# What is the equation of the line normal to  f(x)=-x/sqrt((x-5)(x+2) at  x=6?

Jan 8, 2016

The normal line is the line perpendicular to the tangent line. Take the derivative, plug in your value to get the slope of the tangent, and then take the negative reciprocal to get the slope of the perpendicular.

#### Explanation:

To take the derivative, I suggest using the product rule, so rewrite the function:
$f \left(x\right) = - x {\left[\left(x - 5\right) \left(x + 2\right)\right]}^{-} \left(\frac{1}{2}\right)$

so

$f ' \left(x\right) = - 1 {\left[{x}^{2} - 3 x - 10\right]}^{- \frac{1}{2}} + \left(- x\right) \left[\left(- \frac{1}{2}\right) {\left({x}^{2} - 3 x - 10\right)}^{- \frac{3}{2}} \cdot \left(2 x - 3\right)\right]$

$f ' \left(6\right) = - 1 \cdot {8}^{- \frac{1}{2}} + \left(- 6 \cdot - \frac{1}{2}\right) \cdot {8}^{- \frac{3}{2}} \cdot 9$
$f ' \left(6\right) = - \left(\frac{1}{2 \sqrt{2}}\right) + \frac{27}{16 \sqrt{2}}$
$f ' \left(6\right) = \frac{- 8 + 27}{16 \cdot \sqrt{2}}$
$f ' \left(6\right) = \frac{19}{16 \sqrt{2}}$

so the slope of the normal line (perpendicular to tangent) is

$- \frac{16 \sqrt{2}}{19}$