We find the equation of the tangent line, for this we find the value of the function and its derivative at the point of tangency.
Derivative of the function:
y ‘(x) = 3 * x² + 4 * x – 4.
Its value at the point of contact:
y ‘(- 2) = 3 * 4 – 4 * 2 – 4 = 12 – 8 – 4 = 0.
The value of the function at the touch point:
y (-2) = -8 + 8 + 8 – 3 = 5.
The general equation of the tangent line is f (x) = y ‘(x0) * (x – x0) + y (x0).
Therefore, in our case f (x) = 5.
Answer: the equation of the tangent line is f (x) = 5.
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