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

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(0, 4)
(1, 2)
(2, 2)
(3, 4)

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

Solution: To find horizontal tangents, set the derivative y' = 3x^2 - 6x = 0. This gives x = 0 and x = 2. The point (1, 2) has a horizontal tangent.

Steps: 9

Step 1: Start with the curve equation y = x^3 - 3x^2 + 4.
Step 2: Find the derivative of the curve, which represents the slope of the tangent line. The derivative y' = 3x^2 - 6x.
Step 3: Set the derivative equal to zero to find where the tangent is horizontal: 3x^2 - 6x = 0.
Step 4: Factor the equation: 3x(x - 2) = 0.
Step 5: Solve for x: This gives us x = 0 and x = 2.
Step 6: Now, find the corresponding y-values for these x-values using the original curve equation.
Step 7: For x = 0: y = 0^3 - 3(0^2) + 4 = 4, so the point is (0, 4).
Step 8: For x = 2: y = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0, so the point is (2, 0).
Step 9: The points (0, 4) and (2, 0) both have horizontal tangents.

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

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