Problem 52 - College Algebra with Corequisite Support

Question

For the following exercises, write an equation for a rational function with the given characteristics. Vertical asymptotes at \(x=-4\) and \(x=-1\), \(x\) -intercepts at (1,0) and \((5,0), y\) -intercept at (0,7)

Step-by-Step Solution

Answer

The rational function is \( f(x) = \frac{28}{5} \cdot \frac{(x - 1)(x - 5)}{(x + 4)(x + 1)} \).

Step 1: Determine the Vertical Asymptotes

Vertical asymptotes occur at the points where the denominator of the rational function is zero (but not the numerator). Since the vertical asymptotes are given at \(x = -4\) and \(x = -1\), the denominator should have factors \((x + 4)\) and \((x + 1)\). Thus, the denominator can be expressed as \((x + 4)(x + 1)\).

Step 2: Determine the x-intercepts

The \(x\)-intercepts are the points where the numerator equals zero. Given \(x\)-intercepts at (1,0) and (5,0), the numerator should have factors \((x - 1)\) and \((x - 5)\). The numerator can, therefore, be expressed as \((x - 1)(x - 5)\).

Step 3: Form the Base Rational Function

Combine the factors from steps 1 and 2 to form the base rational function as follows: \[ f(x) = \frac{(x - 1)(x - 5)}{(x + 4)(x + 1)} \] However, this function needs to pass through the \(y\)-intercept given as (0,7).

Step 4: Adjust with the y-intercept

To find the constant factor \(k\) that ensures the \(y\)-intercept is at (0,7), substitute \(x = 0\) and \(f(x) = 7\) into the equation: \[ 7 = k \cdot \frac{(0 - 1)(0 - 5)}{(0 + 4)(0 + 1)} \] Simplifying, \[ 7 = k \cdot \frac{5}{4} \] Solving for \(k\), \[ k = \frac{28}{5} \].

Step 5: Write the Final Equation

The final rational function, incorporating all the characteristics, is given by: \[ f(x) = \frac{28}{5} \cdot \frac{(x - 1)(x - 5)}{(x + 4)(x + 1)} \].

Key Concepts

Vertical Asymptote

A vertical asymptote in a rational function represents a value that the function cannot equal, leading to the "infinite" behavior of the graph as it approaches this value. In simpler terms, it's a vertical line on the graph where the function shoots up to infinity or down to negative infinity, depending on the function's behavior.

• Vertical asymptotes are found by setting the denominator of the rational function to zero, and solving for the variable.
• For example, in our exercise, we have vertical asymptotes at \(x = -4\) and \(x = -1\). This means that the function's denominator includes the factors \((x + 4)\) and \((x + 1)\).

Bearing this in mind, remember vertical asymptotes are not to be confused with removable discontinuities, where a factor might cancel out with a similar factor in the numerator.

x-intercept

The x-intercept in a rational function is where the graph crosses the x-axis. At these points, the output of the function is zero.

• This happens when the numerator of a rational function equals zero, because dividing zero by any number results in zero.
• In our specific case, the x-intercepts are given at the points \((1,0)\) and \((5,0)\). This means the numerator includes the factors \((x - 1)\) and \((x - 5)\).

It's important that these intercepts are roots of the numerator, making the value of the function zero exactly at these x-values. This is crucial for sketching the graph and understanding the point where the graph "touches" or "crosses" the x-axis.

y-intercept

To find the y-intercept of a rational function, we look at where the graph crosses the y-axis, which is when the input \(x\) is zero.

• This involves substituting \(x = 0\) into the rational function, and solving for \(f(x)\).
• In the described exercise, the y-intercept is at \((0, 7)\), meaning when \(x = 0\), \(f(x)\) gives us the value 7.

It's a simple concept, but knowing the y-intercept helps graph the function and gives a concrete starting position, ensuring that you visualize how the graph interacts with the coordinate axes.

Denominator of a Rational Function

Understanding the denominator is crucial for rational functions as it dictates much of the behavior of the graph.

• The denominator determines not only the vertical asymptotes (where it equals zero) but also influences the domain of the function, marking the x-values that are excluded because they make the function undefined.
• In our case, factors of the denominator are \((x + 4)\) and \((x + 1)\), setting the values \(x = -4\) and \(x = -1\) where the function will not exist.

Keep in mind, simplifying any common factors with the numerator can change the layout of asymptotes and intercepts, so always watch for possible reduction in the function to reveal "holes" or cancellations that could adjust the graph's look. This clear view helps in accurately sketching the function's graph, providing insights into its behavior.