Question

How do you find the equation of the line tangent to the curve `y=x^3` at the point (-1,-1)?

Answer 1

First you evaluate the derivative of your function `y'(x)`:
Then you evaluate the derivative at `x_0=-1`, or `y'(-1)`, which is the slope `m` of your tangent line.
Finally find the equation of the line using:
`y-y_0=m(x-x_0)`
With: `x_0=-1` and `y_0=-1`
Try by yourself and if it doesn' work look below for the maths:
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Solution:
`y'(x)=3x^2`
`y'(-1)=3=m`
Line: `y=3x+2`