Thus, $x=-4$ or $x=-2$. The zeros of $f(x)$ are thus $x=-4$, $x=-2$ and $x=0$.

Find the derivative of $f(x)=x^3+6x^2+8x$:

Then use the quadratic formula to find the maxima and minima of $f(x)$:

Thus, $x\approx -3.15$ or $x\approx -0.85$.

To find the points, you need to find their corresponding $y$-values. You find these by putting the $x$-values you found back into the main function $f(x)$:

You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart. Notice that the derivative can be factorized as:

First, you find the second derivative of $f(x)$ by differentiating $f^{\prime }(x)=3x^2+12x+8$:

Let $f^{″}(x)=0$ and solve the equation:

Enter this $x$-value into the original function $f(x)$ to find the $y$-coordinate for the inflection point:

By making a sign chart for the second derivative, you can see where the graph of $f(x)$ is concave and where it is convex. Notice that $f^{″}(x)$ can be factorized as $f^{″}(x)=6(x+2)$: