Question

Asymptote(s) of the function $f\left(x\right)=\frac{x^3+2x^2-9}{2x^3-8x+3}$ is/are

• A. $x=\infty$
• B. $x=0$
• C. $y=\frac{1}{2}$
• D. $y=0$

Answer 1

Answer: (C)

$y=\frac{1}{2}$

$y=\frac{x^3+2x^2-9}{2x^3-8x+3}$

because maximum degree (i.e. 3) is equal in both denominator and numerator, hence there is a horizontal asymptote.

$y=\frac{x^3(1+\frac{2}{x}-\frac{9}{x^3})}{x^3(2-\frac{8}{x^2}+\frac{3}{x^3})}$

Now $y=\frac{(1{+}_{x\to \infty }^{lim}\frac{2}{x}{-}_{x\to \infty }^{lim}\frac{9}{x^3})}{(2{-}_{x\to \infty }^{lim}\frac{8}{x^2}{+}_{x\to \infty }^{lim}\frac{3}{x^3})}$

Hence $y=\frac{1}{2}$ is a horizontal asymptote.