Math Problem Statement

The figure below shows the graph of a rational function f(x). It has vertical asymptotes x = 1 and x = 6, and horizontal asymptote y = 0. The graph has x-intercept 4, and it passes through the point (0, 2). The equation for f(x) has one of the five forms shown below. Choose the appropriate form for f(x), and then write the equation.

Solution

We are given the graph of a rational function f(x)f(x) with several key characteristics:

  1. Vertical asymptotes at x=1x = 1 and x=6x = 6
  2. Horizontal asymptote at y=0y = 0
  3. x-intercept at x=4x = 4
  4. Point (0, 2) lies on the graph

The problem asks us to select the appropriate form for f(x)f(x) from five options, then determine the equation.

Step-by-step reasoning:

  • Asymptotes: The vertical asymptotes at x=1x = 1 and x=6x = 6 imply factors of (x1)(x - 1) and (x6)(x - 6) in the denominator.

  • x-intercept at x=4x = 4: The function has an x-intercept at 4, so the numerator must have a factor of (x4)(x - 4), since setting x=4x = 4 should make f(x)=0f(x) = 0.

  • Horizontal asymptote at y=0y = 0: Since the horizontal asymptote is y=0y = 0, the degree of the numerator must be less than or equal to the degree of the denominator. This implies that the form of f(x)f(x) should be of the type a(x4)(x1)(x6)\frac{a(x - 4)}{(x - 1)(x - 6)}.

Given these conditions, the appropriate form for f(x)f(x) appears to be:

f(x)=a(x4)(x1)(x6)f(x) = \frac{a(x - 4)}{(x - 1)(x - 6)}

To find the constant aa:

We are told that the graph passes through the point (0,2)(0, 2). Substituting x=0x = 0 and f(0)=2f(0) = 2 into the equation:

2=a(04)(01)(06)2 = \frac{a(0 - 4)}{(0 - 1)(0 - 6)}

Simplifying:

2=a(4)(1)(6)=4a6=2a32 = \frac{a(-4)}{( -1)(-6)} = \frac{-4a}{6} = \frac{-2a}{3}

Solving for aa:

2=2a3a=32 = \frac{-2a}{3} \quad \Rightarrow \quad a = -3

Final equation:

Thus, the equation for f(x)f(x) is:

f(x)=3(x4)(x1)(x6)f(x) = \frac{-3(x - 4)}{(x - 1)(x - 6)}

Would you like a more detailed explanation, or do you have any further questions?

Related questions:

  1. How do vertical and horizontal asymptotes affect the shape of rational functions?
  2. Why does the factor (x4)(x - 4) in the numerator give the x-intercept at x=4x = 4?
  3. How does the degree of the numerator compared to the denominator determine the horizontal asymptote?
  4. What happens to the graph of a rational function when the sign of aa changes?
  5. How do you determine the constant aa when given a point on the graph?

Tip:

For rational functions, the number and location of vertical asymptotes come from the roots of the denominator, while the horizontal asymptote is determined by the degrees of the numerator and denominator.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Vertical Asymptotes
Horizontal Asymptotes
X-Intercepts

Formulas

f(x) = a(x - intercept) / [(x - asymptote1)(x - asymptote2)]

Theorems

The degree of the numerator relative to the denominator determines the horizontal asymptote.

Suitable Grade Level

Grades 10-12