Math Problem Statement

Find the horizontal asymptote of f(x) = (2x + x^3 + 4)/(-5x^3 - 4x^2 + 3).

Solution

To find the horizontal asymptote of the function

f(x)=2x+x3+45x34x2+3,f(x) = \frac{2x + x^3 + 4}{-5x^3 - 4x^2 + 3},

we need to examine the degrees of the polynomials in the numerator and denominator.

  1. Degree of the numerator: The highest power of xx in the numerator 2x+x3+42x + x^3 + 4 is x3x^3, so the degree of the numerator is 3.
  2. Degree of the denominator: The highest power of xx in the denominator 5x34x2+3-5x^3 - 4x^2 + 3 is also x3x^3, so the degree of the denominator is also 3.

When the degrees of the numerator and denominator are the same, the horizontal asymptote is found by dividing the leading coefficients of the highest degree terms from both the numerator and the denominator.

  • The leading coefficient in the numerator is 11 (from x3x^3).
  • The leading coefficient in the denominator is 5-5 (from 5x3-5x^3).

Thus, the horizontal asymptote is:

y=15=15.y = \frac{1}{-5} = -\frac{1}{5}.

So, the horizontal asymptote is:

y=15.y = -\frac{1}{5}.

Would you like more details or have any questions?

Here are five related questions:

  1. How do you find the horizontal asymptote when the degree of the numerator is less than that of the denominator?
  2. What happens if the degree of the numerator is greater than the degree of the denominator?
  3. Can rational functions have oblique (slant) asymptotes?
  4. How do vertical asymptotes differ from horizontal asymptotes?
  5. What are some real-world applications of asymptotes in mathematical modeling?

Tip: Always check the degrees of the numerator and denominator when finding horizontal asymptotes, as this relationship determines whether one exists and where it is located.

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Math Problem Analysis

Mathematical Concepts

Calculus
Rational Functions
Asymptotes

Formulas

Horizontal asymptote formula: y = (leading coefficient of numerator)/(leading coefficient of denominator)

Theorems

Degree comparison of polynomials in rational functions

Suitable Grade Level

Grades 11-12