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#1
a. On the same axes sketch the curves with equations Y = 6/X & Y = 1 + X ( I've done this part )

b. The curves intersect at the points A and B, find the coordinates of A and B I've done this part also and have the values A ( -3, -2 ) and B ( 2 , 3).

c. The curve C with equation Y = X ^2 + p X + q, where p and q are integers passes through A and B, find the values of p and q.

Thanks
Vandomo
0
5 years ago
#2
(Original post by van.domo)
a. On the same axes sketch the curves with equations Y = 6/X & Y = 1 + X ( I've done this part )

b. The curves intersect at the points A and B, find the coordinates of A and B I've done this part also and have the values A ( -3, -2 ) and B ( 2 , 3).

c. The curve C with equation Y = X ^2 + p X + q, where p and q are integers passes through A and B, find the values of p and q.

Thanks
Vandomo
Substitute your points into C and you will have a pair of simultaneous equations.
0
5 years ago
#3
If you substitute x and y with the values you found for points A and B you should get the following system of equations:

9-3p+q=-2
4+2p+q=3

Which has as solutions (p,q)=(2,-5).
So the curve we seek is y=x2+2x-5.

Hope this helps, and in case I am mistaken please correct me (I am a GCSE equivalent student).
0
5 years ago
#4
(Original post by kleov.maths)
If you substitute x and y with the values you found for points A and B you should get the following system of equations:

9-3p+q=-2
4+2p+q=3

Which has as solutions (p,q)=(2,-5).
So the curve we seek is y=x2+2x-5.

Hope this helps, and in case I am mistaken please correct me (I am a GCSE equivalent student).

You are not permitted to post full solutions. My previous post gave an appropriate amount of help.
0
#5
(Original post by kleov.maths)
If you substitute x and y with the values you found for points A and B you should get the following system of equations:

9-3p+q=-2
4+2p+q=3

Which has as solutions (p,q)=(2,-5).
So the curve we seek is y=x2+2x-5.

Hope this helps, and in case I am mistaken please correct me (I am a GCSE equivalent student).
Thanks very much, you are right. I did the same thing by solving the problem by simultaneous equations.

Regards
Dominic G
0
5 years ago
#6
(Original post by van.domo)
Thanks very much, you are right. I did the same thing by solving the problem by simultaneous equations.

Regards
Dominic G
I am glad I could help.
To Mr M: I am sorry, I should have read the rules first. It will not happen again.
0
#7
Thanks very much for your help. It's appreciated very much.

Regards
Dominic G
0
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