Find an equation of the tangent line to the curve y = x^3-3x+1 at the point (1,-1)

Question:

Find an equation of the tangent line to the curve {eq}y = x^3-3x+1 {/eq} at the point (1,-1)

The Tangent Line to a Curve:

The tangent of the curve {eq}f(x) {/eq} is {eq}y=mx+b {/eq} at a point {eq}(x_{1}, y_{1}). {/eq}

The slope {eq}(m) {/eq} is the derivative of the function at that same point. We need to find out {eq}m, b {/eq} and plug in the formula to get the desired solution.

Answer and Explanation:

First, we need to compute the slope of the given curve:

{eq}m=\dfrac{\ dy }{\ dx}= \dfrac{\ d }{\ dx} \left(x^3 - 3x +1 \right) = 3x^2-3 {/eq}

Plu...

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from PSAT Prep: Tutoring Solution

Chapter 10 / Lesson 13
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