Question

How do you find the equation of the line tangent to `y=x^3-x` at (0,0)?

Answer 1

x + y = 0. The origin is a POI. The tangent crosses the curve. See all these features, in the Socratic graph.

Explanation:

`y'=3x^2-1, y''=6x and y'''=6 ne 0`.

At, (0, 0), y'=-1, y''=0. So, the origin is a POI.

The curve-crossing tangent at the POI (0, 0) is

`y-0=-1(x-0)`, giving

x + y = 0.

graph{(y-x^3+x)(y+x)=0 [-10, 10, -5, 5]}