Question

standard 9 homework
score: 2/27 answered: 2/20
question 3
for each function, determine the horizontal asymptote.
$f(x)=\frac{x^{2}+1}{x^{3}+2}$ has select an answer
$g(x)=\frac{x^{3}+1}{x^{2}+2}$ has select an answer
$h(x)=\frac{x^{2}+1}{x^{2}+2}$ has select an answer
question help: video message instructor post to forum

Explanation:

Step1: Recall the rules for horizontal asymptotes

For a rational function \(y = \frac{f(x)}{g(x)}=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}\), where \(n\) is the degree of the numerator and \(m\) is the degree of the denominator:

• If \(n
• If \(n=m\), the horizontal asymptote is \(y=\frac{a_n}{b_m}\).
• If \(n>m\), there is no horizontal asymptote (the function has an oblique or no - asymptote depending on the case).

Step2: Analyze \(f(x)=\frac{x^{2}+1}{x^{3}+2}\)

For \(f(x)\), the degree of the numerator \(n = 2\) and the degree of the denominator \(m=3\). Since \(n

Step3: Analyze \(g(x)=\frac{x^{3}+1}{x^{2}+2}\)

For \(g(x)\), the degree of the numerator \(n = 3\) and the degree of the denominator \(m = 2\). Since \(n>m\), there is no horizontal asymptote.

Step4: Analyze \(h(x)=\frac{x^{2}+1}{x^{2}+2}\)

For \(h(x)\), the degree of the numerator \(n = 2\) and the degree of the denominator \(m=2\). The leading coefficient of the numerator \(a_n = 1\) and the leading coefficient of the denominator \(b_m=1\). Using the formula \(y=\frac{a_n}{b_m}\), we get \(y = 1\).

Answer:

• \(f(x)=\frac{x^{2}+1}{x^{3}+2}\) has \(y = 0\) as the horizontal asymptote.
• \(g(x)=\frac{x^{3}+1}{x^{2}+2}\) has no horizontal asymptote.
• \(h(x)=\frac{x^{2}+1}{x^{2}+2}\) has \(y = 1\) as the horizontal asymptote.