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secondary 4 | A Maths
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Date Posted: 1 year ago
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i)
y = –x²+3x–5 → (1)
dy/dx = –2x+3 → (2)
Sub x = 2 into (1)
y = –(2)²+3(2)–5 = –3
Sub x = 2 into (2)
dy/dx = –2(2)+3 = –1
Equation of tangent at P: y = –x+c
When x = 2 and y = –3
–3 = –2+c
c = –1
Equation of tangent at P: y = –x–1
When x = 0
y = –0–1 = –1
When y = 0
0 = –x–1
x = –1
coordinates of A = (–1, 0)
coordinates of B = (0, –1)
Area of AOB = ½(1)(1) = ½units²
ii)
Gradient of normal at Q = –1
Gradient of tangent at Q = –1/–1 = 1
Sub dy/dx = 1 into (2)
1 = –2x+3
x = 1
Sub x = 1 into (1)
y = –(1)²+3(1)–5 = –3
coordinates of Q = (1, –3)
y = –x²+3x–5 → (1)
dy/dx = –2x+3 → (2)
Sub x = 2 into (1)
y = –(2)²+3(2)–5 = –3
Sub x = 2 into (2)
dy/dx = –2(2)+3 = –1
Equation of tangent at P: y = –x+c
When x = 2 and y = –3
–3 = –2+c
c = –1
Equation of tangent at P: y = –x–1
When x = 0
y = –0–1 = –1
When y = 0
0 = –x–1
x = –1
coordinates of A = (–1, 0)
coordinates of B = (0, –1)
Area of AOB = ½(1)(1) = ½units²
ii)
Gradient of normal at Q = –1
Gradient of tangent at Q = –1/–1 = 1
Sub dy/dx = 1 into (2)
1 = –2x+3
x = 1
Sub x = 1 into (1)
y = –(1)²+3(1)–5 = –3
coordinates of Q = (1, –3)
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