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

Question

Emily Ammerman · Accepted Answer

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

From the power rule, we see that #d/dxx^n=nx^(n-1)#. The chain rule tells us that #d/dxu^n=n u^(n-1)*(du)/dx#.

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 #d/dx(2x^2-x)=4x-1#, so:

#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 #x=-1#. To write the equation a line, we need to know two things: a point that the line passes through and the slope of the line.

The line will pass through the point on the curve #f# at #x=-1#.

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

So the normal line passes through the point #(-1,sqrt3)#.

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 #f'(-1)#.

Perpendicular slopes are opposite reciprocals, so the slope of the normal line is #-1/(f'(-1))#.

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 #m# passing through point #(x_1,y_1)# can be expressed as:

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

The normal line has slope #(2sqrt3)/5# and passes through #(-1,sqrt3)#, so the normal line is:

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

Luke Alvarez · Answer

undefined The equation of the normal line of f(x)=sqrt(2x^2-x) at x=-1 is y = -2x - 1.