Math Problem Statement

Draw the graph

Given the rational function R(x) = (x^3 + 4x^2 + 3x) / (x^2 + 3x + 2), determine its domain, intercepts, asymptotes, and sketch the graph.

Solution

The problem provides the rational function:

R(x)=x3+4x2+3xx2+3x+2R(x) = \frac{x^3 + 4x^2 + 3x}{x^2 + 3x + 2}

We are tasked with the following:

  • (a) Determine its domain and intercepts.
  • (b) Find all possible asymptotes.
  • (c) Sketch the graph of R(x)R(x).

(a) Domain and Intercepts

  1. Domain: The domain of R(x)R(x) consists of all real numbers except where the denominator is zero. We need to find where the denominator is zero: x2+3x+2=0x^2 + 3x + 2 = 0 Factorizing the quadratic: (x+1)(x+2)=0(x + 1)(x + 2) = 0 So, x=1x = -1 and x=2x = -2 are excluded from the domain.

    Therefore, the domain is R{1,2}\mathbb{R} \setminus \{-1, -2\}.

  2. Intercepts:

    • y-intercept: Set x=0x = 0 in R(x)R(x): R(0)=03+4(0)2+3(0)02+3(0)+2=0R(0) = \frac{0^3 + 4(0)^2 + 3(0)}{0^2 + 3(0) + 2} = 0 So, the y-intercept is at (0,0)(0, 0).
    • x-intercept: Set the numerator equal to zero: x3+4x2+3x=0    x(x2+4x+3)=0x^3 + 4x^2 + 3x = 0 \implies x(x^2 + 4x + 3) = 0 Factor the quadratic: x(x+1)(x+3)=0x(x + 1)(x + 3) = 0 Thus, the x-intercepts are x=0,1,3x = 0, -1, -3. So the intercepts are (0,0),(1,0),(3,0)(0, 0), (-1, 0), (-3, 0).

(b) Asymptotes

  1. Vertical Asymptotes: These occur where the denominator is zero and the numerator is not zero. From the factorization of the denominator, vertical asymptotes are at x=1x = -1 and x=2x = -2.

  2. Horizontal Asymptote: To find horizontal asymptotes, compare the degrees of the numerator and denominator. Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote. However, there is an oblique asymptote since the degree of the numerator is exactly one higher than that of the denominator. We can perform polynomial long division to find the oblique asymptote.

Let's calculate the oblique asymptote using polynomial division.It seems like I can’t do more advanced data analysis right now. Please try again later. If you'd like, I can still walk through the process or assist with any other questions!

Here are a few follow-up questions to deepen your understanding:

  1. How do vertical asymptotes affect the graph's behavior near x=1x = -1 and x=2x = -2?
  2. What is the importance of horizontal asymptotes in determining the long-term behavior of rational functions?
  3. How does an oblique asymptote differ from a horizontal asymptote?
  4. Can you factor the numerator and denominator of a rational function easily by hand?
  5. How would you use polynomial division to find the oblique asymptote in this example?

Tip: When graphing rational functions, always analyze intercepts and asymptotes before sketching the curve, as they guide the overall shape.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Polynomial Division
Asymptotes
Intercepts

Formulas

x^3 + 4x^2 + 3x = 0
x^2 + 3x + 2 = 0
Vertical Asymptotes: Denominator = 0
Oblique Asymptote: Polynomial Long Division

Theorems

Vertical Asymptote Theorem
Oblique Asymptote Theorem
Zero-Product Property

Suitable Grade Level

Grades 11-12