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secondary 4 | A Maths
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pls help
Date Posted: 1 year ago
Views: 199
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y = px³+q → (1)
dy/dx = 3px² → (2)
2y+5x–9 = 0
y = 9/2–5/2x → (3)
dy/dx when x=–1 = –1/(–5/2) = 2/5
Sub x=–1 into (2) and dy/dx = 2/5
2/5 = 3p(–1)²
p = 2/15
Sub (3) into (1) and x=–1
9/2–5/2(–1) = 2/15(–1)³+q
q = 107/15
dy/dx = 3px² → (2)
2y+5x–9 = 0
y = 9/2–5/2x → (3)
dy/dx when x=–1 = –1/(–5/2) = 2/5
Sub x=–1 into (2) and dy/dx = 2/5
2/5 = 3p(–1)²
p = 2/15
Sub (3) into (1) and x=–1
9/2–5/2(–1) = 2/15(–1)³+q
q = 107/15
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