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

Jun 20, 2018

Equation of normal is $y = \frac{27}{50} x + \frac{2093}{225}$

Explanation:

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

$f \left(- 2\right) = \frac{2}{- 2 - 1} ^ 2 - 2 \cdot \left(- 2\right) + 4 = \frac{2}{9} + 4 + 4 = \frac{74}{9}$

The point is $\left(- 2 , \frac{74}{9}\right)$ at which normal to be drawn.

${f}^{'} \left(x\right) = - \frac{4}{x - 1} ^ 3 - 2$ . Slope of tangent at $x = - 2$ is

${f}^{'} \left(- 2\right) = - \frac{4}{- 2 - 1} ^ 3 - 2 = \frac{4}{27} - 2 = - \frac{50}{27}$

Slope of normal at $x = - 2$ is $m = \frac{27}{50}$

Equation of normal at point $\left(- 2 , \frac{74}{9}\right)$ is

$y - \frac{74}{9} = \frac{27}{50} \left(x + 2\right)$ or

$y = \frac{27}{50} x + \frac{27}{25} + \frac{74}{9}$ or

$y = \frac{27}{50} x + \frac{2093}{225}$ [Ans]