To find the point(s) on the curve y = x^3 + 6x^2 - 15x + 1 at which the gradient is zero, we need to find where the derivative of the curve is zero.
Taking the derivative of y with respect to x, we get:
y' = 3x^2 + 12x - 15
To find where the gradient is zero, we need to solve the equation y' = 0:
3x^2 + 12x - 15 = 0
Dividing both sides by 3, we get:
x^2 + 4x - 5 = 0
Factoring the quadratic equation, we get:
(x + 5)(x - 1) = 0
So the solutions are x = -5 and x = 1.
To find the corresponding points on the curve, we substitute each value of x back into the equation y = x^3 + 6x^2 - 15x + 1:
When x = -5, y = (-5)^3 + 6(-5)^2 - 15(-5) + 1 = -99
When x = 1, y = 1^3 + 6(1)^2 - 15(1) + 1 = -7
Therefore, the points on the curve at which the gradient is zero are (-5, -99) and (1, -7).