# Find the coordinates of the stationary points of the curve y = 2x^3 + 3x^2 - 12x + 1

This is a typical A/S level question, which can be broken down into three key steps.

1) **Differentiation: **Remember that at a stationary point, the gradient of the curve is zero. Therefore, we first have to find the gradient, which means differentiating. (Multiply the coefficient of x by the power, and then reduce the power by 1)

y = 2x^{3} + 3x^{2 }- 12x + 1

dy/dx = 6x^{2} + 6x - 12

2) **Solve Quadratic: **To find where the gradient is zero (dy/dx = 0), we must solve the quadratic equation,

dy/dx = 6x^{2} + 6x - 12 = 0.

This quadratic can be most easily solved by factorisation (alternative methods are completing the square and using the quadratic formula).

Firstly we can take 6 out as a common factor, which results in 6(x^{2 }+ x - 2) = 0. We can then factorise the terms in brackets which results in 6(x+2)(x-1) = 0.

This gives the two solutions x = -2, x = 1.

3) **Substitution: **So far, we have only found the x coordinate of the stationary point, but to answer the question fully we need the y coordinate as well. We do this by substituting x = -2, and x = 1 into the original equation of the curve, y = 2x^{3} + 3x^{2 }- 12x + 1

When x = -2, we can calcuclate that y = 21. Similarly when x = 1, y = -6.

Thus the coordinates of the stationary points are (-2, 21) and (1, -6).