Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=\frac{x^4}{4}-\frac{x^3}{3}-4x^2+12x+20$, we have $y'=x^3-x^2-8x+12=x^3+27-x^2-8x-15=(x+3)(x^2-3x+9)-(x+3)(x+5)=(x+3)(x^2-4x+4)=(x+3)(x-2)^2$ and $y''=3x^2-2x-8=(3x+4)(x-2)$
Step 2. The inflection points can be found at zeros of $y''$ where there is a sign (concavity) change. We have $..(+)..(-\frac{4}{3})..(-)..(2)..(+)..$ and we can identify two inflection points at $x=-\frac{4}{3}$ and $x=2$.
Step 3. The extrema can be found at zeros of $y'$ with sign changes in $y'$: $..(-)..(-3)..(+)..(2)..(+)..$ and we can identify only a local minimum at $(-3,-22.75)$.
Step 4. Graph the function and its derivatives as shown in the figure.
Step 5. The intersection of $y'$ with the x-axis indicates possible extrema positions.
Step 6. The intersection of $y''$ with the x-axis indicates possible inflection points.
Step 7. Positive $y'$ indicates increasing region and negative $y'$ indicates decreasing region of the function.
Step 8. Positive $y''$ indicates concave up region and negative $y''$ indicates concave down region of the function.
