answer for this question is 3.85 units squared
First we'll require the derivative (slope) function.
f'(x) = 6x + 1
Next evaluate the derivative function @ x = -2
f'(-2) = 6(-2) + 1
f'(-2) = -11
Thus the slope of the tangent line @ x = -2 is m = -11
Now we can use either y = mx + b or y - y1 = m(x - x1) to find the tangent equation.
y - 4 = -11(x - (-2))
y - 4 = -11(x + 2)
y - 4 = -11x - 22
y = -11x - 22 + 4
Thus
y = -11x - 18 or 11x + y + 18 = 0 is the equation of the required tangent.