What is the equation of the normal line of $f(x)=(x+2)/(x-4)$ at $x=3$ ?

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Answer 1 · 2016-04-05T05:52:51

Equation of the normal line is $x-2y-13=0$

At $x=3$ , $f(3)=(3+2)/(3-4)=5/-1=-5$

Hence normal is desired at point $(3,-5)$ on the curve.

Slope of tangent is given by $f'(x)$

= $((x-4)*d/dx(x+2)-(x+2)*d/dx(x-4))/(x-4)^2$

= $((x-4)*1-(x+2)*1)/(x-4)^2=(x-4-x-2)/(x-4)^2=-2/(x-4)^2$

Hence slope of tangent at $(3,-5)$ is $-2/(3-4)^2=-2$ .

Hence slope of normal would be $-1/-2=1/2$

And equation of the normal line using point slope form is

$(y-(-5))=1/2(x-3)$ or

$2*(y+5)=(x-3)$ or $x-2y-13=0$

What is the equation of the normal line of f(x)=(x+2)/(x-4)f(x)=(x+2)/(x-4) at x=3x=3?