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secondary 4 | A Maths
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secondary 4 chevron_right A Maths chevron_right Singapore

Pls help

Date Posted: 1 year ago
Views: 175
Eric Nicholas K
Eric Nicholas K
1 year ago
(a)

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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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Part b is too difficult for me to write on the chat box, so here you go.
Shams
Shams
1 year ago
Thanks
Shams
Shams
1 year ago
Thanks