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

Practice Questions

Q1
Find the point on the curve y = x^3 - 3x^2 + 4 that has a horizontal tangent. (2023)
  1. (0, 4)
  2. (1, 2)
  3. (2, 2)
  4. (3, 4)

Questions & Step-by-Step Solutions

Find the point on the curve y = x^3 - 3x^2 + 4 that has a horizontal tangent. (2023)
  • 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.
  • Finding Derivatives – Understanding how to differentiate a function to find its slope.
  • Horizontal Tangents – Identifying points where the derivative equals zero, indicating a horizontal tangent.
  • Evaluating Points on a Curve – Calculating the y-coordinate of the curve at specific x-values to find corresponding points.
Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely