Find the horizontal asymptote of each function.
$f\left(x\right)=\frac{x^2+4x}{2x^3+8x^2}$

• A. Horizontal Asymptote at $y=0$
• B. Horizontal Asymptote at $y=\frac12$
• C. Horizontal Asymptote at $y=2$

Answer: (A)

Verified step by step guidance

1. Identify the degrees of the polynomial in the numerator and the denominator. The degree of the numerator is 2 (from x^2), and the degree of the denominator is 3 (from 2x^3).
2. Compare the degrees of the numerator and the denominator. Since the degree of the numerator (2) is less than the degree of the denominator (3), the horizontal asymptote is y = 0.
3. If the degrees were equal, the horizontal asymptote would be the ratio of the leading coefficients. However, in this case, the degree of the numerator is less than the degree of the denominator.
4. Recall that when the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is always y = 0.
5. Thus, the horizontal asymptote for the given function f(x) = (x^2 + 4x) / (2x^3 + 8x^2) is y = 0.