# What is the equation of the line that is normal to f(x)= x-sqrt( 3x+2)  at  x=4 ?

$1.669 x + y - 6.935 = 0$

#### Explanation:

$f \left(x\right) = x - \sqrt{3 x + 2}$
Let y=f(x)
$y = x - \sqrt{3 x + 2}$
At x=4
$y = 4 - \sqrt{3 \times 4 + 2} = 0.258$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{1}{2 \sqrt{3 x + 2}} \left(3\right)$
$| \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = 4} = 1 - \frac{3}{2 \sqrt{3 \times 4 + 2}} = 0.599$

The tangent and normal are orthogonal to each other
Slope of the tangent is 0.599
m1m2=-1gives
Slope of the normal is -1/0.599=-1.669
The normal passes through the point (4,0.258)
Equation of the line passing through a point having a slope is given by

$\frac{y - 0.258}{x - 4} = - 1.669$
$\frac{y - 0.258}{x - 4} + 1.669 = 0$
Simplifying

$y - 0.258 + 1.669 \left(x - 4\right) = 0$

$y - 0.258 + 1.669 x - 6.676 = 0$
Expressing in the standard form ax+by+c=0
we have
$1.669 x + y - 6.935 = 0$