Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function $y=\frac{4x}{x^2+4}$ as $(-\infty,\infty)$ and this function is even.
Step 2. Take derivatives to get $y'=\frac{4(x^2+4)-4x(2x)}{(x^2+4)^2}=-\frac{4(x^2-4)}{(x^2+4)^2}$ and $y''=-\frac{8x(x^2+4)^2-8(x^2-4)(x^2+4)(2x)}{(x^2+4)^4}==\frac{8x(x^2-12)}{(x^2+4)^3}$.
Step 3. We can test possible critical points at $x=\pm2$. Check the signs of $y'$ across $x=\pm2$: $..(-)..(-2)..(+)..(2)..(-)..$; thus the function decreases on $(-\infty, -2), (2,\infty)$ and increases on $(-2,2)$, and there is a minimum at $y(-2)=-1$ and a maximum at $y(2)=1$..
Step 4. Let $y''=0$, we can find the roots as $x=0,\pm2\sqrt 3$. Check concavity across these roots with signs of $y''$:$..(-)..(-2\sqrt 3)..(+)..(0)..(-)..(2\sqrt 3)..(+)..$, thus the function is concave down on $(-\infty,-2\sqrt 3),(0,2\sqrt 3)$ and concave up on $(-2\sqrt 3,0), (2\sqrt 3,\infty)$, and $x=0,\pm2\sqrt 3$ 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 a y-intercept at $y(0)=0$.
Step 7. Based on the above results, we can graph the function as shown in the figure.
