Answer
$ y(0)=4$ is a local maximum and $y(-4)=0, y(4)=0$ is local minima; absolute minimum is at $y(-4)=y(4)=0$ and the absolute maximum is at $y(0)=4$; no inflection points within the domain; concave down.
See graph.
Work Step by Step
Step 1. Given the function $y=\sqrt{16-x^2}=(16-x^2)^{1/2}, -4\leq x\leq4$, we have $y'=\frac{1}{2}(16-x^2)^{-1/2}(-2x)=-x(16-x^2)^{-1/2} $ and $y''=-(16-x^2)^{-1/2}+\frac{1}{2}x(16-x^2)^{-3/2}(-2x)=\frac{x^2-16}{\sqrt {(16-x^2)^3}}-\frac{x^2}{\sqrt {(16-x^2)^3}}=\frac{-16}{\sqrt {(16-x^2)^3}}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=-4, 0, 4$ as critical points with $y(-4)=0, y(0)=4, y(4)=0$
Step 3. Examine the sign change of $y'$ across the critical points: $(-4)..(+)..(0)..(-)..(4)$; we can identify $ y(0)=4$ as a local maximum and $y(-4)=0, y(4)=0$ as local minima. The absolute minimum is at $y(-4)=y(4)=0$ and the absolute maximum is at $y(0)=4$ .
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We do not have inflection points within the domain. The function is concave down in $(-4,4)$ because $y''\lt0$.
Step 5. See graph.
