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# c2 june 05 - read this if you sat it watch

1. q1 said, find the co-ords of the stationary point on the curve with equation y = ....
what was the equation?
2. Does anyone remember?
3. any1?
4. (Original post by manps)
q1 said, find the co-ords of the stationary point on the curve with equation y = ....
what was the equation?
You're Mr Maths, you should know!
5. 2x^2 +/- something

hmm it had stationery point (-3,18) ? or (3,18) ?
6. think stationery points we're (3,-18) so..

y = 2x^2 + ax + c

dy/dx = 4x + a

when dy/dx = 0, x=3

4x + a = 0
12 + a = 0
a=-12

y=2x^2 - 12x + c

when x=3,y=-18

y=2(3)^2 - 12(3) + c
-18 = 18 - 36 + C
c=0

y = 2x^2 - 12x sound familiar?
7. (Original post by Revelation)
think stationery points we're (3,-18) so..

y = 2x^2 + ax + c

dy/dx = 4x + a

when dy/dx = 0, x=3

4x + a = 0
12 + a = 0
a=-12

y=2x^2 - 12x + c

when x=3,y=-18

y=2(3)^2 - 12(3) + c
-18 = 18 - 36 + C
c=0

y = 2x^2 - 12x sound familiar?
nah ui can rmbr there being a 24 somewhere along the line of deffrentiating. i think it might hav bin whats on ur mark scheme now, manap. Im pretty sure because i forgot to work out the y co-ordinate. silly mistake.
8. (Original post by Revelation)
think stationery points we're (3,-18) so..

y = 2x^2 + ax + c

dy/dx = 4x + a

when dy/dx = 0, x=3

4x + a = 0
12 + a = 0
a=-12

y=2x^2 - 12x + c

when x=3,y=-18

y=2(3)^2 - 12(3) + c
-18 = 18 - 36 + C
c=0

y = 2x^2 - 12x sound familiar?
All I can remember is 12x and a number 2. So that looks right...
9. yes i think that is right. i remember 3, -18 as well.
10. (Original post by manps)
q1 said, find the co-ords of the stationary point on the curve with equation y = ....
what was the equation?
y=12x² - 12x
11. (Original post by Revelation)
think stationery points we're (3,-18) so..

y = 2x^2 + ax + c

dy/dx = 4x + a

when dy/dx = 0, x=3

4x + a = 0
12 + a = 0
a=-12

y=2x^2 - 12x + c

when x=3,y=-18

y=2(3)^2 - 12(3) + c
-18 = 18 - 36 + C
c=0

y = 2x^2 - 12x sound familiar?
thats right i think, i'm fairly confident the differential was 4x-12
12. did we have to give the equation in our answer, I thought we just had to find the stationary point cooridinates.????
13. (Original post by boxexpansion)
did we have to give the equation in our answer, I thought we just had to find the stationary point cooridinates.????
yes, just the coordinates were needed. ppl here we just trying to figure out the initial question which is:
y = 2x2 - 12x

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