Math Problem Statement

Find (a) all horizontal, and (b) all vertical asymptotes of the function f (x) = |4x + 3| (2x^2 + 1)/(x^2 − 16) (2x + 1)

Solution

To find the asymptotes of the given function f(x)=4x+3(2x2+1)(x216)(2x+1)f(x) = \frac{|4x + 3|(2x^2 + 1)}{(x^2 - 16)(2x + 1)}, let's analyze step by step:

(a) Horizontal Asymptotes

Horizontal asymptotes are determined by the end behavior of f(x)f(x) as x±x \to \pm\infty.

  1. Analyze the degree of numerator and denominator:

    • Numerator: 4x+3(2x2+1)|4x + 3|(2x^2 + 1)
      • The dominant term as x±x \to \pm\infty is 4x2x2=8x3|4x| \cdot 2x^2 = 8x^3.
    • Denominator: (x216)(2x+1)(x^2 - 16)(2x + 1)
      • The dominant term as x±x \to \pm\infty is x22x=2x3x^2 \cdot 2x = 2x^3.

    Since the degree of both the numerator and the denominator is 3, the horizontal asymptote is given by the ratio of the leading coefficients.

    limx±f(x)=8x32x3=4.\lim_{x \to \pm\infty} f(x) = \frac{8x^3}{2x^3} = 4.

    Horizontal Asymptote: y=4y = 4.


(b) Vertical Asymptotes

Vertical asymptotes occur at points where the denominator is zero and the numerator is nonzero.

  1. Find where the denominator is zero: The denominator is (x216)(2x+1)(x^2 - 16)(2x + 1).

    • x216=0x^2 - 16 = 0 gives x=±4x = \pm 4.
    • 2x+1=02x + 1 = 0 gives x=12x = -\frac{1}{2}.
  2. Check if the numerator is nonzero at these points:

    • At x=4x = 4: Numerator 4(4)+3(2(4)2+1)=19(33)0|4(4) + 3|(2(4)^2 + 1) = |19|(33) \neq 0.
    • At x=4x = -4: Numerator 4(4)+3(2(4)2+1)=13(33)0|4(-4) + 3|(2(-4)^2 + 1) = |-13|(33) \neq 0.
    • At x=12x = -\frac{1}{2}: Numerator 4(12)+3(2(12)2+1)=1(1.5)0|4(-\frac{1}{2}) + 3|(2(-\frac{1}{2})^2 + 1) = |1|(1.5) \neq 0.

    Since the numerator is nonzero at these points, they are vertical asymptotes.

    Vertical Asymptotes: x=4,x=4,x=12x = 4, x = -4, x = -\frac{1}{2}.


Final Answer:

  • Horizontal Asymptote: y=4y = 4.
  • Vertical Asymptotes: x=4,x=4,x=12x = 4, x = -4, x = -\frac{1}{2}.

Would you like further explanation on how these asymptotes are determined?


Related Questions:

  1. How do horizontal asymptotes differ for rational functions with different degrees in the numerator and denominator?
  2. What happens to the behavior of f(x)f(x) near x=±4x = \pm 4 and x=12x = -\frac{1}{2}?
  3. How does the absolute value 4x+3|4x + 3| affect the symmetry of the function?
  4. Can a function have more than one horizontal asymptote? Why or why not?
  5. What are the oblique asymptotes, and when do they occur?

Tip:

For rational functions, always analyze the highest degree terms of the numerator and denominator for horizontal asymptotes. It simplifies the computation.

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Math Problem Analysis

Mathematical Concepts

Asymptotes
Rational Functions
Absolute Value
End Behavior of Functions

Formulas

Horizontal Asymptote: Limit of f(x) as x approaches infinity or negative infinity
Vertical Asymptote: Points where the denominator is zero and the numerator is non-zero

Theorems

Degree Analysis for Horizontal Asymptotes
Zero of Denominator for Vertical Asymptotes

Suitable Grade Level

Grades 11-12