# What is the equation of the line normal to f(x)=(x-1)^2-2x+5 at x=-2?

$x - 8 y + 146 = 0$

#### Explanation:

The slope $f ' \left(x\right)$ of tangent to the curve: $f \left(x\right) = {\left(x - 1\right)}^{2} - 2 x + 5$ is given by differentiating given function w.r.t. $x$ as follows

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({\left(x - 1\right)}^{2} - 2 x + 5\right)$

$= 2 \left(x - 1\right) - 2$

$= 2 x - 4$

hence the slope of tangent at $x = - 2$ is

$f ' \left(- 2\right) = 2 \left(- 2\right) - 4$

$= - 8$

hence the slope of normal to the curve at the same point $x = - 2$ is

$= - \frac{1}{- 8}$

$= \frac{1}{8}$

Now, substituting $x = - 2$ in the given function: $f \left(x\right) = {\left(x - 1\right)}^{2} - 2 x + 5$ we get $y$ coordinate of the point as follows

$f \left(- 2\right) = {\left(- 2 - 1\right)}^{2} - 2 \left(- 2\right) + 5$

$= 18$

Hence, the equation of normal to the curve at $\left(- 2 , 18\right) \setminus \equiv \left({x}_{1} , {y}_{1}\right)$ & having a slope $m = \frac{1}{8}$ is given as

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 18 = \frac{1}{8} \left(x - \left(- 2\right)\right)$

$8 y - 144 = x + 2$

$x - 8 y + 146 = 0$