What is the equation of the line normal to $f(x)=2/(x-1)^2-2x+4$ at $x=-2$ ?

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Answer 1 · 2018-06-20T12:54:47

Equation of normal is $y = 27/50 x+ 2093/225$

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

$f(-2) =2/(-2-1)^2-2 *(-2) +4 = 2/9+4+4=74/9$

The point is $(-2, 74/9)$ at which normal to be drawn.

$f^'(x) =-4/(x-1)^3-2$ . Slope of tangent at $x=-2$ is

$f^'(-2) =-4/(-2-1)^3-2 = 4/27-2= -50/27$

Slope of normal at $x=-2$ is $m= 27/50$

Equation of normal at point $(-2, 74/9)$ is

$y - 74/9 = 27/50 (x+2)$ or

$y = 27/50 x+ 27/25 +74/9$ or

$y = 27/50 x+ 2093/225$ [Ans]

What is the equation of the line normal to f(x)=2/(x-1)^2-2x+4 f(x)=2/(x-1)^2-2x+4 at x=-2 x=-2?