y=3x^2-x^3
P= (1,2)
y=
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.
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)
$.