Find the point on the curve y = x^3 - 3x^2 + 4 where the tangent is horizontal. (2023)

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Q: Find the point on the curve y = x^3 - 3x^2 + 4 where the tangent is horizontal. (2023)

Solution: To find horizontal tangents, set dy/dx = 0. dy/dx = 3x^2 - 6x = 0 gives x = 0 and x = 2. At x = 1, y = 2.

Steps: 10

Step 1: Start with the curve equation y = x^3 - 3x^2 + 4.
Step 2: Find the derivative of y with respect to x, which is dy/dx. This tells us the slope of the tangent line.
Step 3: Calculate the derivative: dy/dx = 3x^2 - 6x.
Step 4: To find where the tangent is horizontal, set dy/dx = 0. This means we want to solve the equation 3x^2 - 6x = 0.
Step 5: Factor the equation: 3x(x - 2) = 0.
Step 6: Set each factor to zero: 3x = 0 gives x = 0, and x - 2 = 0 gives x = 2.
Step 7: Now we have two x-values: x = 0 and x = 2. We need to find the corresponding y-values for these x-values.
Step 8: For x = 0, substitute into the original equation: y = 0^3 - 3(0^2) + 4 = 4.
Step 9: For x = 2, substitute into the original equation: y = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0.
Step 10: The points where the tangent is horizontal are (0, 4) and (2, 0).