Math Problem Statement

solve

solve f(x) = (-4x + 6) / (-x + 2); draw vertical and horizontal asymptotes and plot points.

Solution

To solve the given function f(x)=4x+6x+2f(x) = \frac{-4x + 6}{-x + 2}, we will follow the steps mentioned in the problem:

Step 1: Identify the Vertical and Horizontal Asymptotes

  • Vertical Asymptote: This occurs when the denominator equals zero. Set x+2=0-x + 2 = 0, which gives x=2x = 2. Therefore, there is a vertical asymptote at x=2x = 2.
  • Horizontal Asymptote: Since the degrees of the numerator and denominator are the same, divide the leading coefficients: Horizontal Asymptote: y=41=4\text{Horizontal Asymptote: } y = \frac{-4}{-1} = 4 So the horizontal asymptote is y=4y = 4.

Step 2: Plot Two Points on Each Piece of the Graph

To ensure accurate plotting, substitute a few values for xx on either side of the vertical asymptote (x=2x = 2).

For x<2x < 2:

  1. Choose x=1x = 1: f(1)=4(1)+6(1)+2=4+61+2=21=2f(1) = \frac{-4(1) + 6}{-(1) + 2} = \frac{-4 + 6}{-1 + 2} = \frac{2}{1} = 2 Point: (1,2)(1, 2)

  2. Choose x=0x = 0: f(0)=4(0)+6(0)+2=62=3f(0) = \frac{-4(0) + 6}{-(0) + 2} = \frac{6}{2} = 3 Point: (0,3)(0, 3)

For x>2x > 2:

  1. Choose x=3x = 3: f(3)=4(3)+6(3)+2=12+63+2=61=6f(3) = \frac{-4(3) + 6}{-(3) + 2} = \frac{-12 + 6}{-3 + 2} = \frac{-6}{-1} = 6 Point: (3,6)(3, 6)

  2. Choose x=4x = 4: f(4)=4(4)+6(4)+2=16+64+2=102=5f(4) = \frac{-4(4) + 6}{-(4) + 2} = \frac{-16 + 6}{-4 + 2} = \frac{-10}{-2} = 5 Point: (4,5)(4, 5)

Step 3: Plot the Graph

  • Draw the vertical asymptote x=2x = 2.
  • Draw the horizontal asymptote y=4y = 4.
  • Plot the points:
    • (1,2)(1, 2), (0,3)(0, 3) for x<2x < 2.
    • (3,6)(3, 6), (4,5)(4, 5) for x>2x > 2.
  • Connect the points smoothly in a way that the graph approaches the asymptotes.

If needed, I can assist further by guiding you through the plotting or creating a digital graph. Let me know!


Questions to Expand Understanding

  1. How can you verify vertical asymptotes for other rational functions?
  2. What is the general rule to find horizontal asymptotes based on degrees of numerator and denominator?
  3. Can there be slant asymptotes in rational functions? If yes, when do they occur?
  4. How do transformations of the basic rational function 1x\frac{1}{x} affect its asymptotes?
  5. What are the implications of having a hole in the graph of a rational function?

Tip:

For rational functions, always simplify the expression first to check for any removable discontinuities (holes) before identifying the asymptotes.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Asymptotes
Graphing

Formulas

Vertical Asymptote: Denominator = 0
Horizontal Asymptote: Leading coefficient rule (degree comparison)

Theorems

Vertical Asymptote Theorem
Horizontal Asymptote Rule (Equal Degrees)

Suitable Grade Level

Grades 9-12