若直線y=5x+k與拋物線y=x^2+3x+4相切,求k
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若直線y=5x+k與拋物線y=x^2+3x+4相切,求k
2008-10-17 5:50 pm
回答 (3)
2008-10-26 6:01 pm
As follow AS~~~
圖片參考：http://www.photo-host.org/img/143338screenhunter_01_oct._26_09.56.gif
2008-10-26 10:07:57 補充：
更正:
x^2+3x-(k+1)=0________________________(*)
(*)判別式=0
(3)^2 -4(1)[-(k+1)]=0
9+4k+4=0
4k=-13
k=-13/4
圖片參考：http://www.photo-host.org/img/143338screenhunter_01_oct._26_09.56.gif
2008-10-26 10:07:57 補充：
更正:
x^2+3x-(k+1)=0________________________(*)
(*)判別式=0
(3)^2 -4(1)[-(k+1)]=0
9+4k+4=0
4k=-13
k=-13/4
2008-10-17 10:42 pm
Sub. y = 5x + k into the curve, we get
5x + k = x^2 + 3x + 4
x^2 - 2x + (4-k) = 0
For the line to be a tangent to the parabola, delta = 0, that is
4 - 4(4-k) = 0
1 - 4 + k = 0
k = 3.
5x + k = x^2 + 3x + 4
x^2 - 2x + (4-k) = 0
For the line to be a tangent to the parabola, delta = 0, that is
4 - 4(4-k) = 0
1 - 4 + k = 0
k = 3.
2008-10-17 7:22 pm
微分 y=x^2+3x+4,
dy/dx = 2x + 3
直線斜率=5
設切點為(x',y')
=> 2x' + 3 =5 => x' = 1
y' = x'^2 + 3x' + 4 = 1^2 + 3*1 +4 = 8
所以 y' = 5x' + k
=> 8 = 5*1 +k
=> k =3
dy/dx = 2x + 3
直線斜率=5
設切點為(x',y')
=> 2x' + 3 =5 => x' = 1
y' = x'^2 + 3x' + 4 = 1^2 + 3*1 +4 = 8
所以 y' = 5x' + k
=> 8 = 5*1 +k
=> k =3
收錄日期: 2021-04-23 20:38:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081017000051KK00343