Applications of Differentiation
2007-05-11 5:49 am
It is given that the curve y=x^2-2x+4. Find the points on the curve at which the tangent to the curve passes through the origin.
回答 (2)
2007-05-11 6:10 am
✔ 最佳答案
kaihong411 錯鬼左有兩點(2,4) 同(-2,12)
2007-05-11 13:23:33 補充：
設果點為(m,n)dy/dx = 2x - 2tangent (y - n) / (x - m) = 2m - 2經(0,0) 即 n/m = 2m - 2n = 2m^2 - 2m --------------- (1)(m,n) 在曲線上面, 即 n = m^2 - 2m + 4 ----------------- (2)由(1)同(2)得 2m^2 - 2m = m^2 - 2m + 4m^2 = 4m = 2 或 m = -2m = 2 時, n = 4m = -2 時, n = 12
2007-05-11 9:11 pm
y = x^2 - 2x + 4
dy/dx = 2x - 2
let the point(s) be (a,b) on the curve
b = a^2 - 2a + 4
dy/dx|(x = a) = 2a - 2
(b - 0) / (a - 0) = 2a - 2
b = 2a^2 - 2a
2a^2 - 2a = a^2 - 2a + 4
a^2 - 4 = 0
(a + 2)(a - 2) = 0
a = -2 or a = 2
dy/dx = 2x - 2
let the point(s) be (a,b) on the curve
b = a^2 - 2a + 4
dy/dx|(x = a) = 2a - 2
(b - 0) / (a - 0) = 2a - 2
b = 2a^2 - 2a
2a^2 - 2a = a^2 - 2a + 4
a^2 - 4 = 0
(a + 2)(a - 2) = 0
a = -2 or a = 2
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