Final answer:
The points on the curve y = x³ + 6x² - 15x + 1 where the gradient is zero are found by setting the derivative 3x² + 12x - 15 equal to zero, giving the points (1, -7) and (-5, 101).
Step-by-step explanation:
To find the points on the curve y = x³ + 6x² - 15x + 1 where the gradient is zero, we must calculate the derivative of the function and set it equal to zero. The derivative y' (or f'(x)) represents the gradient of the curve at any point x. So, let's find the derivative:
y' = 3x² + 12x - 15
To find the points where the gradient is zero, we solve y' = 0:
3x² + 12x - 15 = 0
Divide by 3 to simplify:
x² + 4x - 5 = 0
This is a quadratic equation, and we can solve for x using the quadratic formula:
∂ = b² - 4ac
In our equation, a = 1, b = 4, and c = -5, so:
∂ = (4)² - 4(1)(-5) = 16 + 20 = 36
The solutions for x are:
x = (-b ± √∂) / (2a)
x = (-4 ± √36) / 2
x = (-4 ± 6) / 2
The two x values that make the gradient zero are:
To find the corresponding y values, we plug these x values back into the original equation y = x³ + 6x² - 15x + 1:
- For x = 1: y = 1³ + 6(1)² - 15(1) + 1 = 1 + 6 - 15 + 1 = -7
- For x = -5: y = (-5)³ + 6(-5)² - 15(-5) + 1 = -125 + 150 + 75 + 1 = 101
The points on the curve where the gradient is zero are (1, -7) and (-5, 101).