Competition - post a photo you've taken of nature >>

Urgent help needed on Maths question - A level

AQASUX

The curve y + x^3 + px + q has a minimum point at (4,-11). Find the coordinates of the maximum point of the curve

Scroll to see replies

Reply 1

brianeverit

Original post by AQASUX

The curve y + x^3 + px + q has a minimum point at (4,-11). Find the coordinates of the maximum point of the curve

Putting the coordinates of the minimum point into the equatuion will give you a simple equation involving p and q.
Then, because there is a minimum point when x=4 you know that the derivative must be zero when x=4

Reply 2

AQASUX

So how would the steps go? Sorry if I'm being thick. I've got -75=p+q but where from there?

Reply 3

TenOfThem

Original post by AQASUX

So how would the steps go? Sorry if I'm being thick. I've got -75=p+q but where from there?

Can I ask how you got that?

Reply 4

nickk_harrisonn

Original post by AQASUX

The curve y + x^3 + px + q has a minimum point at (4,-11). Find the coordinates of the maximum point of the curve

Well I'm guessing that you mean y=x^3+px+q instead.
Input (4,-11) into the equation AND into the derivative of the equation and you will have two simultaneous equations which can be solved for p and q.
Then input p into the derivative and replace dy/dx with 0 and you can solve to find the x-coordinate of the maximum point.
Then you can input p, q and the x-coordinate into the original equation to find the y-coordinate of the maximum point.
And Eureka! You have the coordinates of the other maximum point!

Reply 5

AQASUX

Original post by TenOfThem

Can I ask how you got that?

-11=4^3 + p(4) + q [substituting in the x and y values from minimum]

-11=64+4p+q

-75=4p+q

Reply 6

AQASUX

Original post by TenOfThem

Can I ask how you got that?

sorry i put p instead of 4p

Reply 7

nickk_harrisonn

Original post by AQASUX

-11=4^3 + p(4) + q [substituting in the x and y values from minimum]

-11=64+4p+q

-75=4p+q

That's good, but then use the derivative (dy/dx = 3x^2 +p) to find p (dy/dx = 0 and x = 4)

Reply 8

AQASUX

Original post by nickk_harrisonn

Would the derivative be 3x^2 + p -1

Reply 9

TenOfThem

Original post by AQASUX

-11=4^3 + p(4) + q [substituting in the x and y values from minimum]

-11=64+4p+q

-75=4p+q

Just to check - did you realise your OP says y+ rather than y=

But yes, now that you have the 4p

You were told to differentiate and use the fact that dy/dx = 0 at the turning point

This gives another equation in p

Reply 10

TenOfThem

Original post by AQASUX

Would the derivative be 3x^2 + p -1

Where does your -1 come from?

Reply 11

AQASUX

Original post by TenOfThem

Where does your -1 come from?

oh yeah lol i got confused for a minute. Is it just 3x^2 + p. And yeah i realised the original post was wrong haha

Reply 12

TenOfThem

Original post by AQASUX

oh yeah lol i got confused for a minute. Is it just 3x^2 + p. And yeah i realised the original post was wrong haha

At the turning point that =0

Reply 13

AQASUX

Original post by TenOfThem

At the turning point that =0

But then when i come to do the simultaneous equation i get the two as:

-17 + 4p + q
and 3x^2 +p

Correct?

This gives three variables though

Reply 14

AQASUX

*-75 = 4p+ q sorry im tired

heyyy

Original post by AQASUX

But then when i come to do the simultaneous equation i get the two as:

and 3x^2 +p

Correct?

This gives three variables though

This =0 at the turning point - you know what x is at the turning point

Reply 17

AQASUX

Original post by TenOfThem

This =0 at the turning point - you know what x is at the turning point

So x is 4 at the turning point. So 3(4)^2 +p=0. So 48+p=0. So p = -48 ? I can do that without simultaneous equations but is it right. If it is ill just work q out from the original and use the derivative to find the maximum

Reply 18

TenOfThem

Original post by AQASUX

p is correct

not sure what you mean by this

yes you need to put p into the other equation to find q

Reply 19

AQASUX

Original post by TenOfThem

p is correct

not sure what you mean by this

yes you need to put p into the other equation to find q

People were telling me to do simultaneous equations before.
So,
y= x^3 + 48x + q
-11 = 64 + 192 + q
245 = q
y=x^3 + 48x + 245
dy/dx = 3x^2 + 48
0 = 3x^2 + 48
0=x^2 + 16
using quadratic formula x = +/- squareroot4