# What is the equation of the normal line of #f(x)=sqrt(2x^2-x)# at #x=-1#?

##### 1 Answer

#### Explanation:

From the power rule, we see that

So then:

#f(x)=sqrt(2x^2-x)=(2x^2-x)^(1/2)#

Then:

#f'(x)=1/2(2x^2-x)^(-1/2)*d/dx(2x^2-x)#

Through the power rule we see that

#f'(x)=1/2 1/(2x^2-x)^(1/2)*(4x-1)#

#f'(x)=(4x-1)/(2sqrt(2x^2-x))#

Now that we have the derivative, we need to write the equation of the normal line at

The line will pass through the point on the curve

#f(-1)=sqrt(2(-1)^2-(-1))=sqrt3#

So the normal line passes through the point

The normal line is perpendicular to the tangent line. The tangent line's slope is found through the value of the derivative at a point, which here is

Perpendicular slopes are opposite reciprocals, so the slope of the normal line is

First:

#f'(-1)=(4(-1)-1)/(2sqrt(2(-1)^2-(-1)))=(-5)/(2sqrt3)#

Thus the slope of the normal line is:

#-1/(f'(-1))=(2sqrt3)/5#

A line with slope

#y-y_1=m(x-x_1)#

The normal line has slope

#y-sqrt3=(2sqrt3)/5(x+1)#