What is the equation of the line that is normal to f(x)= (2-x)/sqrt( 2x+2)  at  x=2 ?

$y = \setminus \sqrt{6} \left(x - 2\right)$

Explanation:

The given function

$f \left(x\right) = \setminus \frac{2 - x}{\setminus \sqrt{2 x + 2}}$

setting $x = 2$ in the given function, we get $y$-coordinate of point

$y = f \left(2\right)$

$= \setminus \frac{2 - 2}{\setminus \sqrt{2 \setminus \cdot 2 + 2}} = 0$

Now, differentiating above equation w.r.t. $x$ we get slope $\frac{\mathrm{dy}}{\mathrm{dx}}$ of tangent as follows

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} f \left(x\right)$

$= \setminus \frac{\setminus \sqrt{2 x + 2} \left(- 1\right) - \left(2 - x\right) \setminus \frac{1}{\setminus \sqrt{2 x + 2}}}{{\left(\setminus \sqrt{2 x + 2}\right)}^{2}}$

$= \setminus \frac{- \left(2 x + 2\right) - \left(2 - x\right)}{{\left(2 x + 2\right)}^{\frac{3}{2}}}$

$= \setminus \frac{- x - 4}{{\left(2 x + 2\right)}^{\frac{3}{2}}}$

setting $x = 2$ in above equation, we get slope of tangent at $\left(2 , 0\right)$

$= \setminus \frac{- 2 - 4}{{\left(2 \setminus \cdot 2 + 2\right)}^{\frac{3}{2}}}$

$= - \frac{1}{\setminus} \sqrt{6}$

hence the slope $m$ of normal at the same point

$= - \frac{1}{- \frac{1}{\setminus} \sqrt{6}} = \setminus \sqrt{6}$

hence the equation of normal at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(2 , 0\right)$ & having slope $m = \setminus \sqrt{6}$ is given by following formula

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

$y - 0 = \setminus \sqrt{6} \left(x - 2\right)$

$y = \setminus \sqrt{6} \left(x - 2\right)$