Write the equation of the tangent to the graph of the function y = x ^ 2 (2-3x) at the point Xo = -1

The equation of the tangent to the graph of the function y = f (x) at the point x0 is determined by the formula y = f (x0) + f ‘(x0) (x-x0), where f (x0) is the value of the function f (x) at the point x0, f ‘(x0) is the value of the derivative of the function f (x) at the point x0.
1) First, let’s simplify the expression: y = x ^ 2 (2-3x) = 2x ^ 2-3x ^ 3;
2) Find the derivative: f ‘(x) = 4x-9x ^ 2;
3) Find the value of the derivative of the function f (x) at the point x0: f ‘(x0) = 4 * (x0) -9 * (x0) ^ 2 = 4 * (- 1) -9 * (- 1) ^ 2 = -4-9 * 1 = -1-9 = -13;
4) Find the value of the function f (x) at the point x0: f (x0) = 2 * (x0) ^ 2-3 * (x0) ^ 3 = 2 * (- 1) ^ 2-3 * (- 1) ^ 3 = 2 * 1-3 * (- 1) = 2 + 3 = 5;
5) Substituting the found values of f (x0) and f ‘(x0) into the tangent equation, we get: y = f (x0) + f’ (x0) (x-x0) = 5-13 * (x + 1) = 5 -13x-13 = -13x-8.