Find an equation of the line tangent to the graph of $f(x)=2x^3$ 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 $f^{\prime }(x)=6x^2$.

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

Step 4 ：Substituting $x=-2$ into $f^{\prime }(x)=6x^2$ gives $f^{\prime }(-2)=6(-2{)}^2=24$. 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 $\boxed{{\displaystyle y=24x+32}}$.

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.