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

1. ## 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=

2. 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.

3. I still don't understand

4. 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)
$
.

5. is it y=3x-1

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