Problem 55 - College Algebra

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

Step 1: Understand the Characteristics

We are given that the function has a vertical asymptote at \(x=-1\), a double zero at \(x=2\), and a \(y\)-intercept at \((0,2)\). A rational function can be represented as \(f(x) = \frac{p(x)}{q(x)}\) where \(p(x)\) and \(q(x)\) are polynomials.

Step 2: Construct Denominator for Vertical Asymptote

To have a vertical asymptote at \(x=-1\), the denominator must have a factor of \((x+1)\). Thus, let \(q(x) = (x+1)\).

Step 3: Construct Numerator for Double Zero

A double zero at \(x=2\) means the numerator will have a factor of \((x-2)^2\). Thus, start with \(p(x) = a(x-2)^2\) where \(a\) is a constant to be determined.

Step 4: Determine the Constant for y-intercept

Substitute \(x=0\) in \(f(x) = \frac{a(x-2)^2}{x+1}\) and set it equal to 2 (the given \(y\)-intercept): \(f(0) = \frac{a(0-2)^2}{0+1} = 2\). Simplifying yields \(\frac{4a}{1} = 2\) which gives \(a = \frac{1}{2}\).

Step 5: Combine Results into a Rational Function

Substitute \(a\) back into \(p(x)\). The complete rational function is \(f(x) = \frac{\frac{1}{2}(x-2)^2}{x+1}\).

Key Concepts

Vertical Asymptote

A vertical asymptote in a rational function is a line where the function itself shoots up to infinity or dives down to negative infinity. This typically happens because of division by zero in the function's denominator. To identify a vertical asymptote, look at the polynomial in the denominator and find the values of x that make it zero.

In our specific case, the vertical asymptote is given at \(x = -1\). This means the denominator of the rational function must include a factor \((x + 1)\). Thus, the rational function would "break down" or become undefined when \(x = -1\), resulting in the vertical asymptote.

Double Zero

A double zero in a rational function means that the zero not only makes the function equal zero at that x-value but does so with a multiplicity greater than one. In simple terms, the graph of the function doesn't just touch the x-axis but actually bounces off of it at that zero.

For a double zero at \(x = 2\), the numerator of the rational function must include a factor of \((x - 2)^2\). The square indicates the double nature, ensuring that when \(x = 2\), the numerator equals zero, confirming it as a zero with multiplicity of two.

Y-intercept

The y-intercept of a rational function is the point where the graph crosses the y-axis. This occurs when \(x = 0\). To find the y-intercept, you simply substitute \(x = 0\) into the function and solve for \(y\).

Given that the y-intercept is \((0, 2)\), when inserting \(x = 0\) into the equation, the result should equal 2. In our function, this was achieved by adjusting the constant \(a\) in the numerator, leading us to find \(a = \frac{1}{2}\) after substitution.

Numerator and Denominator in Rational Functions

The numerator and denominator play crucial roles in shaping the graph of a rational function. In our specific rational function \(f(x) = \frac{\frac{1}{2}(x-2)^2}{x+1}\), understanding these components is key.

The **numerator** \(\frac{1}{2}(x-2)^2\) decides where the function equals zero or has x-intercepts. The double zero at \(x=2\) is enforced here.

The **denominator** \(x+1\) determines the position of vertical asymptotes. It's factors highlight where the function is undefined, thus creating those gaps in the graph where the line may shoot up or down infinitely. Together, the numerator and denominator in a rational function dictate its overall behavior and defining features.