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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

 Dec 1, 2018
edited by Guest  Dec 1, 2018
 #1
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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)

 

 

cool cool cool

 Dec 1, 2018
 #2
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ye i also did the part where u use f(x')= 12 but i dont get where the x-1 come from

Guest Dec 1, 2018
 #3
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ye i also did the part where u use f(x')= 12 but i dont get where the x-1 come from

laugh

 Dec 2, 2018

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