Find the gradient?
2007-11-25 11:08 am
Find the gradient of the curve y = [ 2 - x ] / x^2 at the point where the curve touches the x axis/
回答 (2)
2007-11-25 11:15 am
✔ 最佳答案
gradient of any curve is given by its first derivative i.e. dy/dx.dy/dx = [(-1)x^2 - 2x(2-x)] / (x^2)^2
= [x^2-4x]/x^4
= [x-4]/x^3
at the x axis, y = 0 , putting this into y = [ 2 - x ] / x^2 , we obtain x=2
put this value into dy/dx
-->dy/dx= [2-4]/(2)^3 = -1/4
參考: Myself
2007-11-25 12:03 pm
y = [ 2 - x ] / x^2
when y = 0 . . . x = 2
gradient
y ' = [ x^2 (-1) - (2-x)2x ] / x^4
y ' = [ 2^2 (-1) - (2-2)2(2) ] / 2^4
y ' = - 4 / 16
y ' = - 1/4
when y = 0 . . . x = 2
gradient
y ' = [ x^2 (-1) - (2-x)2x ] / x^4
y ' = [ 2^2 (-1) - (2-2)2(2) ] / 2^4
y ' = - 4 / 16
y ' = - 1/4
收錄日期: 2021-05-03 12:24:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071125030800AAf6VdG