Problem 14 - Precalculus: Functions and Graphs

Question

Sketch the graph of \(f\) $$f(x)=\frac{(5 x+3)(x+1)}{(3 x-7)(x+1)}$$

Step-by-Step Solution

Answer

The graph features a vertical asymptote at \( x = \frac{7}{3} \), a horizontal asymptote at \( y = \frac{5}{3} \), a y-intercept at \( y = -\frac{3}{7} \), and an x-intercept at \( x = -\frac{3}{5} \).

Step 1: Identify the Domain

The function given is \( f(x) = \frac{(5x+3)(x+1)}{(3x-7)(x+1)} \). To find the domain, identify any values of \( x \) that cause the function to be undefined. This occurs when the denominator is zero. Therefore, set the denominator equal to zero: \((3x-7)(x+1) = 0\). Solving these equations, we find that \( x = \frac{7}{3} \) and \( x = -1 \). Thus, the domain is all real numbers except \( x = \frac{7}{3} \) and \( x = -1 \).

Step 2: Simplify the Function

The given function is \( f(x) = \frac{(5x+3)(x+1)}{(3x-7)(x+1)} \). Notice that \( (x+1) \) appears in both the numerator and the denominator, meaning it can be simplified away, assuming \( x eq -1 \). Simplifying the expression, it becomes \( f(x) = \frac{5x+3}{3x-7} \). Remember the domain exclusions include \( x = -1 \) due to the original function.

Step 3: Find Vertical Asymptotes

Vertical asymptotes occur where the function is undefined and not canceled by factors in the numerator. From the simplified form \( f(x) = \frac{5x+3}{3x-7} \), the denominator is zero when \( x = \frac{7}{3} \). Thus, \( x = \frac{7}{3} \) is a vertical asymptote. \( x = -1 \) is not a vertical asymptote since it's canceled, but it's still an undefined point.

Step 4: Find Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. Both the numerator (\(5x+3\)) and the denominator (\(3x-7\)) are first-degree polynomials. Thus, the horizontal asymptote is given by the ratio of the leading coefficients: \( y = \frac{5}{3} \).

Step 5: Calculate Intercepts

For the y-intercept, substitute \( x = 0 \) into the simplified expression \( f(x) = \frac{5x+3}{3x-7} \), giving \( f(0) = \frac{3}{-7} = -\frac{3}{7} \). For x-intercepts, set the numerator equal to zero: \( 5x+3 = 0 \). Solving, \( x = -\frac{3}{5} \) is the x-intercept.

Step 6: Sketch the Graph

Start by plotting the identified intercepts: \((-\frac{3}{5}, 0)\) and \((0, -\frac{3}{7})\). Plot the vertical asymptote \( x = \frac{7}{3} \) as a dashed vertical line and the horizontal asymptote \( y = \frac{5}{3} \) as a dashed horizontal line. The undefined point at \( x = -1 \) is indicated with an open circle. Sketch the curve accordingly, ensuring it approaches the asymptotes correctly, increasing or decreasing without crossing them.

Key Concepts

Domain of a Function

When dealing with rational functions like \( f(x) = \frac{(5x+3)(x+1)}{(3x-7)(x+1)} \), determining the domain is crucial. The domain of a function represents all the possible input values (in this case, x-values) for which the function is defined.
For rational functions, any values of \( x \) that result in a zero in the denominator make the function undefined. To find these values, we set the denominator equation to zero:

• \((3x-7)(x+1) = 0\)

Solving this gives \( x = \frac{7}{3} \) and \( x = -1 \). These are the values excluded from the domain. Hence, the domain of \( f(x) \) is all real numbers except \( x = \frac{7}{3} \) and \( x = -1 \). When sketching the function, these values indicate discontinuities or, in some cases, vertical asymptotes or removable discontinuities.

Asymptotes

Asymptotes are lines that a graph approaches but never touches. They provide insight into the behavior of a function at extreme x-values or around undefined points. For rational functions, we commonly deal with two types: vertical and horizontal asymptotes.
To find vertical asymptotes, consider the values that make the denominator zero and are not simplified away. In our function \( f(x) = \frac{5x+3}{3x-7} \), the vertical asymptote is at \( x = \frac{7}{3} \) because at this point, the denominator equals zero and the factor was not canceled out.
Horizontal asymptotes depend on the degrees of the numerator and denominator polynomials. If both have the same degree, the horizontal asymptote is the ratio of their leading coefficients. For \( f(x) = \frac{5x+3}{3x-7} \), the horizontal asymptote is \( y = \frac{5}{3} \). This shows that as \( x \) becomes very large, the function approaches \( y = \frac{5}{3} \) but never actually reaches it.

Intercepts

Intercepts are the points where a graph crosses the axes. Understanding intercepts can simplify the process of sketching the function graph.
The y-intercept is found by substituting \( x = 0 \) into the function. This gives the point where the graph crosses the y-axis. For \( f(x) = \frac{5x+3}{3x-7} \), substituting \( x = 0 \) results in \( f(0) = -\frac{3}{7} \). Thus, the y-intercept is at

• \((0, -\frac{3}{7})\)

To find x-intercepts, set the numerator equal to zero and solve for \( x \). This will provide points where the graph crosses the x-axis. For this function, solving \( 5x+3 = 0 \) gives \( x = -\frac{3}{5} \), signifying an x-intercept at

• \((-\frac{3}{5}, 0)\)

These intercepts serve as essential anchor points for accurately sketching the graph.