(x%2B3)(2-x)(1-x)%5E(2))%2F(x%5E(4)(x%2B6)(x-9)(2x%5E(2)%2B4x%2B9)).jpeg "((2x-1)(x+3)(2-x)(1-x)^(2))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9))")
Analyzing the Rational Function: ((2x-1)(x+3)(2-x)(1-x)^(2))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9))
This article aims to break down and analyze the given rational function:
((2x-1)(x+3)(2-x)(1-x)^(2))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9))
We will explore its key features, including:
- Domain: The set of all possible input values (x-values) for which the function is defined.
- Zeros (or roots): The x-values where the function equals zero.
- Vertical Asymptotes: Lines that the function approaches as x approaches a specific value.
- Horizontal Asymptote: A line that the function approaches as x approaches positive or negative infinity.
- Holes: Points where the function is undefined but can be made continuous by simplifying the function.
Domain
The domain of a rational function is restricted where the denominator is equal to zero. We need to find the values of x that make the denominator zero:
- x^(4) = 0 --> x = 0 (multiplicity 4)
- x + 6 = 0 --> x = -6
- x - 9 = 0 --> x = 9
- 2x^(2) + 4x + 9 = 0 --> This quadratic has no real roots (its discriminant is negative)
Therefore, the domain of the function is all real numbers except for x = 0, -6, and 9.
Zeros (Roots)
The zeros of the function occur where the numerator is equal to zero:
- 2x - 1 = 0 --> x = 1/2
- x + 3 = 0 --> x = -3
- 2 - x = 0 --> x = 2
- (1 - x)^(2) = 0 --> x = 1 (multiplicity 2)
Therefore, the function has zeros at x = 1/2, -3, 2, and 1.
Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero, and the numerator is not zero. We already found the zeros of the denominator:
- x = 0 (multiplicity 4)
- x = -6
- x = 9
The numerator is not zero at any of these points. Therefore, the function has vertical asymptotes at x = 0, -6, and 9.
Horizontal Asymptote
To find the horizontal asymptote, we need to consider the degree of the numerator and denominator:
- Degree of numerator: 5 (from (2x-1)(x+3)(2-x)(1-x)^(2))
- Degree of denominator: 7 (from x^(4)(x+6)(x-9)(2x^(2)+4x+9))
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
Holes
Holes occur when a factor in the numerator and denominator cancel out. Notice that we can simplify the function by canceling out (1-x) from the numerator and denominator:
((2x-1)(x+3)(2-x)(1-x)^(2))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9)) = ((2x-1)(x+3)(2-x)(1-x))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9))
This simplification implies a hole at x = 1. To find the y-coordinate of the hole, substitute x = 1 into the simplified function:
y = ((2(1)-1)(1+3)(2-1))/(1^(4)(1+6)(1-9)(2(1)^(2)+4(1)+9)) = -2/432
Therefore, the function has a hole at (1, -2/432).
Summary
The function ((2x-1)(x+3)(2-x)(1-x)^(2))/(x^(4)(x+6)(x-9)(2x^(2)+4x+9)) has the following key features:
- Domain: All real numbers except for x = 0, -6, and 9.
- Zeros: x = 1/2, -3, 2, and 1.
- Vertical Asymptotes: x = 0, -6, and 9.
- Horizontal Asymptote: y = 0.
- Hole: (1, -2/432).
By understanding these features, we can gain a comprehensive picture of the behavior of this rational function.