How do you find the equation of a line tangent to the function #y=1/x# at x=-1?

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Answer 1 · 2016-12-26T19:39:01

#y=-x-2#

To find the eqn of the tangent you need

1) the gradient #m# at the point in question.

2) The coordinates# (x_1,y_1)# at the point in question.

3 Use # y-y_1=m(x-x_1)#

1) gradient at # x=-1#

#y=1/x=>y=x^-1#

#(dy)/(dx)=-x^-2=-1/x^2#

#m=((dy)/(dx))_(x=-1)=-1/(-1)^2=-1#

2) Coordinates# (x_1,y_1)#

#y=1/-1=-1#

# (x_1,y_1)=(-1,-1)#

3) #y-y_1=m(x-x_1)#

#y--1=-1(x--1)#

#y+1=-x-1#

#y=-x-2#

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