Problem 44 - A Graphical Approach to College Algebra

Question

Sketch a graph of each rational function. Your graph should include all asymptotes. Do not use a calculator. $$f(x)=\frac{16 x^{2}-9}{x^{2}-9}$$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = 3\), \(x = -3\); horizontal asymptote at \(y = 16\); x-intercepts at \(x = \pm \frac{3}{4}\).

Step 1: Identify the Vertical Asymptotes

The function \(f(x) = \frac{16x^2 - 9}{x^2 - 9}\) is undefined when the denominator is zero. Set the denominator equal to zero and solve for \(x\):\[x^2 - 9 = 0\]\[x^2 = 9\]\[x = \pm 3\]Thus, the vertical asymptotes occur at \(x = 3\) and \(x = -3\).

Step 2: Determine the Horizontal Asymptote

For rational functions where the degrees of the numerator and denominator are the same, the horizontal asymptote is found by the leading coefficients of the numerator and denominator. Here both the numerator and denominator are degree 2.The leading coefficient of the numerator is 16 and the denominator is 1.Thus, the horizontal asymptote is \(y = \frac{16}{1} = 16\).

Step 3: Analyze the Behavior Around Vertical Asymptotes

Near the vertical asymptotes, the behavior of the graph can be determined by checking the sign of the function slightly to the left and right of each asymptote. Consider points such as \(x = 2.9\) and \(x = 3.1\), and also \(x = -2.9\) and \(x = -3.1\), to identify where the graph goes to \(+\infty\) or \(-\infty\).

Step 4: Check for Intercepts

To find the x-intercepts, set the numerator equal to zero:\[16x^2 - 9 = 0\]\[16x^2 = 9\]\[x^2 = \frac{9}{16}\]\[x = \pm \frac{3}{4}\]The x-intercepts are \(x = \frac{3}{4}\) and \(x = -\frac{3}{4}\).There is no y-intercept since the function is undefined at \(x = 0\).

Step 5: Sketch the Graph

With all critical points and asymptotes known:- Draw vertical asymptotes at \(x = 3\) and \(x = -3\).- Draw the horizontal asymptote at \(y = 16\).- Plot the intercepts at \(x = \frac{3}{4}\) and \(x = -\frac{3}{4}\).- Sketch the graph approaching the asymptotes and passing through the intercepts, considering the sign analysis in step 3.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are key features of rational functions. They occur at values of \(x\) where the function is undefined because the denominator equals zero.
For our function, \(f(x) = \frac{16x^2 - 9}{x^2 - 9}\), we find the vertical asymptotes by setting the denominator \(x^2 - 9\) equal to zero. Solving \(x^2 - 9 = 0\) gives \(x = \pm 3\).
This means our function has vertical asymptotes at \(x = 3\) and \(x = -3\). These asymptotes divide the graph into sections where the behavior of the function needs separate analysis. Remember, at these lines, the graph will spike up to infinity (\(\infty\)) or down to negative infinity (\(-\infty\)).
To understand how the graph behaves around these lines, pick values of \(x\) just to the left and right of \(x = 3\) and \(x = -3\). This will show if the graph heads towards \(+\infty\) or \(-\infty\).

• Set the denominator to zero to find vertical asymptotes.
• For \(x^2 - 9 = 0\), solve for \(x\).
• Vertical asymptotes occur at \(x = \pm 3\).

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of the rational function as \(x\) approaches infinity (large positive or negative values).
In our function \(f(x) = \frac{16x^2 - 9}{x^2 - 9}\), both the numerator and the denominator have a degree of 2. When this happens, you find the horizontal asymptote by dividing the leading coefficients.
The leading coefficient of the numerator is 16, and for the denominator, it is 1. Thus, the horizontal asymptote is \(y = \frac{16}{1} = 16\).
This asymptote tells us that as \(x\) becomes very large in the positive or negative direction, the value of \(f(x)\) approaches 16.
Horizontal asymptotes give us insights into the end behavior of a function, showing the trend that the function values follow.

• Check the degree of the numerator and denominator.
• If they are the same, divide the leading coefficients for the horizontal asymptote.
• For \(f(x) = \frac{16x^2 - 9}{x^2 - 9}\), the horizontal asymptote is \(y = 16\).

Intercepts

Intercepts are points where the graph of a function crosses the axes. For rational functions, these are identified by setting the numerator equal to zero (for x-intercepts) or analyzing when \(x = 0\) (for y-intercepts).
In the function \(f(x) = \frac{16x^2 - 9}{x^2 - 9}\), to find the x-intercepts, set the numerator equal to zero: \(16x^2 - 9 = 0\). Solve this equation to obtain \(x = \pm \frac{3}{4}\). Thus, the graph intersects the x-axis at these points.
For the y-intercept, calculate \(f(0)\). However, because the denominator \(x^2 - 9\) equals \(-9\) when \(x = 0\), the function seems undefined, so no y-intercept exists.
Understanding the intercepts helps in sketching the function and knowing exactly where it will touch the axes, providing crucial points for drawing the graph.

• Find x-intercepts by solving \(16x^2 - 9 = 0\).
• The solutions are \(x = \pm \frac{3}{4}\).
• No y-intercept for our function as it is undefined at \(x = 0\).