# What is the equation of the line tangent to f(x)= sqrt(3-2x)  at x=-2?

Sep 26, 2016

Find the first derivative to get the slope of the tangent line. Then use the point slope form to get the general equation of the tangent line.

#### Explanation:

First, find the first derivative because this will give you the slope of the tangent line.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(3 - 2 x\right)}^{- \frac{1}{2}} \left(- 2\right)$

Or simply:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(3 - 2 x\right)}^{- \frac{1}{2}}$

Substituting $x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(3 + 4\right)}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(- 7\right)}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(7\right)}^{- \frac{1}{2}} = \frac{1}{7} ^ \left(\frac{1}{2}\right) = \frac{1}{\sqrt{7}}$#

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{7}}{7}$

The slope of the tangent line is $\frac{\sqrt{7}}{7}$.

You have $x = - 2$.

Solving for $y$:

$y = \sqrt{3 - 2 \left(- 2\right)}$
$y = \sqrt{3 + 4}$
$y = \sqrt{7}$

Using the point-slope form:

$y - \sqrt{7} = \frac{\sqrt{7}}{7} \left(x - \left(- 2\right)\right)$
$y - \sqrt{7} = \frac{\sqrt{7}}{7} \left(x + 2\right)$

You can put it in slope-intercept form by solving for $y$:

$y = \frac{\sqrt{7}}{7} x + \frac{9 \sqrt{7}}{7}$