Question

f(x)=\frac{3}{x^{2}}
(a) sketch the graph of f.

Explanation:

Step1: Analyze domain

The function $f(x)=\frac{3}{x^{2}}$ is undefined when $x = 0$. So the domain is all real - numbers except $x=0$, i.e., $(-\infty,0)\cup(0,\infty)$.

Step2: Analyze sign

Since $x^{2}>0$ for all $x
eq0$ and the numerator $3>0$, then $f(x)=\frac{3}{x^{2}}>0$ for all $x
eq0$.

Step3: Analyze asymptotes

As $x
ightarrow0^{+}$ or $x
ightarrow0^{-}$, $x^{2}
ightarrow0^{+}$ and $\frac{3}{x^{2}}
ightarrow+\infty$, so $x = 0$ (the y - axis) is a vertical asymptote.
As $x
ightarrow\pm\infty$, $x^{2}
ightarrow+\infty$ and $\frac{3}{x^{2}}
ightarrow0$, so $y = 0$ (the x - axis) is a horizontal asymptote.

Step4: Analyze symmetry

$f(-x)=\frac{3}{(-x)^{2}}=\frac{3}{x^{2}}=f(x)$, so the function is even and symmetric about the y - axis.

Step5: Plot key points

When $x = 1$, $y = 3$; when $x=-1$, $y = 3$; when $x = 2$, $y=\frac{3}{4}$; when $x=-2$, $y=\frac{3}{4}$.

Based on the above analysis, the correct graph is the one that has two symmetric branches above the x - axis, approaching the x - axis as $x
ightarrow\pm\infty$ and approaching $+\infty$ as $x
ightarrow0$. The second graph (from the left in the image) is the correct one.

Answer:

The graph with two symmetric branches above the x - axis, having a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 0$.