The function y=2x^2+ 1/x has a tangent line drawn at the
point (1,3). Find the equation of this tangent line.
Maths question (20 points)
2007-01-25 12:24 am
回答 (2)
2007-01-25 12:34 am
y = 2x2 + x-1
dy/dx = 4x - x-2
At (1, 3), dy/dx = 4 - 1 = 3
(y - 3) / (x-1) = 3
y - 3 = 3x - 3
3x - y = 0
so!
2007-01-24 16:35:30 補充：
dy/dx 是 curve 的斜率再用 point-slope form 去求公式
dy/dx = 4x - x-2
At (1, 3), dy/dx = 4 - 1 = 3
(y - 3) / (x-1) = 3
y - 3 = 3x - 3
3x - y = 0
so!
2007-01-24 16:35:30 補充：
dy/dx 是 curve 的斜率再用 point-slope form 去求公式
參考: me!
2007-01-25 12:30 am
Ｆｉｒｓｔ，ｕｓｅｙ ｅｑｕ ｉｎｔｏ １．
y = 2x2 + x-1
dy/dx = 4x - x-2
At (1, 3), dy/dx = 4 - 1 = 3
(y - 3) / (x-1) = 3
y - 3 = 3x - 3
3x - y = 0
Ｓo,the solution is3x - y = 0
y = 2x2 + x-1
dy/dx = 4x - x-2
At (1, 3), dy/dx = 4 - 1 = 3
(y - 3) / (x-1) = 3
y - 3 = 3x - 3
3x - y = 0
Ｓo,the solution is3x - y = 0
收錄日期: 2021-04-23 20:02:58
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