# Find the equation of the tangent line to the following curve at the given point P. Il

• Feb 12th 2008, 11:51 AM
plstevens
Find 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 PM
mr fantastic
Quote:

Originally Posted by plstevens
y=3x^2-x^3

P= (1,2)

y=

The general equation of a line is $y - y_1 = m(x - x_1)$ where m is the gradient and $(x_1, y_1)$ is a known point on the line.

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

To get m, find $\frac{dy}{dx}$ of $y = 3x^2 - x^3$ and evaluate at x = 1.
• Feb 12th 2008, 05:47 PM
plstevens
I still don't understand
• Feb 12th 2008, 05:56 PM
mr fantastic
Quote:

Originally Posted by plstevens
I still don't understand

What exactly don't you understand?

I've said that in $
y - y_1 = m(x - x_1)
$
, $
(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 $x_1 = 1$ and $y_1 = 2$ into $
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 $
y - y_1 = m(x - x_1)
$
.
• Feb 12th 2008, 06:34 PM
plstevens
is it y=3x-1
• Feb 12th 2008, 08:00 PM
mr fantastic
Quote:

Originally Posted by plstevens
is it y=3x-1

(Clapping)