Find the equation of the tangent to the curve with equation y = x 3 +2x 2 -3x + 2at at the point where x = 1.

Find the equation of the tangent to the curve with equation y = x^3 +2x^2 -3x + 2at at the point where x = 1.

The derivative (slope) at any point on the curve is given by:

y ' = 3x^2 + 4x - 3

And at x = 1 , f(1) = y = (1)^3 + 2(1)^2 - 3(1) + 2 =  2

And the slope at x = 1  ..... =  3(1)^2 + 4(1) - 3  = 4

And the equation of the tangent line at this point is given by

y - 2 = 4(x - 1)

y = 4x - 4 + 2

y = 4x - 2

Here's a graph of the situation :   https://www.desmos.com/calculator/a5uden7nol