Question 25761
As you probably know, the slope of the tangent at the given point (x = -4) is given by the value of the 1st derivative of the function {{{y = x^2+8x+12}}} at that point (x = -4).
{{{dy/dx = 2x + 8}}} Substitute x = -4 and solve.
{{{dy/dx = 2(-4) + 8}}}
{{{dy/dx = 0}}} So, the slope of the tangent is 0 which means that it's a horizontal line. To find the point (x, y) of tangency, you'll substitute x = -4 into the original equation.

{{{y = (-4)^2 + 8(-4) + 12}}} 
{{{y = 16 - 32 + 12}}}
{{{y = -4}}}

The point of tangency is (-4, -4) and the equation of the line passing through that point with a slope of m = 0 is:
 Using the point-slope form:{{{y-y1 = m(x-x1)}}}
{{{y -(-4) = 0(x-(-4))}}} Simplify.
{{{y + 4 = 0}}}
{{{y = -4}}} This is the equation of the tangent at x = -4

Here's what it looks like in a graph:
{{{graph(300,200,-10,2,-6,5,x^2+8x+12,-4)}}}