作业帮本人在国外读a levelfind the equation of the tangent to the curve y=x2-2x which is prependicular to the line y=x-3说错了 是这一题 find the equation of the normal to the curve at te point with the given x-coordinate y=1-x2 where x=0
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本人在国外读a levelfind the equation of the tangent to the curve y=x2-2x which is prependicular to the line y=x-3说错了 是这一题 find the equation of the normal to the curve at te point with the given x-coordinate y=1-x2 where x=0
本人在国外读a level
find the equation of the tangent to the curve y=x2-2x which is prependicular to the line y=x-3
说错了 是这一题 find the equation of the normal to the curve at te point with the given x-coordinate y=1-x2 where x=0
本人在国外读a levelfind the equation of the tangent to the curve y=x2-2x which is prependicular to the line y=x-3说错了 是这一题 find the equation of the normal to the curve at te point with the given x-coordinate y=1-x2 where x=0
我也在英国读书
gradient=dy/dx=-2x
when x=0,dy/dx=0,gradient of the curve=0(parallel to the x-axis)
the gradient of the normal will be parallel to the y-axis
& because it passes the point (0,1)
therefore the equation of the normal is y=0
垂直于y=x-3
所以所求直线不妨设为y=-x+t
与y=x2-2x相切，所以-x+t=x2-2x的判别式为0
即1+4t=0，t=-1/4
所以所求直线为y=-x-1/4