Question

Number of asymptotes, the function $f\left(x\right)=\frac{x^3-27}{2x^2-4}$ has

• A. $1$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (B)

$3$

$y=(\frac{x^3-27}{2x^2-4})$

for vertical asymotote ,find the point where denominator equal to $0$

hence $x=\sqrt{2}$ and $x=-\sqrt{2}$

now put them on equ($1$),we will get $y=\infty$

hence $x=\sqrt{2}$ and $x=-\sqrt{2}$ are two vertical asymptote of the function $y=(\frac{x^3-27}{2x^2-4})$

$horizontal$ $asymptote$

line $y=L$ is a horizontal asymptote of fuction if either ${}_{x\to \infty }^{lim}f\left(x\right)=L$ and ${}_{x\to -\infty }^{lim}f\left(x\right)=L$

where $L$ is A finite number

${}_{x{\to }_{-}^{+}\infty }^{lim}(\frac{x^3-27}{2x^2-4}){=}_{-}^{+}\infty$

hence there is no horizonatal asymptote .

$slant$ $asymptote$

do polynomial long division ,

$y=(\frac{x^3-27}{2x^2-4})=\frac{x}{2}+\frac{2x-27}{2x^2-4}$

the rational polynomial approaches 0 as the variable approaches infinity

hence $y=\frac{x}{2}$ is a slant asymptote

total 3 asymptote