Write the equation of the tangent to the graph of the function: F (x) = 2x ^ 3-5 at the point x0 = -2.

Let us compose the equation of the tangent tangent according to the formula y = f (x0) + f ′ (x0) * (x – x0).
To do this, find f (x0), that is, f (-2) and get:
f (-2) = 2 * (-2) ^ 3 – 5 = 2 * (-2) * (-2) – 5 = -4 * (-2) – 5 = 8 – 5 = 3.
Next, we find the derivative of this function:
f ′ (x) = (2x ^ 3 – 5) ′ = 3x ^ 2.
Therefore, the value of the derivative of this function at the point x0 = -2:
f ′ (-2) = 3 * (-2) ^ 2 = 3 * (-2) * (-2) = -6 * (-2) = 18.
Then the equation of the tangent to the graph of the function f (x) = 2x ^ 3 – 5 at the point x0 = -2 will have the form:
y = 3 + 18 * (x – (-2)) = 3 + 18 * (x + 2) = 3 + 18 * x + 18 * 2 = 3 + 18 * x + 36 = 39 + 18x.
Answer: y = 39 + 18x.