Problem 19 - Calculus Early Transcendentals

Question

Graphing rational functions Use the guidelines of this section to make a complete graph of \(f\) $$f(x)=\frac{x^{2}+12}{2 x+1}$$

Step-by-Step Solution

Answer

Answer: The vertical asymptote is at \(x = -\frac{1}{2}\), the horizontal asymptote is at \(y = \frac{1}{2}\), there is no x-intercept, and the y-intercept is at \((0, 12)\).

Step 1: Find the vertical asymptotes

To find any vertical asymptotes, we need to find the values of \(x\) for which the denominator is equal to 0. In this case, we have: $$2x + 1 = 0$$ Solving for \(x\), we obtain: $$x = -\frac{1}{2}$$ So we have a vertical asymptote at \(x = -\frac{1}{2}\).

Step 2: Find the horizontal asymptotes

To find any horizontal asymptotes, we compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at \(y=0\). If the degrees are equal, the horizontal asymptote is at \(y=\frac{a}{b}\), where \(a\) and \(b\) are the leading coefficients of the numerator and the denominator, respectively. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. In this case, both the numerator and the denominator have the same degree (2 and 1, respectively), so the horizontal asymptote is at: $$y = \frac{1}{2}$$

Step 3: Identify any holes

A rational function has a hole at the value of \(x\) if the denominator and the numerator have a common factor at that value of \(x\). In this case, the numerator and the denominator do not have any common factors, so there are no holes in the function.

Step 4: Find the intercepts

To find the \(x\)-intercept, set the function equal to 0 and solve for \(x\): $$f(x) = \frac{x^2+12}{2x+1} = 0$$ Since the numerator must be equal to 0 for the function to equal 0, we find the \(x\)-intercept: $$x^2 + 12 = 0$$ However, there are no real solutions to this equation, so there is no \(x\)-intercept. To find the \(y\)-intercept, set \(x = 0\): $$f(0) = \frac{0^2 + 12}{2(0) + 1} = \frac{12}{1} = 12$$ So the \(y\)-intercept is at \((0, 12)\).

Step 5: Plot points and sketch the graph

We can now plot a few points around the vertical and horizontal asymptotes to see how the function behaves. For example, we can choose points close to \(x = -\frac{1}{2}\) such as \(x=-1\) and \(x=0\), as well as points above and below the horizontal asymptote, such as \(x = -2\) and \(x = 1\). Using these points, we can sketch a complete graph of the rational function \(f(x) = \frac{x^2 + 12}{2x + 1}\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes are important features of rational functions, as they indicate where the function approaches infinity. To find a vertical asymptote, you need to determine the values of \(x\) that make the denominator zero, because division by zero is undefined in mathematics. This is generally done by solving the equation of the denominator equal to zero.

For instance, in the given function \(f(x) = \frac{x^2+12}{2x+1}\), the denominator is \(2x + 1\). Setting it equal to zero and solving gives:\(2x + 1 = 0\) which simplifies to \(x = -\frac{1}{2}\). Thus, the function has a vertical asymptote at \(x = -\frac{1}{2}\).

Vertical asymptotes tell you that as \(x\) approaches \(-\frac{1}{2}\), the function \(f(x)\) will increase or decrease without bound. You can use this information to understand the behavior of the graph near these points and anticipate how it will stretch towards infinity.

Horizontal Asymptotes

Horizontal asymptotes describe the end behavior of a rational function. They help you understand how the function behaves as \(x\) approaches infinity or negative infinity.

The location of the horizontal asymptote depends on the degrees of the polynomial in the numerator and denominator.

• If the degree of the numerator is less than that of the denominator, the horizontal asymptote is the x-axis \( (y = 0)\).
• If the numerator and denominator have the same degree, the horizontal asymptote is \(y = \frac{a}{b}\), where \(a\) and \(b\) are leading coefficients of the numerator and denominator respectively.
• If the numerator's degree is greater than the denominator's, there is no horizontal asymptote.

In our example, \(f(x) = \frac{x^2+12}{2x+1}\), the numerator (\(x^2\)) and denominator (\(2x\)) both have degrees that are effectively equal once compared (degree 2 for numerator, 1 when not considering the constant). Thus, the horizontal asymptote is found at \(y = \frac{1}{2}\).

This means that as \(x\) increases or decreases towards infinity, the value of the function \(f(x)\) will get closer and closer to \(y = \frac{1}{2}\).

Intercepts of Rational Functions

Intercepts show where the graph of the rational function crosses the x-axis or y-axis.

**Finding the x-intercept:** To find the x-intercept, you set \(f(x)\) to zero and solve for \(x\). This means solving \(x^2 + 12 = 0\). For our function, there are no real solutions to this equation, indicating that the graph of \(f(x)\) does not cross the x-axis. Therefore, it has no x-intercepts.

**Finding the y-intercept:** The y-intercept occurs at \(x = 0\). By substituting \(x = 0\) into the function, you find \(f(0) = \frac{12}{1} = 12\). Thus, the y-intercept is at the point \((0, 12)\).

Intercepts are useful for identifying key points where the graph interacts with the axes and can be starting points for sketching the graph of the function. Observing these intersections helps in understanding the overall layout and symmetry of the graph.