Question

for each function, determine the long run behavior.$\frac{-9x^{3}+1}{-6x^{3}+2}$ has$\frac{x^{3}+1}{x^{4}+2}$ has$\frac{0x^{4}+1}{-10x^{3}+2}$ hasselect an answerno horizontal asymptotea horizontal asymptote at y=0a horizontal asymptote at y=3/2question help: video read written example

Explanation:

Step1: Analyze first rational function

For $f(x)=\frac{-9x^3 + 1}{-6x^3 + 2}$, take the limit as $x\to\pm\infty$ by dividing numerator and denominator by $x^3$:
$$\lim_{x\to\pm\infty}\frac{-9 + \frac{1}{x^3}}{-6 + \frac{2}{x^3}} = \frac{-9}{-6} = \frac{3}{2}$$

Step2: Analyze second rational function

For $g(x)=\frac{x^3 + 1}{x^4 + 2}$, divide numerator and denominator by $x^4$:
$$\lim_{x\to\pm\infty}\frac{\frac{1}{x} + \frac{1}{x^4}}{1 + \frac{2}{x^4}} = \frac{0 + 0}{1 + 0} = 0$$

Step3: Analyze third rational function

For $h(x)=\frac{0x^4 + 1}{-10x^3 + 2}=\frac{1}{-10x^3 + 2}$, check limits:
$$\lim_{x\to+\infty}\frac{1}{-10x^3 + 2}=0^-,\quad \lim_{x\to-\infty}\frac{1}{-10x^3 + 2}=0^+$$
But the degree of the denominator (3) is greater than the numerator (0), so the horizontal asymptote is $y=0$.

Answer:

1. $\frac{-9x^3 + 1}{-6x^3 + 2}$ has: a Horizontal asymptote at y=3/2
2. $\frac{x^3 + 1}{x^4 + 2}$ has: a Horizontal asymptote at y=0
3. $\frac{0x^4 + 1}{-10x^3 + 2}$ has: a Horizontal asymptote at y=0