8. Let $$ f(x)=\frac{4 x-1}{x^{2}+2 x+1} $$. What is the equation of the horizontal asymptote of $$ f(x) $$ ? From what direction does approach the asymptote as $$ x $$ tends towards $$ -\infty $$ ?

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1. The function is given by  
   $$
   f(x)=\frac{4x-1}{x^2+2x+1}.
   $$

2. Notice that the denominator factors as  
   $$
   x^2+2x+1 = (x+1)^2.
   $$
   However, for finding the horizontal asymptote, we only need to compare the degrees of the numerator and denominator.

3. The degree of the numerator is $$1$$ and the degree of the denominator is $$2$$. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is  
   $$
   y=0.
   $$

4. To determine from which direction $$f(x)$$ approaches $$0$$ as $$x$$ tends towards $$-\infty$$, examine the leading behavior for large negative $$x$$. For large $$|x|$$, the terms of lower order become negligible, and the function approximates to  
   $$
   f(x) \approx \frac{4x}{x^2} = \frac{4}{x}.
   $$

5. As $$x \to -\infty$$, $$\frac{4}{x}$$ approaches $$0$$. However, since $$x$$ is negative, $$\frac{4}{x}$$ will be negative. Thus, $$f(x)$$ approaches $$0$$ from below.

6. In summary, the horizontal asymptote of $$f(x)$$ is  
   $$
   y=0,
   $$
   and the function approaches this asymptote from the negative side as $$x$$ tends towards $$-\infty$$.
The horizontal asymptote of $$ f(x) $$ is $$ y=0 $$. As $$ x $$ approaches $$-\infty$$, $$ f(x) $$ approaches $$ 0 $$ from below.

8. Let \( f(x)=\frac{4 x-1}{x^{2}+2 x+1} \). What is the equation of the horizontal asymptote of \( f(x) \) ? From what direction does approach the asymptote as \( x \) tends towards \( -\infty \) ?

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