Given that \(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\) and y = 6 when x = 3, find the equation for y.

  • A \(x^{3} - 4x - 9\)
  • B \(x^{3} - 4x + 9\)
  • C \(x^{3} + 4x - 9\)
  • D \(x^{3} + 4x + 9\)
waec 2010furthermathematics

The correct answer is A. \(x^{3} - 4x - 9\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4\)

\(y = \int (3x^{2} - 4) \mathrm {d} x = x^{3} - 4x + c\)

y = 6 when x = 3

\(6 = 3^{3} - 4(3) + c \implies 6 = 27 - 12 + c\)

\(c = 6 - 15 = -9\)

\(y = x^{3} - 4x - 9\)

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