Problem 56 - College Algebra

Question

For the following exercises, write an equation for a rational function with the given characteristics. Vertical asymptote at \(x=3\) double zero at \(x=1, y\) -intercept at \((0,4)\)

Step-by-Step Solution

Answer

The rational function is \( f(x) = \frac{-12(x-1)^2}{x-3} \).

Step 1: Identify Characteristics Needed

To write an equation for the rational function, note the required features: a vertical asymptote at \(x = 3\), a double zero at \(x = 1\), and a \(y\)-intercept at \((0,4)\).

Step 2: Determine the Denominator

A vertical asymptote occurs where the denominator is zero but the numerator is not. So, we need a factor of \((x-3)\) in the denominator.

Step 3: Determine the Numerator

A double zero at \(x=1\) means the numerator must include \((x-1)^2\). Start with the numerator as \((x-1)^2\).

Step 4: Include y-intercept Condition

The condition \(f(0)=4\) helps determine the constant multiplier. Substituting \(x=0\) into preliminary function: \[ f(x) = \frac{a(x-1)^2}{x-3} \]Substitute \(x=0\) and solve for \(a\): \[4 = \frac{a(0-1)^2}{0-3} \Rightarrow 4 = \frac{a}{3} \Rightarrow a = -12\].

Step 5: Formulate the Function

Thus, the rational function that satisfies all conditions is \[ f(x) = \frac{-12(x-1)^2}{x-3} \].

Key Concepts

Vertical Asymptote

When studying rational functions, it's essential to understand the concept of a vertical asymptote. A vertical asymptote is a vertical line that a function approaches but never touches or crosses. This occurs because at the value of the asymptote, the denominator of the rational function becomes zero, making the function undefined at that point.
When given a vertical asymptote at a specific point, such as at \(x = 3\), it means that the denominator must have a corresponding factor. In our case, the denominator requires a term of \( (x - 3) \). This factor ensures that the function is undefined at \(x = 3\), creating the vertical asymptote.
It’s crucial to remember that while the denominator leads to vertical asymptotes, these points are not in the domain of the function. Thus, always check these values to avoid them when evaluating the function.

Double Zero

The term 'double zero' refers to a root of a function that is repeated or has a multiplicity of two. In simpler terms, it is a point where the function touches the x-axis and turns, rather than crossing it. For example, if a rational function has a double zero at \(x = 1\), it means the numerator must include the factor \( (x-1)^2 \).
The exponent of two indicates that the zero is counted twice or that the x-axis is 'touched' but not crossed. This characteristic affects the graph’s shape, making it appear that the curve flattens out at this zero rather than cutting through the axis.In our function, the numerator's form, \( (x - 1)^2 \), directly reflects the double zero at \(x = 1\). Having such a zero impacts how the numerator behaves and helps dictate the condition of the overall function.

Y-Intercept

The y-intercept of a function is a point where the graph crosses the y-axis. At this point, the value of \( x \) is zero. Finding the y-intercept is vital as it gives insight into the starting point of the function's curve on the vertical axis.
The y-intercept can be found by substituting \(x = 0\) into the equation of the function. In the exercise, the given y-intercept is \( (0, 4) \). So, ensuring that the function meets this specification means we must adjust or solve for parameters in the function, such as adjusting a coefficient.
In our solution, setting \( f(0) = 4 \) allowed us to determine the necessary multiplier for our function equation. This step ensures the graph aligns correctly and accurately displays the y-intercept in accordance with the given problem.

Numerator and Denominator in Functions

The numerator and denominator are the two fundamental parts of any rational function. Understanding their roles is pivotal when constructing and analyzing these functions.
The numerator contains the factors responsible for the zeros of the function. In this exercise, the numerator \( (x - 1)^2 \) is specifically designed to address the requirement of a double zero at \(x = 1\). The exponent indicates that the zero is doubled and affects how the function behaves at that point.
On the other hand, the denominator is primarily responsible for the vertical asymptotes. Here, the factor \( (x - 3) \) in the denominator creates a vertical asymptote at \(x = 3\), implying that the function becomes undefined there. The interplay of these components is crucial for determining the function's overall behavior, especially how it approaches both zeros and asymptotes.