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Maths a level differentiation

4) A curve has the equation y= 2x^3 - 6x
a) show that the curve crosses the x-axis at the origin and points ( square root of 3, 0) and (- square root of 3, 0)

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(edited 7 years ago)
Original post by imaan2121
4) A curve has the equation y= 2x^3 - 6x
a) show that the curve crosses the x-axis at the origin and points ( square root of 3, 0) and (- square root of 3, 0)


What have you tried?
Reply 2
Original post by Mr M
What have you tried?


I have factorised out 2x
y= 2x(x^2-3)
and then got y=2x(x-square root of 3)(x+square root of 3)
giving me x= 0 x= +square root of 3 x= - square root of 3
but when I substitute these values into y they don't get 0
Original post by imaan2121
I have factorised out 2x
y= 2x(x^2-3)
and then got y=2x(x-square root of 3)(x+square root of 3)
giving me x= 0 x= +square root of 3 x= - square root of 3
but when I substitute these values into y they don't get 0


Really? What do they give you then?
Reply 4
Original post by Mr M
Really? What do they give you then?


oops, my mistake!
Reply 5
Original post by imaan2121
4) A curve has the equation y= 2x^3 - 6x
a) show that the curve crosses the x-axis at the origin and points ( square root of 3, 0) and (- square root of 3, 0)


I was going to give you the full solution, but I then realised that might not be the best thing to do. Let me give you a few hints, though:

- you are proving an intersection at the origin, so set the equation equal to 0
- factorise by taking out x

Feel free to ask if you can't go any further.

Good luck with your maths A-Level :smile:

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