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

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Answer 1 · 2016-09-26T12:35:02

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.

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#