Problem 509 - Precalculus

Question

For the following exercises, write an equation for a rational function with the given characteristics. Vertical asymptote at \(x=-1,\) Double zero at \(x=2, y\) -intercept at \((0,2)\)

Step-by-Step Solution

Answer

The equation is \(f(x)=\frac{1}{2} \frac{(x-2)^2}{x+1}\).

Step 1: Understand the Rational Function Components

A rational function is of the form \( f(x) = \frac{p(x)}{q(x)} \), where both \(p(x)\) and \(q(x)\) are polynomials. Given the features mentioned, we need to determine these polynomials.

Step 2: Determine the Vertical Asymptote

A vertical asymptote occurs at values of \(x\) that make the denominator zero but not the numerator. Since we have a vertical asymptote at \(x = -1\), the factor \((x + 1)\) must be in the denominator. Thus, \(q(x) = (x+1)\).

Step 3: Determine the Double Zero

A double zero at \(x = 2\) means \(x = 2\) is a root of the numerator with a multiplicity of 2. Therefore, the numerator must include \((x - 2)^2\). Thus, \(p(x) = (x-2)^2\).

Step 4: Adjust to Meet the Y-Intercept Condition

The y-intercept of a function occurs when \(x = 0\). We need the function \(f(x)\) to satisfy \(f(0) = 2\). Substituting \(x = 0\) into \(f(x) = \frac{(x-2)^2}{x+1}\), gives \(f(0) = \frac{(0-2)^2}{0+1} = \frac{4}{1} = 4\), but we need 2. To adjust, multiply \(p(x)\) by \(\frac{1}{2}\). The new function is \(f(x) = \frac{1}{2} \frac{(x-2)^2}{x+1}\).

Step 5: Verify the Rational Function

Check the function \(f(x) = \frac{1}{2} \frac{(x-2)^2}{x+1}\). Verify vertical asymptote at \(x = -1\), confirming \(x+1\) in denominator causes division by zero. Confirm the double root at \(x = 2\) by checking \((x-2)^2\). Lastly, ensure \(f(0) = 2\), as previously calculated. All conditions are satisfied.

Key Concepts

Vertical Asymptote

A vertical asymptote in a rational function represents a value for which the function tends to infinity. This occurs when the denominator polynomial, which is part of the divisor, equals zero, while the numerator does not equal zero at this same point. In our exercise, the vertical asymptote is at \(x = -1\). Thus, in our function world's logic, the denominator must include the factor \((x + 1)\). To break it down:

• A vertical asymptote is typically written as \(x = a\).
• The denominator of the rational function includes \((x - a)\).
• Ensure the numerator does not cancel this factor.

With \(x = -1\) as the vertical asymptote, the denominator becomes \((x + 1)\) to meet this criteria. No term in the numerator should cancel out this factor at \(x = -1\). This creates an infinite behavior of the function near \(x = -1\).

Double Zero

When you hear the term "double zero," it refers to a root of the numerator polynomial where the value of the rational function becomes zero with a multiplicity greater than one. In simpler terms, it's where the graph touches or bounces off the x-axis instead of just crossing it. In our case, this happens at \(x = 2\). This concept looks like this in practice:

• A single zero would be \((x - 2)\).
• A double zero means we have \((x - 2)^2\), showing it repeats.

Thus, the numerator for our function includes the term \((x - 2)^2\). This assures that at \(x = 2\), the function does not merely touch zero but is zero itself twice, indicating a bounce instead of a crossing on the graph.

Y-Intercept

The y-intercept of a rational function is simply the point where the graph intersects the y-axis. This occurs when \(x = 0\). For the function’s y-intercept to be the point \((0, 2)\), the function must satisfy \(f(0) = 2\). Let's delve into how we adjust our rational function to ensure this condition is met.For our solved rational function:

• First, substitute \(x = 0\) into the function \(f(x) = \frac{1}{2} \frac{(x-2)^2}{x+1}\).
• Calculate \(f(0) = \frac{1}{2} \frac{(0-2)^2}{0+1} = \frac{1}{2} \cdot \frac{4}{1} = 2\).

By ensuring that the numerator and denominator are adjusted appropriately through multiplying by a constant in this case, the y-intercept correctly sits at \((0, 2)\). This adjustment confirms that our function hugs the y-axis at the right place.

Numerator and Denominator Polynomials

In any rational function, the numerator and denominator are key players in determining the traits of the function. They dictate features such as zeros, asymptotes, and intercepts. A rational function can be expressed in the form \( f(x) = \frac{p(x)}{q(x)} \) where \(p(x)\) and \(q(x)\) are polynomials.Our task for the rational equation involves:

• The numerator: Contains \((x - 2)^2\) for the double zero at 2.
• The denominator: Contains \((x + 1)\) to ensure a vertical asymptote at \(x = -1\).

The beauty of rational functions is in their flexibility and the control they provide. By carefully selecting terms for \(p(x)\) and \(q(x)\), specific characteristics like zeros and asymptotes can be precisely controlled to model real-world behaviors or desired graphical features. This functional manipulation provides the foundation for our rational function modeling.