# What is the equation of the line that is normal to #f(x)= xsqrt( 3x+2) # at # x=1 4 #?

##### 1 Answer

#### Explanation:

First, find the point of tangency.

#f(14)=14sqrt(42+2)=14sqrt44=28sqrt11#

The function and normal line will pass through the point

First, find the slope of the tangent line by differentiating the function.

#f(x)=x(3x+2)^(1/2)#

To differentiate, we will have to use the product rule.

#f'(x)=(3x+2)^(1/2)d/dx(x)+xd/dx(3x+2)^(1/2)#

The two derivatives present are:

#d/dx(x)=1#

Via the chain rule:

#d/dx(3x+2)^(1/2)=1/2(3x+2)^(-1/2)d/dx(3x+2)#

#=3/2(3x+2)^(-1/2)#

Plugging these both back in to find

#f'(x)=(3x+2)^(1/2)*1+x*3/2(3x+2)^(-1/2)#

#f'(x)=sqrt(3x+2)+(3x)/(2sqrt(3x+2))#

Getting a common denominator:

#f'(x)=(2(3x+2))/(2sqrt(3x+2))+(3x)/(2sqrt(3x+2))#

#f'(x)=(9x+4)/(2sqrt(3x+2))#

The slope of the tangent line is

#f'(14)=(126+4)/(2sqrt(42+2))=130/(2sqrt44)=65/(2sqrt11)#

The slope of the normal line is the opposite reciprocal of the slope of the tangent line, since the two lines are perpendicular.

The opposite reciprocal of

The equation of a line with slope

#y-28sqrt11=-(2sqrt11)/65(x-14)#

Graphed are the function and the normal line:

graph{(y-28sqrt11+(2sqrt11)/65(x-14))(y-xsqrt(3x+2))=0 [-72.4, 227.8, -13.4, 136.7]}