find the horizontal and vertical asymptotes for the function.\nf(x) = \\frac{x^{3}}{5x^{3}-40}\nvertical…

Explanation:

Step1: Find vertical asymptote

Set the denominator equal to 0. So, $5x^{3}-40 = 0$.

Step2: Solve for x

First, factor out 5: $5(x^{3}-8)=0$. Then, since $x^{3}-8=(x - 2)(x^{2}+2x + 4)$ and $5\neq0$, we have $x - 2=0$, so $x = 2$.

Step3: Find horizontal asymptote

Since the degree of the numerator and denominator are the same (both 3), divide the leading - coefficients. The leading coefficient of the numerator is 1 and of the denominator is 5. So, $y=\frac{1}{5}$.

Answer:

Vertical asymptote: $x = 2$ Horizontal asymptote: $y=\frac{1}{5}$