the straight line y=8x+ktouches the parabola y=x^2+20 ar a point,
find the value of k
點find??
我想知個方法
thx
我想問F.4數學(english)(聯立方程)
2007-06-03 7:03 am
回答 (3)
2007-06-03 7:50 am
✔ 最佳答案
PUT 8x+k=x^2+20i.e. 0=x^2-8x+20-k ---(1)
since the straight line touches the curve
so delta of (1) =0
i.e. 0=b^2-4ac
0=64-4(20-k)
64-80+4k=0
4k-16=0
4k=16
k=4
參考: meself
2007-06-03 7:16 am
8x + kt = x^2 + 20
x^2 - 8x + (20 - k) = 0 ----- (1)
solve delta=0
找出x
(應該得1個答案)
sub into (1)
找出k :)
x^2 - 8x + (20 - k) = 0 ----- (1)
solve delta=0
找出x
(應該得1個答案)
sub into (1)
找出k :)
參考: 自己的腦
2007-06-03 7:16 am
首先要將 y=x^2+20 代入y=8x+k
x^2+20=8x+k
x^2-8x+(20-k)=0
跟住用b^2-2ac
(-8)^2-2(1)(20-k)=0
64-40-2k=0
k=12//
x^2+20=8x+k
x^2-8x+(20-k)=0
跟住用b^2-2ac
(-8)^2-2(1)(20-k)=0
64-40-2k=0
k=12//
參考: 自己
收錄日期: 2021-04-23 21:11:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070602000051KK05188