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]}