What is the equation of the line tangent to the function f (x)= 3 x^2+ x at the point(-2,4 )?

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.