Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function $y=\frac{8}{x^2+4}$ as $(-\infty,\infty)$ and this function is even.
Step 2. Take derivatives to get $y'=-\frac{16x}{(x^2+4)^2}$ and $y''=-\frac{16(x^2+4)^2-32x(x^2+4)(2x)}{(x^2+4)^4}=-\frac{16(3x^2-4)}{(x^2+4)^3}$.
Step 3. We can test a possible critical point at $x=0$. Check the signs of $y'$ across $x=0$: $..(+)..(0)..(-)..$; thus the function increases on $(-\infty, 0)$ and decreases on $(0,\infty)$, and there is a maximum at $y(0)=2$.
Step 4. Let $y''=0$; we can find the roots as $x_1=-\frac{2\sqrt 3}{3}, x_2=\frac{2\sqrt 3}{3}$. Check concavity across $x_1, x_2$ with signs of $y''$:$..(+)..(x_1)..(-)..(x_2)..(+)..$, thus the function is concave down on $(x_1,x_2)$ and concave up on $(-\infty,x_1),(x_2,\infty)$, and $x_1, x_2$ are inflection points.
Step 5. As $x\to\pm\infty, y\to0$, we can identify a horizontal asymptote as $y=0$.
Step 6. The function has no x-intercepts but has a y-intercept at $y(0)=2$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
