Equate the tangent to the graph of the function y = x ^ 3 – 2x ^ 2 – 3x + 5 at the point with the abscissa x = -2.

The equation of the tangent to the graph of the function at the point x0 has the form: y = f (x0) + (f (xo)) ‘(x-x0).
Let’s find the derivative of the function:
(f (x)) ‘= (x ^ 3 – 2x ^ 2 – 3x + 5)’ = 3 * x ^ 2-4x-3;
(f (-2)) ‘= 3 * (- 2) ^ 2-4 (-2) -3 = 3 * 4 + 8-3 = 17;
since f (-2) = (- 2) ^ 3-2 (-2) ^ 2-3 (-2) + 5 = -8-8 + 6 + 5 = -5;
We get the required equation:
y = -5 + 17 (x + 2).