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secondary 4 | A Maths
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just nd help w part 1 thx!
Date Posted: 2 years ago
Views: 464
y'=9/(x²-10x+25)
(i)
y=0
x+4 = 5-x
x = 0.5
y' = 9/(x²-10x+25) = ⁴⁄₉ (Ans)
(ii)
x = 0
y' = 9/(x²-10x+25) = ⁹⁄₂₅ (Ans)
(i)
y=0
x+4 = 5-x
x = 0.5
y' = 9/(x²-10x+25) = ⁴⁄₉ (Ans)
(ii)
x = 0
y' = 9/(x²-10x+25) = ⁹⁄₂₅ (Ans)
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Date Posted:
2 years ago
Y=(x + 4)/(5-x)
dy/dx = (1-2x)/(5-x)^2
At x-axis, y=0
So, (x+4)/(5-x)=0
So, x+4=0, x=-4
When, x=-4, dy/dx = [1-2(-4)]/[5-(-4)]^2 =9/81=1/9
dy/dx = (1-2x)/(5-x)^2
At x-axis, y=0
So, (x+4)/(5-x)=0
So, x+4=0, x=-4
When, x=-4, dy/dx = [1-2(-4)]/[5-(-4)]^2 =9/81=1/9
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