Show an equation of the tangent to the curve 𝑦 = 𝑥^5 − 3𝑥^2 +𝑥 + 5 at the point (−1, 0) is given by 𝑦 = 12(𝑥 +1) by differentiation

Guest Dec 1, 2018

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Guest
Dec 1, 2018

#1**+1 **

Show an equation of the tangent to the curve 𝑦 = 𝑥^5 − 3𝑥^2 +𝑥 + 5 at the point (−1, 0) is given by 𝑦 = 12(𝑥 +1) by differentiation

y' = 5x^4 - 6x + 1

The slope of the tangent line at (-1,0) is

y' (-1) = 5(-1)^4 - 6(-1) + 1 = 12

The equation of the tangent line to (-1,0) is

y = 12 ( x - 1) + 0

y = 12(x - 1)

CPhill Dec 1, 2018