What is the equation of the normal line of f(x)=(x-9)^2/(x-1) at x=-2?

The normal line

$y = \frac{9 x}{55} - \frac{6601}{165}$

Explanation:

The given curve is

$y = {\left(x - 9\right)}^{2} / \left(x - 1\right)$
at the point $x = - 2$

Let us solve for the point $\left({x}_{1} , {y}_{1}\right)$ first

Let ${x}_{1} = - 2$

${y}_{1} = {\left({x}_{1} - 9\right)}^{2} / \left({x}_{1} - 1\right) = {\left(- 2 - 9\right)}^{2} / \left(- 2 - 1\right) = \frac{121}{-} 3 = - \frac{121}{3}$

Our point $\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - \frac{121}{3}\right)$

Solve for the slope of the normal line ${m}_{n} = - \frac{1}{m}$

$m = \frac{d}{\mathrm{dx}} \left({\left(x - 9\right)}^{2} / \left(x - 1\right)\right) = \frac{\left(x - 1\right) \cdot 2 \left(x - 9\right) - {\left(x - 9\right)}^{2} \cdot 1}{x - 1} ^ 2$

substitute $x = - 2$

$m = \frac{\left(- 2 - 1\right) \cdot 2 \left(- 2 - 9\right) - {\left(- 2 - 9\right)}^{2} \cdot 1}{- 2 - 1} ^ 2 = - \frac{55}{9}$

${m}_{n} = - \frac{1}{m}$

${m}_{n} = - \frac{1}{- \frac{55}{9}}$

${m}_{n} = \frac{9}{55}$

Let us solve for the normal line using ${m}_{n}$ and $\left({x}_{1} , {y}_{1}\right)$

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

$y - \left(- \frac{121}{3}\right) = \frac{9}{55} \left(x - \left(- 2\right)\right)$

$y = \frac{9 x}{55} - \frac{6601}{165}$

Kindly see the graphs of $y = {\left(x - 9\right)}^{2} / \left(x - 1\right)$ and the normal line $y = \frac{9 x}{55} - \frac{6601}{165}$

God bless....I hope the explanation is useful.