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secondary 4 | A Maths

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Since the curve has a critical point at (-1, 3), dy/dx will be zero when x = -1.

So, 0 = [k (-1)³ - 4] / (-1)²

0 = -k - 4

k = -4

Then, we have

dy/dx = (-4x³ - 4) / x²

We can break this into individual fractions.

dy/dx = -4x³/x² - 4/x²

dy/dx = -4x + 4 times x^-2

Integrating with respect to x,

y = -4x² divided by 2 + 4 times x^-1 divided by -1 + some constant which we call c

y = -2x² - 4/x + c

We know that (-1, 3) is on the curve, so we substitute x = -1 and y = 3 into this equation.

3 = -2 (-1)² - 4/(-1) + c

3 = -2 + 4 + c

3 = 2 + c

1 = c

So, the equation of the curve is y = -2x² - 4/x + 1.

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