Question

for each function, determine the long run behavior\\(\frac{x^2 + 1}{x^3 + 2}\\) has select an answer\\(\frac{x^2 + 1}{x^2 + 2}\\) has select an answer\\(\frac{x^3 + 1}{x^2 + 2}\\) has select an answer question help: message instructor

Explanation:

To determine the long - run behavior of a rational function \(y=\frac{f(x)}{g(x)}\) (where \(f(x)\) and \(g(x)\) are polynomials), we compare the degrees of the numerator polynomial \(f(x)\) and the denominator polynomial \(g(x)\). Let the degree of \(f(x)\) be \(n\) and the degree of \(g(x)\) be \(m\).

For the function \(\frac{x^{2}+1}{x^{3}+2}\)

• The degree of the numerator \(f(x)=x^{2} + 1\) is \(n = 2\) (the highest power of \(x\) in the numerator).
• The degree of the denominator \(g(x)=x^{3}+2\) is \(m=3\) (the highest power of \(x\) in the denominator).
• Since \(n

For the function \(\frac{x^{2}+1}{x^{2}+2}\)

• The degree of the numerator \(f(x)=x^{2}+1\) is \(n = 2\).
• The degree of the denominator \(g(x)=x^{2}+2\) is \(m = 2\).
• When \(n=m\), we look at the ratio of the leading coefficients. The leading coefficient of the numerator (the coefficient of \(x^{2}\)) is \(a = 1\), and the leading coefficient of the denominator (the coefficient of \(x^{2}\)) is \(b=1\). The long - run behavior is given by \(y=\frac{a}{b}\). So, as \(x\to\pm\infty\), \(\frac{x^{2}+1}{x^{2}+2}\to\frac{1}{1}=1\). In the long - run, the function approaches the horizontal line \(y = 1\).

For the function \(\frac{x^{3}+1}{x^{2}+2}\)

• The degree of the numerator \(f(x)=x^{3}+1\) is \(n = 3\).
• The degree of the denominator \(g(x)=x^{2}+2\) is \(m = 2\).
• Since \(n>m\) (the degree of the numerator is greater than the degree of the denominator), the function has a slant (oblique) asymptote. We can perform polynomial long - division of \(x^{3}+1\) by \(x^{2}+2\):
• Divide \(x^{3}\) by \(x^{2}\) to get \(x\). Multiply \(x^{2}+2\) by \(x\) to get \(x^{3}+2x\).
• Subtract \(x^{3}+2x\) from \(x^{3}+1\): \((x^{3}+1)-(x^{3}+2x)=- 2x + 1\).
• So, \(\frac{x^{3}+1}{x^{2}+2}=x-\frac{2x - 1}{x^{2}+2}\). As \(x\to\pm\infty\), \(\frac{2x - 1}{x^{2}+2}\to0\) (because the degree of the numerator of \(\frac{2x - 1}{x^{2}+2}\) is \(1\) and the degree of the denominator is \(2\), and \(1<2\)). So, as \(x\to\pm\infty\), \(\frac{x^{3}+1}{x^{2}+2}\to x\). In the long - run, the function behaves like the linear function \(y=x\).

Final Answers

• For \(\frac{x^{2}+1}{x^{3}+2}\): Approaches \(y = 0\) (the \(x\) - axis) as \(x\to\pm\infty\)
• For \(\frac{x^{2}+1}{x^{2}+2}\): Approaches \(y = 1\) as \(x\to\pm\infty\)
• For \(\frac{x^{3}+1}{x^{2}+2}\): Approaches \(y=x\) as \(x\to\pm\infty\)

Answer:

To determine the long - run behavior of a rational function \(y=\frac{f(x)}{g(x)}\) (where \(f(x)\) and \(g(x)\) are polynomials), we compare the degrees of the numerator polynomial \(f(x)\) and the denominator polynomial \(g(x)\). Let the degree of \(f(x)\) be \(n\) and the degree of \(g(x)\) be \(m\).

For the function \(\frac{x^{2}+1}{x^{3}+2}\)

• The degree of the numerator \(f(x)=x^{2} + 1\) is \(n = 2\) (the highest power of \(x\) in the numerator).
• The degree of the denominator \(g(x)=x^{3}+2\) is \(m=3\) (the highest power of \(x\) in the denominator).
• Since \(n

For the function \(\frac{x^{2}+1}{x^{2}+2}\)

• The degree of the numerator \(f(x)=x^{2}+1\) is \(n = 2\).
• The degree of the denominator \(g(x)=x^{2}+2\) is \(m = 2\).
• When \(n=m\), we look at the ratio of the leading coefficients. The leading coefficient of the numerator (the coefficient of \(x^{2}\)) is \(a = 1\), and the leading coefficient of the denominator (the coefficient of \(x^{2}\)) is \(b=1\). The long - run behavior is given by \(y=\frac{a}{b}\). So, as \(x\to\pm\infty\), \(\frac{x^{2}+1}{x^{2}+2}\to\frac{1}{1}=1\). In the long - run, the function approaches the horizontal line \(y = 1\).

For the function \(\frac{x^{3}+1}{x^{2}+2}\)

• The degree of the numerator \(f(x)=x^{3}+1\) is \(n = 3\).
• The degree of the denominator \(g(x)=x^{2}+2\) is \(m = 2\).
• Since \(n>m\) (the degree of the numerator is greater than the degree of the denominator), the function has a slant (oblique) asymptote. We can perform polynomial long - division of \(x^{3}+1\) by \(x^{2}+2\):
• Divide \(x^{3}\) by \(x^{2}\) to get \(x\). Multiply \(x^{2}+2\) by \(x\) to get \(x^{3}+2x\).
• Subtract \(x^{3}+2x\) from \(x^{3}+1\): \((x^{3}+1)-(x^{3}+2x)=- 2x + 1\).
• So, \(\frac{x^{3}+1}{x^{2}+2}=x-\frac{2x - 1}{x^{2}+2}\). As \(x\to\pm\infty\), \(\frac{2x - 1}{x^{2}+2}\to0\) (because the degree of the numerator of \(\frac{2x - 1}{x^{2}+2}\) is \(1\) and the degree of the denominator is \(2\), and \(1<2\)). So, as \(x\to\pm\infty\), \(\frac{x^{3}+1}{x^{2}+2}\to x\). In the long - run, the function behaves like the linear function \(y=x\).

Final Answers

• For \(\frac{x^{2}+1}{x^{3}+2}\): Approaches \(y = 0\) (the \(x\) - axis) as \(x\to\pm\infty\)
• For \(\frac{x^{2}+1}{x^{2}+2}\): Approaches \(y = 1\) as \(x\to\pm\infty\)
• For \(\frac{x^{3}+1}{x^{2}+2}\): Approaches \(y=x\) as \(x\to\pm\infty\)