Question

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

Answer 1

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

Explanation:

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)`