Question

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

Answer 1

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.

`dy/dx= 1/2(3-2x)^(-1/2) (-2)`

Or simply:

`dy/dx=-(3-2x)^(-1/2)`

Substituting `x`:

`dy/dx = -(3+4)^(-1/2)`

`dy/dx = - (-7)^(-1/2)`

`dy/dx = (7)^(-1/2) = 1/7^(1/2) = 1/sqrt7`#

`dy/dx = sqrt(7)/7`

The slope of the tangent line is `sqrt(7)/7`.

You have `x= -2`.

Solving for `y`:

`y = sqrt(3-2(-2))`
`y = sqrt (3+4)`
`y = sqrt (7)`

Using the point-slope form:

`y-sqrt(7)= sqrt(7)/7(x-(-2))`
`y-sqrt(7) = sqrt (7)/7(x+2)`

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

`y = sqrt7/7x+(9sqrt7)/7`