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

Practice Questions

Q1
Find the point on the curve y = x^3 - 3x^2 + 4 where the tangent is horizontal. (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 where the tangent is horizontal. (2023)
  • 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).
  • Differentiation – Understanding how to find the derivative of a function to determine the slope of the tangent line.
  • Finding Critical Points – Identifying points where the derivative equals zero to find horizontal tangents.
  • Evaluating Functions – Calculating the value of the original function at specific x-values to find corresponding y-values.
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