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)
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 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
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
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