Problem 507 - Precalculus

Question

For the following exercises, write an equation for a rational function with the given characteristics. Vertical asymptotes at \(x=-4\) and \(x=-5, x\) -intercepts at \((4,0)\) and \((-6,0),\) Horizontal asymptote at \(y=7\)

Step-by-Step Solution

Answer

The rational function is \(f(x) = 7\frac{(x-4)(x+6)}{(x+4)(x+5)}\).

Step 1: Determine the denominator

Since the rational function has vertical asymptotes at \(x = -4\) and \(x = -5\), this means the denominator of the function will be \((x+4)(x+5)\).

Step 2: Determine the numerator

The \(x\)-intercepts of the function are at \((4,0)\) and \((-6,0)\), which means the numerator will be \((x-4)(x+6)\).

Step 3: Include the Horizontal Asymptote Condition

Given the horizontal asymptote at \(y = 7\), the constants in both the numerator and denominator must ensure the limit behavior of the function reaches 7. Therefore, we modify the function to \(7\frac{(x-4)(x+6)}{(x+4)(x+5)}\), which maintains the horizontal asymptote of \(y=7\).

Step 4: Write the Final Equation

By combining the numerator and denominator along with the horizontal asymptote condition, the rational function is: \(f(x) = 7\frac{(x-4)(x+6)}{(x+4)(x+5)}\).

Key Concepts

Vertical Asymptotes

In the world of rational functions, vertical asymptotes are the mysterious lines where the function "shoots off" towards infinity. This happens when there are values of \(x\) that cause the denominator of the function to become zero. To locate these lines, simply set the denominator equal to zero and solve for \(x\).

For the exercise at hand, we have vertical asymptotes at \(x = -4\) and \(x = -5\). This tells us that when \(x\) reaches these values, the function becomes undefined and the graph has a vertical line where it cannot "cross over." This will be the vertical asymptotes 'guiding' our function. They help us understand how the function behaves around these problematic points.

• The vertical asymptotes occur where the denominator is zero.
• In our function, \((x+4)\) and \((x+5)\) are the terms in the denominator causing these asymptotes.
• We avoid having \(f(x)\) cross at \(x = -4\) and \(x = -5\) because they are not part of the domain of the function.

Horizontal Asymptote

Horizontal asymptotes in rational functions provide a snapshot of how the function behaves as \(x\) either zooms to infinity or plunges towards negative infinity. They indicate the value that \(f(x)\) approaches as \(x\) grows larger and larger or diminishes beyond bounds.

In our exercise, the horizontal asymptote is given as \(y=7\). This means that as \(x\) moves to ±∞, the function values will settle around 7. A horizontal asymptote is typically determined by the degrees of the polynomial in the numerator and the denominator. When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.

• The horizontal asymptote is determined as \(y = \frac{7}{1} = 7\) because we adjusted the leading coefficients to achieve this balance.
• This pattern helps to predict the end behavior of the graph of the rational function.

X-Intercepts

The x-intercepts of a rational function are fundamental points where the graph crosses the x-axis. At these points, the output of the function is zero. To find x-intercepts, set the numerator equal to zero and solve for \(x\). The denominator should not be zero at these points, ensuring these values are true solutions.

In this particular exercise, the x-intercepts are known to be at \((4,0)\) and \((-6,0)\). To achieve these x-intercepts, the numerator of the function, \((x-4)(x+6)\), is structured such that it becomes zero at these points.

• The numerator \((x-4)\) equals zero at \(x=4\).
• The numerator \((x+6)\) equals zero at \(x=-6\).
• These solutions do not make the denominator zero, thus truly achieving the x-intercepts.