Answer
See graph and explanations.
Work Step by Step
Step 1. Rewrite the function as $y=x^2+\frac{1}{x^2}; x\ne0$, we can identify a vertical asymptote at $x=0$ and the domain can be found to be all real numbers except $x\ne0$.
Step 2. Take derivatives to get $y'=2x-\frac{2}{x^3}$ and $y''=2+\frac{6}{x^4}$. Critical points can be found as $x=0, \pm1$. Check the signs of $y'$ across the critical points: $..(-)..(-1)..(+)..(0)..(-)..(1)..(+)..$; thus the function increases on $(-1,0),(1,\infty)$ and decreases on $(-\infty,-1),(0,1)$. The local and absolute minimum can be found as $y(-1)=y(1)=2$.
Step 3. Check concavity across $x=0$ with signs of $y''$:$..(+)..(0)..(+)..$; thus the function is concave up on
$(-\infty,0)$ and $(0,\infty)$ and there is no inflection point.
Step 4. There is no x- or y-intercept as $x\ne0$ and $y\ne0$.
Step 5. Based on the above results, we can graph the function as shown in the figure.
