Question
Find the horizontal asymptote of f(x)= -3x+5x^3-1/ 4x^3+2x^2+1
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Answers
Answered by
GPT-4o mini
To find the horizontal asymptote of the function
\[
f(x) = \frac{-3x + 5x^3 - 1}{4x^3 + 2x^2 + 1},
\]
we look at the degrees of the numerator and denominator.
1. **Identify the degrees**: The degree of the numerator is 3 (since the highest power of \(x\) in \(-3x + 5x^3 - 1\) is \(x^3\)) and the degree of the denominator is also 3 (since the highest power of \(x\) in \(4x^3 + 2x^2 + 1\) is \(x^3\)).
2. **Equal degrees**: Since the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients.
- The leading coefficient of the numerator (from \(5x^3\)) is \(5\).
- The leading coefficient of the denominator (from \(4x^3\)) is \(4\).
3. **Calculate the horizontal asymptote**:
\[
y = \frac{5}{4}.
\]
Thus, the horizontal asymptote of the function \(f(x)\) is
\[
y = \frac{5}{4}.
\]
\[
f(x) = \frac{-3x + 5x^3 - 1}{4x^3 + 2x^2 + 1},
\]
we look at the degrees of the numerator and denominator.
1. **Identify the degrees**: The degree of the numerator is 3 (since the highest power of \(x\) in \(-3x + 5x^3 - 1\) is \(x^3\)) and the degree of the denominator is also 3 (since the highest power of \(x\) in \(4x^3 + 2x^2 + 1\) is \(x^3\)).
2. **Equal degrees**: Since the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients.
- The leading coefficient of the numerator (from \(5x^3\)) is \(5\).
- The leading coefficient of the denominator (from \(4x^3\)) is \(4\).
3. **Calculate the horizontal asymptote**:
\[
y = \frac{5}{4}.
\]
Thus, the horizontal asymptote of the function \(f(x)\) is
\[
y = \frac{5}{4}.
\]