Problem 39 - Precalculus with Calculus Previews

Question

Find a rational function that satisfies the given conditions. There is no unique answer. vertical asymptotes: \(x=-1, x=2\) horizontal asymptote: \(y=3\) \(x\) -intercept: (3,0)

Step-by-Step Solution

Answer

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

Step 1: Understand the Properties of the Rational Function

The rational function needs vertical asymptotes at \(x = -1\) and \(x = 2\), a horizontal asymptote at \(y = 3\), and an \(x\)-intercept at \(x = 3\). A rational function is of the form \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials.

Step 2: Establish the Form Based on Asymptotes

To have vertical asymptotes at \(x = -1\) and \(x = 2\), \(Q(x)\) must include \((x+1)(x-2)\). The horizontal asymptote \(y=3\) suggests \(f(x) \approx \frac{a}{1}\) as \(x \to \infty\), so \(a = 3\). Thus, start with \(f(x) = \frac{3P(x)}{(x+1)(x-2)}\).

Step 3: Include the X-Intercept

For the \(x\)-intercept at \(x=3\), \(P(x)\) must include \((x-3)\). Let \(P(x) = (x-3)\) to have \(x=3\) as a root. So the function becomes \(f(x) = \frac{3(x-3)}{(x+1)(x-2)}\).

Step 4: Verify the Conditions

Check the function \(f(x) = \frac{3(x-3)}{(x+1)(x-2)}\):1. Vertical asymptotes: Set \((x+1)(x-2) = 0\), gives \(x = -1, 2\).2. Horizontal asymptote: As \(x \to \pm\infty\), behavior of \(f(x)\) approaches \(3\).3. \(x\)-intercept: Set \(y=0\) which gives \(x-3=0\), hence \(x=3\). All conditions are satisfied.

Key Concepts

Vertical Asymptotes

In the world of rational functions, vertical asymptotes are the lines where the function grows unbounded, shooting off towards infinity. Imagine you're driving along a road that suddenly turns into a cliff. This is how your graph behaves near a vertical asymptote. For our function, we have vertical asymptotes at \(x = -1\) and \(x = 2\). That means, as \(x\) approaches these values, the function \(f(x)\) becomes infinitely large or small.
To create these asymptotes, the denominator of the rational function must equal zero at \(x = -1\) and \(x = 2\). Therefore, it includes factors like \((x+1)\) and \((x-2)\). It's important to remember that we only get vertical asymptotes if these factors do not cancel out with factors in the numerator.
Summing it up:

• Vertical asymptotes occur where the denominator is zero.
• They show undefined points where the function heads to infinity.
• For \(f(x) = \frac{3(x-3)}{(x+1)(x-2)}\), vertical asymptotes are at \(x = -1\) and \(x = 2\).

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a function as \(x\) approaches infinity or negative infinity. Think of them as the roads your graph drives on as it stretches out towards the horizon. They show the constant value the function will approach, but never quite reach.
For the problem at hand, we have a horizontal asymptote at \(y = 3\). This aligns with the rational function's end behavior, given that the degrees of the numerator and denominator polynomials are equal, leaving us with the ratio of their leading coefficients. The rational function behaves as \(f(x) \approx \frac{a}{1}\), with \(a = 3\). This gives us a constant value that our function will approach when \(x\) goes to infinity.
Key Takeaways:

• Horizontal asymptotes show the long-term behavior of the function.
• They exist because the function levels out as \(x\) becomes very large or very small.
• For \(f(x) = \frac{3(x-3)}{(x+1)(x-2)}\), the horizontal asymptote is at \(y = 3\).

X-Intercepts

The x-intercept of a function is the point where the graph crosses the x-axis. This is where the output value, or \(y\), becomes zero. To determine the x-intercepts of a rational function, set the numerator equal to zero, since the entire fraction equals zero when the numerator is zero.
In this example, the function has an x-intercept at \((3, 0)\). We make sure this condition is satisfied by structuring \(P(x)\) in the numerator to include \((x - 3)\). This ensures that when we substitute \(x = 3\) into our function, the numerator will become zero, and therefore, the function \(f(x)\) will be zero.
To remember:

• X-intercepts occur where the numerator of the rational function is zero.
• It's the point where the graph crosses the x-axis.
• For \(f(x) = \frac{3(x-3)}{(x+1)(x-2)}\), the x-intercept is at \(x = 3\).