Question

what is the horizontal asymptote of $f(x)=\frac{-2x}{x+1}$?
\\(\circ\\) $y=-2$
\\(\circ\\) $y=-1$
\\(\circ\\) $y=0$
\\(\circ\\) $y=1$

Explanation:

Step1: Divide numerator and denominator by x

For \( f(x)=\frac{-2x}{x + 1} \), divide numerator and denominator by \( x\) (assuming \(x
eq0\)):
\( f(x)=\frac{\frac{-2x}{x}}{\frac{x}{x}+\frac{1}{x}}=\frac{-2}{1+\frac{1}{x}} \)

Step2: Find the limit as \(x

ightarrow\pm\infty\)
As \(x
ightarrow\pm\infty\), \(\frac{1}{x}
ightarrow0\).
So, \(\lim_{x
ightarrow\pm\infty}f(x)=\lim_{x
ightarrow\pm\infty}\frac{-2}{1+\frac{1}{x}}=\frac{-2}{1 + 0}=-2\)

Answer:

\(y = - 2\)