Question

How do you find the equation of the tangent line to the graph of `y=x^3-3x^2+x` at the point (2,-2)?

Answer 1

I would find the derivative of your function, `y'(x)`, evaluate the derivative at the point of interest (`x=2`) and use it as the slope of the line through `(2,-2)`.
Step 1
Derive: `y'(x)=3x^2-6x+1`
Step2:
Evaluate: `y'(2)=12-12+1=1`
Step3:
Tangent: line through `(2,-2)` and slope `1`;
`y-(-2)=1*(x-2)`
`y+2=x-2`
`y=x-4` which is your tangent.

Graphically: