f(x)=\\frac{4x + 1}{x + 2}\nwhat is the horizontal asymptote to this function?

Explanation:

Step1: Divide numerator and denominator by (x)

$$ \begin{align*} f(x)&=\frac{4x + 1}{x + 2}\ &=\frac{\frac{4x}{x}+\frac{1}{x}}{\frac{x}{x}+\frac{2}{x}}\ &=\frac{4+\frac{1}{x}}{1+\frac{2}{x}} \end{align*} $$

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

As (x\to\pm\infty), (\frac{1}{x}\to0) and (\frac{2}{x}\to0). $$ \begin{align*} \lim_{x\to\pm\infty}f(x)&=\lim_{x\to\pm\infty}\frac{4+\frac{1}{x}}{1+\frac{2}{x}}\ &=\frac{4 + 0}{1+0}\ &=4 \end{align*} $$

Answer:

(y = 4)