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secondary 4 | A Maths
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need help with this qn, pls explain too
Date Posted: 3 years ago
Views: 392
Where the curve meets the x-axis,
y = 0
2x - 16/x² = 0
2x = 16/x²
x = 8/x²
x³ = 8
x = 2
P (2. 0)
At P, dy/dx
= 2 + 32/x³
= 2 + 32/2³
= 2 + 4
= 6
This means that the gradient of the curve at P is 6, and combined with the point P (2, 0),
y = 6x + c
0 = 6 (2) + c
0 =12 + c
c = -12
So PQ has the equation y = 6x - 12
y = 0
2x - 16/x² = 0
2x = 16/x²
x = 8/x²
x³ = 8
x = 2
P (2. 0)
At P, dy/dx
= 2 + 32/x³
= 2 + 32/2³
= 2 + 4
= 6
This means that the gradient of the curve at P is 6, and combined with the point P (2, 0),
y = 6x + c
0 = 6 (2) + c
0 =12 + c
c = -12
So PQ has the equation y = 6x - 12
thx :) had the same answer but i was doubting it as the marks allocated for this question is 5m but my working seemed too short for it
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