Equation of Tangent Line

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jpnov

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Sep 28, 2010
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Find an equation of the tangent line to the curve at the point (-1, 3).
y = 3x^3 - 6x

I'm not so sure how to do this. What I did is use

m = lim (x->a) [f(x) - f(a)] / (x-a)

subbed in the x,y vals given

got: lim (x->-1) [3(x^3-2x-1)]/(x+1)

Not sure what to do from here. If I sub in -1 I'll just get 0/0. And after I get the slope, i just need to do y - 3 = m(x-3) right?
 
jpnov said:
Find an equation of the tangent line to the curve at the point (-1, 3).

y = 3x^3 - 6x

got: lim (x->-1) [3(x^3-2x-1)]/(x+1)

Factor x^3 - 2x - 1. Then you can cancel the denominator.

after I get the slope, i just need to do y - 3 = m(x-3) right?

Yes, use the Point-Slope Formula, but correct your value above for the given x.
Click to expand...

Please show your work, if you need specific help.

Cheers ~ Mark 8-)

 
Thanks Mark. How do i factor that o_O

I tried factoring for a while (before posting) and figured that wasn't the right path. What does that factor into? Sorry...
 


Have you learned how to divide one polynomial by another (either by longhand or synthetically) ?

 
A long time ago I think... I'm so rusty on my algebra. This is causing me so much trouble in Calculus.
 


I think there's at least one short video at khanacademy.org on polynomial division.

You could search keywords "polynomial division" at purplemath.com for written lessons.

The factorization is 3(x + 1)(x^2 - x - 1)

 
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