Problem 46 - Precalculus: Functions and Graphs

Question

Find an equation of a rational function \(f\) that satisfles the given conditions. vertical asymptotes: \(x=-2, x=0\) horizontal asymptote: \(y=0\) \(x\) -intercept: \(2 ; f(3)=1\)

Step-by-Step Solution

Answer

The rational function is \( f(x) = \frac{15(x - 2)}{x(x + 2)} \).

Step 1: Identify the Form of the Rational Function

A rational function can generally be written as \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials. Vertical asymptotes occur where \( q(x) = 0 \), and a horizontal asymptote of \( y=0 \) suggests that the degree of \( p(x) \) is less than that of \( q(x) \).

Step 2: Set the Denominator Based on Vertical Asymptotes

Since we have vertical asymptotes at \( x = -2 \) and \( x = 0 \), the denominator \( q(x) \) must be \( (x + 2)x \) or simply \( x(x + 2) \), leading to \( f(x) = \frac{p(x)}{x(x+2)} \).

Step 3: Determine the Numerator from X-intercept and Other Criteria

The function has an \( x \)-intercept at \( x = 2 \), so the numerator must be \( p(x) = k(x - 2) \) for some constant \( k \). Thus, \( f(x) = \frac{k(x - 2)}{x(x + 2)} \).

Step 4: Use Given Point to Find Constant \( k \)

Substitute \( f(3) = 1 \) into the equation to solve for \( k \):-\( f(3) = 1 = \frac{k(3 - 2)}{3(3+2)} = \frac{k \cdot 1}{3 \cdot 5} = \frac{k}{15} \).-So, \( 1 = \frac{k}{15} \Rightarrow k = 15 \).

Step 5: Write Final Equation of the Rational Function

Substitute \( k = 15 \) into the expression for \( f(x) \):\[ f(x) = \frac{15(x - 2)}{x(x + 2)} \].

Key Concepts

Vertical Asymptotes

Vertical asymptotes are a fundamental concept in understanding rational functions. Essentially, they represent the values of \( x \) where the function becomes undefined and the graph blows up towards infinity. For the rational function \( f(x) = \frac{p(x)}{q(x)} \), vertical asymptotes occur where the denominator \( q(x) = 0 \). To find them, set the denominator equal to zero and solve for \( x \).

In our exercise, we have vertical asymptotes at \( x = -2 \) and \( x = 0 \). This tells us that \( q(x) \), the denominator, must be such that \( q(x) = (x + 2)x \). When you see these values \(-2\) and \(0\), they inform where the graph of the function will dramatically decrease or increase without limit, providing critical insight into the behavior of the function. Remember, vertical asymptotes help define the boundaries of the rational function's behavior on the graph.

Horizontal Asymptote

Horizontal asymptotes give us an idea of where the graph of a function will settle as \( x \) approaches very large positive or negative values (i.e., towards infinity or negative infinity). For rational functions expressed as \( f(x) = \frac{p(x)}{q(x)} \), the horizontal asymptote is determined by comparing the degrees of the numerator and the denominator polynomial.

If the degree of the numerator \( p(x) \) is less than the degree of the denominator \( q(x) \), the horizontal asymptote will be at \( y = 0 \). This means that as \( x \) tends toward infinity, the value of \( f(x) \) approaches zero. In our specific example, since the degree of the numerator, which is \( 1 \) (from \( p(x) = k(x - 2) \)), is less than the degree of the denominator, which is \( 2 \) (from \( q(x) = x(x + 2) \)), we have a horizontal asymptote at \( y = 0 \).

• This helps understand how the function behaves at extremes.
• It explains the ultimate destiny of the function's graph far from the origin.

X-intercept

The x-intercept of a function is the point where the graph crosses the x-axis. At this point, the value of \( y \) is zero. For rational functions, to find the x-intercept, set the numerator \( p(x) \) equal to zero because the entire fraction equals zero when the numerator is zero (and the denominator is non-zero).

In our problem, the x-intercept is given as \( x = 2 \). This implies that \( p(x) = k(x - 2) \) must equal zero when \( x = 2 \), indicating \( 2 \) is a root of the numerator. So, part of constructing the function was ensuring \( p(x) = k(x - 2) \). By knowing the x-intercept, we are guided to understand that the graph of the function will cross the x-axis exactly at this point.

• Finding the x-intercept involves solving \( p(x) = 0 \).
• The x-intercept provides a check on how the graph will interact with the x-axis.