Find an equation of the line tangent to the graph of $f(x)=2 x^{3}$ at $(-2,-16)$.

Step 1 ：First, we need to find the derivative of the function \(f(x) = 2x^3\). The derivative of a function at a certain point gives the slope of the tangent line at that point.

Step 2 ：The derivative of \(f(x) = 2x^3\) is \(f'(x) = 6x^2\).

Step 3 ：Now we substitute \(x = -2\) into \(f'(x)\) to find the slope of the tangent line at the point \((-2, -16)\).

Step 4 ：Substituting \(x = -2\) into \(f'(x) = 6x^2\) gives \(f'(-2) = 6(-2)^2 = 24\). So the slope of the tangent line at the point \((-2, -16)\) is 24.

Step 5 ：Now we use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line.

Step 6 ：Substituting \((x_1, y_1) = (-2, -16)\) and \(m = 24\) into the equation gives \(y - (-16) = 24(x - (-2))\).

Step 7 ：Simplifying the equation gives \(y + 16 = 24x + 48\).

Step 8 ：Subtracting 16 from both sides of the equation gives \(y = 24x + 32\).

Step 9 ：So the equation of the line tangent to the graph of \(f(x) = 2x^3\) at the point \((-2, -16)\) is \(\boxed{y = 24x + 32}\).

Step 10 ：Finally, we check that the point \((-2, -16)\) satisfies the equation \(y = 24x + 32\). Substituting \(x = -2\) and \(y = -16\) into the equation gives \(-16 = 24(-2) + 32\), which simplifies to \(-16 = -16\). So the point \((-2, -16)\) does satisfy the equation, and our solution is correct.