Problem 36 - 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. $$f(x)=\frac{2 x+1}{(x+2)(x+4)}$$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = -2\) and \(x = -4\), horizontal at \(y = 0\); intercepts: \((-\frac{1}{2}, 0)\) and \((0, \frac{1}{8})\). Draw with these points.

Step 1: Identify the vertical asymptotes

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\): \[(x+2)(x+4) = 0\]This gives two solutions, \(x = -2\) and \(x = -4\). Therefore, the vertical asymptotes are at \(x = -2\) and \(x = -4\).

Step 2: Identify the horizontal asymptote

Since the degrees of the numerator and denominator are different, determine the horizontal asymptote by comparing the degrees. The numerator has degree 1, and the denominator has degree 2. When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

Step 3: Determine the x-intercept

The \(x\)-intercepts can be found by setting the numerator equal to zero and solving for \(x\): \[2x + 1 = 0\]Solving gives \(x = -\frac{1}{2}\). Therefore, the graph intersects the x-axis at \((-\frac{1}{2}, 0)\).

Step 4: Determine the y-intercept

The \(y\)-intercept is found by setting \(x = 0\) in the function. Calculate:\[f(0) = \frac{2(0) + 1}{(0+2)(0+4)} = \frac{1}{8}\]So the \(y\)-intercept is at \((0, \frac{1}{8})\).

Step 5: Sketch the graph

Draw the vertical asymptotes as dashed lines at \(x = -2\) and \(x = -4\). Draw the horizontal asymptote as a dashed line at \(y = 0\). Mark the intercepts found in Steps 3 and 4: \((-\frac{1}{2}, 0)\) and \((0, \frac{1}{8})\). Using these intercepts and asymptotes, sketch the curve taking into account the behavior near the asymptotes and ensuring the curve approaches the asymptotes as \(|x|\) becomes large.

Key Concepts

Understanding Vertical Asymptotes

Vertical asymptotes occur in a rational function where the denominator equals zero, leading to undefined values for \( f(x) \). In the given function \( f(x) = \frac{2x + 1}{(x+2)(x+4)} \), finding vertical asymptotes involves these steps:

• Set the denominator \((x+2)(x+4)\) equal to zero.
• Solve the equation \((x+2)(x+4) = 0\) to find the values of \( x \) that make the function undefined.
• The solutions \( x = -2 \) and \( x = -4 \) are the vertical asymptotes.

Vertical asymptotes are shown as dashed lines on a graph. The curve of a rational function will "approach" these lines but will never touch or cross them. This tendency signifies regions where the function increases or decreases without bound as \( x \) approaches the line.

Understanding Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \( x \) trends towards positive or negative infinity. Unlike vertical asymptotes, these relate to the degrees of the polynomials in the rational function. For the function \( f(x) = \frac{2x + 1}{(x+2)(x+4)} \):

• The numerator degree is 1 and the denominator degree is 2, making the denominator's degree higher.
• When this is the case, the horizontal asymptote is the line \( y = 0 \).

This means as \( x \) grows very large in magnitude (both positive and negative), the value of \( f(x) \) will olook more like 0. The graph visually "flattens out" along the \( y \)-axis at the horizontal asymptote which acts as a boundary it will never cross but only approach.

Finding Intercepts in Graphing

Intercepts are key points to find when graphing as they show where the function crosses the x- or y-axis. **X-Intercept:** To find the x-intercept:

• Set the numerator of the function to zero, \( 2x + 1 = 0 \), and solve for \( x \).
• Solve \( x = -\frac{1}{2} \) to find the x-intercept at \(( -\frac{1}{2}, 0) \).

The x-intercept shows the point where the graph crosses the x-axis. **Y-Intercept:** For the y-intercept:

• Set \( x \) to zero in the function and calculate \( f(0) = \frac{1}{8} \).
• This means the y-intercept is found at \((0, \frac{1}{8}) \).

The y-intercept indicates where the graph crosses the y-axis. Both of these intercepts are critical for constructing the overall shape of the graph, showing where the function directly makes contact with the axes.