Find the points on the curve y = 2x^3 + 3x^2 - 12x + 1 where the tangent is horizontal

That means the slope of the tangent is zero. Look for the points where the derivative

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

x^2 +x -2 = 0
(x+2)(x-1) - 0

x - -2 or +1.

To find the points on the curve where the tangent is horizontal, we need to find the values of x where the derivative of the function is equal to zero.

Step 1: Find the derivative of the function y = 2x^3 + 3x^2 - 12x + 1.

The derivative of y with respect to x is given by:
dy/dx = 6x^2 + 6x - 12.

Step 2: Set the derivative equal to zero and solve for x.

6x^2 + 6x - 12 = 0.

Step 3: Factor the quadratic equation.

2(3x^2 + 3x - 6) = 0.

Step 4: Divide both sides by 2.

3x^2 + 3x - 6 = 0.

Step 5: Divide through by 3 to simplify the equation.

x^2 + x - 2 = 0.

Step 6: Factor the quadratic equation.

(x + 2)(x - 1) = 0.

Step 7: Set each factor equal to zero and solve for x.

x + 2 = 0 or x - 1 = 0.

x = -2 or x = 1.

So, the values of x where the tangent is horizontal are x = -2 and x = 1.

Step 8: Find the corresponding y-values for the points (x, y).

For x = -2:
y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 1
= -16 + 12 + 24 + 1
= 21.

For x = 1:
y = 2(1)^3 + 3(1)^2 - 12(1) + 1
= 2 + 3 - 12 + 1
= -6.

Therefore, the points on the curve y = 2x^3 + 3x^2 - 12x + 1 where the tangent is horizontal are (-2, 21) and (1, -6).

To find the points on the curve where the tangent is horizontal, we need to find the derivative of the function, set it equal to zero, and solve for x. The points where the derivative is zero represent the x-coordinates of the points where the tangent is horizontal.

Step 1: Find the derivative of the function y = 2x^3 + 3x^2 - 12x + 1.
To find the derivative, we take the derivative of each term separately using the power rule. The power rule states that if f(x) = x^n, then f'(x) = nx^(n-1).
So, taking the derivative of each term:
dy/dx = d/dx(2x^3) + d/dx(3x^2) + d/dx(-12x) + d/dx(1)
= 6x^2 + 6x - 12

Step 2: Set the derivative equal to zero and solve for x.
Setting 6x^2 + 6x - 12 = 0
Dividing by 6 to simplify the equation: x^2 + x - 2 = 0
Factoring: (x + 2)(x - 1) = 0
Setting each factor equal to zero:
x + 2 = 0 or x - 1 = 0
x = -2 or x = 1

Step 3: Substitute the x-values into the original equation to find the y-values.
Substituting x = -2 into y = 2x^3 + 3x^2 - 12x + 1:
y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 1
y = -15
So, we have one point (-2, -15) where the tangent is horizontal.

Substituting x = 1 into y = 2x^3 + 3x^2 - 12x + 1:
y = 2(1)^3 + 3(1)^2 - 12(1) + 1
y = -4
So, we have another point (1, -4) where the tangent is horizontal.

Therefore, the points on the curve y = 2x^3 + 3x^2 - 12x + 1 where the tangent is horizontal are (-2, -15) and (1, -4).