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# FSMQ Differentiation Question watch

1. Can anybody help me out on IV and V please?

The point(2,-8) is on the curve y = x^3 - px + q
i) use this information to find a relationship between p and q
ii)find the gradient function dy/dx.

The tangent to this curve at the point (2,-8) is parallel to the x axis.
iii) use this information to find the value of p

IV
)find the co-ordinates of the other point where the tangent is parallel to the x axis.
V) State the co-ordinates of the of the point P where the curve crosses the y axis
2. (Original post by hereNow)
Can anybody help me out on IV and V please?

The point(2,-8) is on the curve y = x^3 - px + q
i) use this information to find a relationship between p and q
ii)find the gradient function dy/dx.

The tangent to this curve at the point (2,-8) is parallel to the x axis.
iii) use this information to find the value of p

IV
)find the co-ordinates of the other point where the tangent is parallel to the x axis.
V) State the co-ordinates of the of the point P where the curve crosses the y axis
Tell me, what is the gradient of the line y= 0?
3. 0 im guessing
4. (Original post by hereNow)
0 im guessing
Yes. So if a line is parallel to it, what is its gradient?
Btw i meant y=0
5. 0 So are you saying that i should equate dy/dx to 0 and solve it?
6. (Original post by hereNow)
0 So are you saying that i should equate dy/dx to 0 and solve it?
Precisely. Welcome to TSR btw
7. Thanks, i tried that before but how am i supposed to solve 3x^2 - p = 0 ??
8. (Original post by hereNow)
Thanks, i tried that before but how am i supposed to solve 3x^2 - p = 0 ??
Part iii) says find p, I assumed you'd done so . . .
9. umm i know this might be a stupid question but can you show me how you got the co-ordinates assuming P = 12.
10. (Original post by hereNow)
umm i know this might be a stupid question but can you show me how you got the co-ordinates assuming P = 12.
3x2 =12
=> x2 =4 so x = 2 or ?
11. i got 2 but the answers have the coordinates as -2,24
12. Joostan Please answer this final question why is it -2 instead of 2
13. X^2 =4, therefore x = -2 or 2. Substitute the "-2" back into the original equation to get the y coordinate. This (-2,y) is the other coordinate because u are already told that tangent at (2,-8) is parallel to the x-axis
14. (Original post by hereNow)
Joostan Please answer this final question why is it -2 instead of 2
Because we already know that 2 is a stationary point from the question which has the point (2,-8) as gradient zero. When you get x^2 = 4, then there are two solutions: 2 and -2, because both square to make 4.

Then you go back to substitute in p= -12 and x = 2 to the original eq to get q = 8, then you substitute in x = -2 to get the y value.

The y-intersection is where x = 0, giving y = 8.

15. (Original post by Gmart)
Because we already know that 2 is a stationary point from the question which has the point (2,-8) as gradient zero. When you get x^2 = 4, then there are two solutions: 2 and -2, because both square to make 4.

Then you go back to substitute in p= -12 and x = 2 to the original eq to get q = 8, then you substitute in x = -2 to get the y value.

The y-intersection is where x = 0, giving y = 8.

p=/= -12
16. (Original post by hereNow)
Joostan Please answer this final question why is it -2 instead of 2
Sorry I had to pick up some lunch
It's -2, because the stationary point at 2 was already given.
As others have already said, you then substitute in the points to the equation to find the y-coordinate
17. (Original post by joostan)
p=/= -12
Is this saying that p doesn't equal -12?

3x^2 + p = 0

??
18. (Original post by Gmart)
Is this saying that p doesn't equal -12?

3x^2 + p = 0

??
19. (Original post by Gmart)
Is this saying that p doesn't equal -12?

3x^2 + p = 0

??
y = x^3 - px + q
dy/dx = 3x^2 - p

3(2)^2 - p = 0
p = 12.

p=/= -12
20. (Original post by joostan)
Whoops, wrote the question down with a plus rather than a minus - my bad

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