Question

the gradient function of a curve is px^2-qx,where p and q are constant .the curve has a turning point at (2,-1).the gradient of the tangent to the curve at the point x=-1 is 9.find the value of p and q.

Answer 1

Can you differentiate y = px^2-qx with respect to x first?

Answer 2

yes.. but why ? ><

Answer 3

Because the words "turning point" imply that the first derivative is 0 at (2, -1)
And "gradient of tangent at x=-1" is the first derivative of the function at x=-1.

So, by finding the first derivative of px^2-qx, we can set two equations, so that we can solve p and q.

Answer 4

can you show me the step ?

Answer 5

Do you know how to differentiate px^2-qx?

Answer 6

yes. dy/dx= 2p-q. isit ?

Answer 7

It... is not.
Check the first term again.

Answer 8

2px-q.

Answer 9

Yup!
So, from " a turning point at (2,-1)" Just sub x=2 and y=-1 into y=px^2-qx. This is the first equation you get. What would that be?

Answer 10

4p-2q=-1 -----(1)

Answer 11

Yes.
Next, for " tangent to the curve at the point x=-1 is 9", we put dy/dx=9 and x=-1 into the first derivative. That is our second equation. What is that?

Answer 12

* for "the gradient of the tangent to the curve at the point x=-1 is 9"

Answer 13

9=-2p-q . ---(2) thn use similtaneous solve it right ?

Answer 14

Exactly :)

Answer 15

thanks alot ! :)

Answer 16

Welcome~