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

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

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

Answer

Explanation:

Step1: Divide numerator and denominator by (x)

$$ \begin{align*} f(x)&=\frac{-2x}{x + 1}\ &=\frac{-2x/x}{(x+1)/x}\ &=\frac{-2}{1+\frac{1}{x}} \end{align*} $$

Step2: Find the limit as (x\rightarrow\pm\infty)

As (x\rightarrow\pm\infty), (\frac{1}{x}\rightarrow0). Then (\lim_{x\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\frac{-2}{1 + \frac{1}{x}}) Substitute (\frac{1}{x}=0) into the expression: (\frac{-2}{1+0}=-2)

Answer:

(y = - 2) (the first option)