Find the equation of the tangent to the curve y= x 3 +3x 2 ​-5x-7 at the point where x = 2?

Find the equation of the tangent to the curve y= x3+3x2-5x-7  at the point where x = 2?

$\begin{array}{rcll} y = f(x) &=& x^3+3x^2-5x-7 \\ f(2) &=& 2^3+3\cdot 2^2-5\cdot 2-7 \\ f(2) &=& 8+12-10-7 \\ f(2) &=& 3 \\\\ y' = f'(x) &=& 3x^2+6x-5\\ f'(2) &=& 3\cdot 2^2+6\cdot 2-5\\ f'(2) &=& 12+12-5\\ f'(2) &=& 19\\ \end{array}$

The equation of the tangent:

$\begin{array}{rcll} y &=& mx+b \\\\ y_p &=& m\cdot x_p + b \qquad x_p = 2 \qquad y_p = 3 \qquad m = f'(2) = 19\\ 3 &=& 19\cdot 2 +b \\ b &=& 3-38 \\ b &=& -35\\\\ y_{\text{tangent}} &=& 19x-35 \end{array}$

First: What's your equation?

Its written in question

Hmm.. This is a little bit confusing for me! 

I knew I could count on Heureka :) <3

Differentiation will be applied i guess?

Told you we could count on him!! :)

Great Job!

Thx guys :)