1) slove the equation of sinx = -0.2 for 0 <= x <= 360 Deg
2) the curve y = x^2 + ax + b has a turning point at (1,3). find the value of a and b.
give me the full solution plz
Maths.. HELP!!!
2007-11-08 7:21 am
回答 (2)
2007-11-08 7:43 am
✔ 最佳答案
1) sinx = -0.2x = 180* + 11.5* or 360* - 11.5*
= 191.5* or 348.5* ( cor. to 1 d.p. )
2) y = x^2 + ax + b
dy / dx = 2x + a
For turning pt, dy / dx = 0, so
2 ( 1 ) + a = 0
a = -2
Then,
3 = 1^2 + ( 1 )( -2 ) + b
b = 4
So a = -2, b = 4.
參考: My Maths Knowledge
2007-11-08 7:45 am
(1) x = -11.53°
x = 348.46° or 191.54°
(2) y=x^2+ax+b
3 = 1^2 + a + b
a+b = 2 -----(1)
dy/dx = 2x + a
slope of tangent at (1,3) = 2 +a
Because this is the second degree curve, the slope at the minimum or maximum is zero
slop of tangent at (1,3) = 0 = 2 + a
a = -2
From (1): -2 + b = 2
=> b = 4
The equation is y = x^2 -2x +4
x = 348.46° or 191.54°
(2) y=x^2+ax+b
3 = 1^2 + a + b
a+b = 2 -----(1)
dy/dx = 2x + a
slope of tangent at (1,3) = 2 +a
Because this is the second degree curve, the slope at the minimum or maximum is zero
slop of tangent at (1,3) = 0 = 2 + a
a = -2
From (1): -2 + b = 2
=> b = 4
The equation is y = x^2 -2x +4
參考: My calculation
收錄日期: 2021-04-21 13:56:34
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