Re: [微積] 求切線
※ 引述《KOREALee (韓國最高)》之銘言：
: Find an equation of the tangent line to the curve
: x-1
: y = ----- that is parallel to the line x - 2y = 2
: x+1
2
y'(x) = m(x) = ---------
(x+1)^2
又 x - 2y = 2 之斜率為 (1/2)
令 m(x) = 1/2, 得 x = -3 and 1
有二切點 (-3,2), (1,0) 帶回點斜式
=> (y-2) = (1/2)(x+3)
(y-0) = (1/2)(x-1)
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