For the function $f(x)=x^{2}+2$, find the equation of the tangent line at $x=-2$.

Step 1 ：First, find the slope of the tangent line by taking the derivative of the function: \(f'(x) = 2x\)

Step 2 ：Next, find the slope at the point \(x=-2\): \(f'(-2) = 2(-2) = -4\)

Step 3 ：Now, find the y-coordinate of the point on the function: \(f(-2) = (-2)^2 + 2 = 4 + 2 = 6\)

Step 4 ：The point on the function is \((-2, 6)\) and the slope of the tangent line is \(-4\)

Step 5 ：Use the point-slope form of a line: \(y - y_1 = m(x - x_1)\)

Step 6 ：Plug in the point and slope: \(y - 6 = -4(x + 2)\)

Step 7 ：Simplify the equation: \(y - 6 = -4x - 8\)

Step 8 ：Add 6 to both sides: \(y = -4x - 2\)

Step 9 ：\(\boxed{y = -4x - 2}\)