Question

identify the graph of $f(x)=\frac{10 - 10x^{2}}{x^{2}}$?

Explanation:

Step1: Simplify the function

First, rewrite $f(x)=\frac{10 - 10x^{2}}{x^{2}}=\frac{10}{x^{2}}-10$.

Step2: Analyze the domain

The function $f(x)$ is undefined when $x = 0$ since division by zero is not allowed. So the domain is $x
eq0$.

Step3: Analyze the vertical - asymptote

As $x\to0^{+}$ or $x\to0^{-}$, $\frac{10}{x^{2}}\to+\infty$, so $y = f(x)=\frac{10}{x^{2}}-10\to+\infty$. Thus, $x = 0$ is a vertical asymptote.

Step4: Analyze the horizontal - asymptote

As $x\to\pm\infty$, $\frac{10}{x^{2}}\to0$, so $y=\frac{10}{x^{2}}-10\to - 10$. Thus, $y=-10$ is a horizontal asymptote.

Step5: Analyze the sign of the function

When $x
eq0$, $x^{2}>0$. If $x
eq0$, $\frac{10}{x^{2}}>0$, and $y=\frac{10}{x^{2}}-10$. When $\frac{10}{x^{2}}>10$ (i.e., $x^{2}<1$ and $x
eq0$ or $- 1<x<0$ or $0<x<1$), $y>0$; when $\frac{10}{x^{2}}<10$ (i.e., $x>1$ or $x < - 1$), $y<0$.

Answer:

The graph has a vertical asymptote at $x = 0$, a horizontal asymptote at $y=-10$, is positive for $-11$ and $x < - 1$.