How do you find the equation of the tangent and normal line to the curve #y=(x+1)/(x-3) at x=2?

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Answer 1 · 2016-08-06T08:57:32

Slope of tangent is #4x+y-5=0# and slope of normal is #x-4y-14=0#

At #x=2#, #y=(2+1)/(2-3)=3/-1=-3#, hence, we have to find equation of the tangent and normal at curve #y=(x+1)/(x-3)# at #(2,-3)#.

For finding equation, we need to find value of derivative of #y=(x+1)/(x-3)# at #(2,-3)#, for which we use quotient rule.

#f'(x)=(dy)/(dx)=((x-3)xx1-(x+1)xx1)/(x-3)^2#

= #(x-3-x-1)/(x-3)^2=-4/(x-3)^2#

and slope of tangent at #x=2# is #f'(2)=-4/(2-3)^2=-4# and slope of normal is #-1/-4=1/4#.

Now using point slope form #(y-y_1)=m(x-x_1)#

Slope of tangent is #(y-(-3))=-4(x-2)# or #y+3=-4x+8# or #4x+y-5=0#

Slope of normal is #(y-(-3))=1/4(x-2)# or #4y+12=x-2# or #x-4y-14=0#

graph{(y-(x+1)/(x-3))(4x+y-5)(x-4y-14)=0 [-16.5, 23.5, -13.68, 6.32]}