Find the equation of the tangent line to the curve at the given point. $y = 1+2x-x^3$ at $(1,2)$

Question

I have the equation $y = 1+2x-x^3$ and the point $(1,2)$.

When I work it out I come up with the derivative of $2-3x^2$.

When I apply the point I come up with a slope of $-1$ and a tangent line of $y=4-x$.

Can someone work it out and confirm my answer or show me where I am going wrong?

We get that the tangent line has equation of shape $y=-x+b$. When $x=1$, we have $y=2$, so $b=3$. — André Nicolas, Oct 05 '13 at 23:07
I cannot keep track of numbers very well. I wrote it incorrectly and it almost cost me the problem. — wolfcall, Oct 05 '13 at 23:19

score 4 · Accepted Answer · answered Oct 05 '13 at 23:14

4

Try using the point-slope formula to obtain the equation of the line, given $m = -1$ and $(x_0, y_0) = (1,2)$, which is given by:

$$y - y_0 = m(x - x_0):$$

$$ y - 2 = -(x - 1) \iff y = -x + 3$$

answered Oct 05 '13 at 23:14

amWhy

Thank you. Thank you. Thank you. This problem has been driving me nuts for 2 days. I had a major brain fart and replaced a 1 with a 2. – wolfcall Oct 05 '13 at 23:18
You're welcome. You're work is fine, up to the equation of the line, so take heart. – amWhy Oct 05 '13 at 23:20
Also don't leave cal homework until last minute and cause me to rush it. And have no one to help me find my mistake. I love the internet. Good to know I understand the concept it is just rushing to rewrite it that caused the error. – wolfcall Oct 05 '13 at 23:28
@amWhy: Also needs a TU! +1 – Amzoti Oct 06 '13 at 00:27

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