Question

7 which of the following functions has a slant asymptote?
a) (f(x)=\frac{x^{5}+1}{x^{4}+3x^{2}+2}) go to station 2
b) (f(x)=\frac{x^{2}+1}{x^{3}-x^{2}-1}) go to station 5
c) (f(x)=\frac{4x^{2}+x + 1}{x^{2}}) go to station 8
d) (f(x)=\frac{x}{x^{2}-1}) go to station 1

Explanation:

Step1: Recall slant - asymptote condition

A rational function $y = \frac{f(x)}{g(x)}$ has a slant asymptote when the degree of the numerator $f(x)$ is exactly one more than the degree of the denominator $g(x)$.

Step2: Analyze option a

For $f(x)=\frac{x^{5}+1}{x^{4}+3x^{2}+2}$, the degree of the numerator is $n = 5$ and the degree of the denominator is $m=4$. Since $n=m + 1$, we can use polynomial long - division. Dividing $x^{5}+1$ by $x^{4}+3x^{2}+2$ gives $x-\frac{3x^{3}+2x - 1}{x^{4}+3x^{2}+2}$. As $x\to\pm\infty$, the rational part $\frac{3x^{3}+2x - 1}{x^{4}+3x^{2}+2}\to0$, and the slant asymptote is $y = x$.

Step3: Analyze option b

For $f(x)=\frac{x^{2}+1}{x^{3}-x^{2}-1}$, the degree of the numerator is $n = 2$ and the degree of the denominator is $m = 3$. Since $n

Step4: Analyze option c

For $f(x)=\frac{4x^{2}+x + 1}{x^{2}}$, the degree of the numerator is $n = 2$ and the degree of the denominator is $m = 2$. Dividing $4x^{2}+x + 1$ by $x^{2}$ gives $4+\frac{x + 1}{x^{2}}$. As $x\to\pm\infty$, the rational part $\frac{x + 1}{x^{2}}\to0$, and the horizontal asymptote is $y = 4$.

Step5: Analyze option d

For $f(x)=\frac{x}{x^{2}-1}$, the degree of the numerator is $n = 1$ and the degree of the denominator is $m = 2$. Since $n

Answer:

a. $f(x)=\frac{x^{5}+1}{x^{4}+3x^{2}+2}$