How do you answer question 4 ??

You're given two coordinates that must satisfy the eq. So plug them in and work off the two equations you got.

Do you solve them simultaneously?

Yeah.

I get -2.4 = b from doing 3.6=ab^2 - 6=ab which is not correct
b should be 0.6. How?

How exactly did you get that...? What you'd put doesn't really work.

I subtracted both equations from each other so that a-a is nothing b^2 -b is b or is the whole of ab squared
I tried with the whole of ab^2 too but that still didn't work