y=3x^2-x^3

P= (1,2)

y=

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- Feb 12th 2008, 11:51 AMplstevensFind the equation of the tangent line to the following curve at the given point P. Il
y=3x^2-x^3

P= (1,2)

y= - Feb 12th 2008, 02:32 PMmr fantastic
The general equation of a line is $\displaystyle y - y_1 = m(x - x_1)$ where m is the gradient and $\displaystyle (x_1, y_1)$ is a known point on the line.

In your question, a known point is (1, 2)

To get m, find $\displaystyle \frac{dy}{dx}$ of $\displaystyle y = 3x^2 - x^3$ and evaluate at x = 1. - Feb 12th 2008, 05:47 PMplstevens
I still don't understand

- Feb 12th 2008, 05:56 PMmr fantastic
What

*exactly*don't you understand?

I've said that in $\displaystyle

y - y_1 = m(x - x_1)

$, $\displaystyle

(x_1, y_1)

$ is a known point.

A known point is (1, 2). It's given to you on a platter.

That means you substitute $\displaystyle x_1 = 1$ and $\displaystyle y_1 = 2$ into $\displaystyle

y - y_1 = m(x - x_1)

$.

Now get the value of m. How I said to in my first reply. Then substitute the value into $\displaystyle

y - y_1 = m(x - x_1)

$. - Feb 12th 2008, 06:34 PMplstevens
is it y=3x-1

- Feb 12th 2008, 08:00 PMmr fantastic