What is the equation of the line tangent to # f(x)=(x-2)/x # at # x=-3 #?

Question

1517 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T15:46:44

#y=2/9x+7/3#

#f(x)=(x-2)/x# , #A=RR#*#=(-oo,0)uu(0,+oo)#

#f'(x)=((x-2)'x-(x-2)(x)')/x^2=(x-(x-2))/x^2=#

#=(x-x+2)/x^2=2/x^2#

#f(-3)=5/3# , #f'(-3)=2/9#

#y-f(-3)=f'(-3)(x+3)# #<=>#

#y-5/3=2/9(x+3)# #<=>#

#y=2/9x+7/3#

enter image source here