Question

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

Answer 1

`y=1/5x+32/5`

Explanation:

The normal line will intersect the curve at `(-2,f(-2))`. Since `f(-2)=(-2)^2-(-2)=6`, we know the normal line passes through the point `(2,6)`.

To find the slope of the normal line, first find the slope of the tangent line at that point by finding the value of the derivative at `x=-2`. Then, since the normal line is perpendicular to the tangent line, take the opposite reciprocal of the slope of the tangent line.

Through the power rule, we see that `f'(x)=2x-1`. We then see that the slope of the tangent line at `x=-2` is `f'(-2)=2(-2)-1=-5`.

The slope of the normal line is then `-1/(-5)=1/5`.

Using the point `(-2,6)` and slope of `1/5`, we can write the equation of the line from `y=mx+b`:

`6=1/5(-2)+b" "=>" "b=32/5`

The normal line is:

`y=1/5x+32/5`

Graphed are `f(x)` and the normal line:

graph{(y-x^2+x)(-y+1/5x+32/5)=0 [-15.61, 12.86, -1.46, 12.77]}