Problem 43 - Precalculus with Calculus Previews

Question

Find the asymptotes and any holes in the graph of the given rational function. Find \(x\) - and \(y\) -intercepts of the graph. Sketch the graph \(f\). $$ f(x)=\frac{x+1}{x\left(x^{2}+4 x+3\right)} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = 0\) and \(x = -3\); hole at \(x = -1\); no \(x\)- or \(y\)-intercepts.

Step 1: Find Vertical Asymptotes

To find the vertical asymptotes of a rational function, we set the denominator equal to zero because vertical asymptotes occur where the function is undefined. Solve the equation \(x(x^2 + 4x + 3) = 0\). This expands to \(x(x+1)(x+3) = 0\). The solutions are \(x = 0\), \(x = -1\), and \(x = -3\). These are potential vertical asymptotes.

Step 2: Check for Holes in the Graph

A hole in the graph occurs if a factor is cancelled from both the numerator and the denominator. Here, the numerator is \(x+1\) and the denominator has \(x+1\) as a factor. Thus, there is a hole at \(x = -1\). Therefore, only \(x = 0\) and \(x = -3\) are vertical asymptotes.

Step 3: Find Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the numerator and denominator. The degree of the numerator \(x+1\) is 1, and the degree of the denominator \(x(x^2+4x+3)\) is 3. Since the degree of the denominator is greater, the horizontal asymptote is \(y = 0\).

Step 4: Find x-intercepts

The \(x\)-intercepts occur where the numerator is zero, \(f(x) = 0\). Solve \(x + 1 = 0\), which gives \(x = -1\). However, since \(x = -1\) corresponds to a hole, there is no real \(x\)-intercept.

Step 5: Find y-intercepts

The \(y\)-intercept occurs where \(x = 0\). Evaluate \(f(0) = \frac{0 + 1}{0(0^2 + 4 \, 0 + 3)}\), which results in division by zero. Thus, there is no real \(y\)-intercept.

Step 6: Sketch the Graph of f(x)

Summarize the findings: vertical asymptotes at \(x = 0\), \(x = -3\), a horizontal asymptote at \(y = 0\), a hole at \(x = -1\), and no real \(x\) or \(y\) intercepts. Create a sketch reflecting these features, ensuring the behavior approaches asymptotes correctly, with the curve breaking at the hole located at \(x = -1\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes are important features of rational functions and occur where the function approaches infinity. They are found by setting the denominator equal to zero and solving for the variable. For the given function, we set the denominator, \(x(x^2+4x+3)\), to zero, revealing potential vertical asymptotes at \(x=0\), \(x=-1\), and \(x=-3\). However, due to the hole, \(x = -1\) is not a vertical asymptote. This means the remaining vertical asymptotes are at \(x = 0\) and \(x = -3\). These lines represent values that the function can never reach because at these points, the function tends to either positive or negative infinity, causing the graph to exhibit a vertical break.

Horizontal Asymptotes

Horizontal asymptotes give insight into the end behavior of rational functions. They show what value the function approaches as \(x\) goes to infinity or negative infinity. To determine horizontal asymptotes, compare the degrees of the numerator and the denominator. In this function, the numerator \(x+1\) has degree 1, and the denominator \(x(x^2+4x+3)\) has degree 3. Because the degree of the denominator is larger, the horizontal asymptote is at \(y = 0\). This tells us that as \(x\) becomes very large or very small, the value of \(f(x)\) will get closer and closer to zero but will never actually reach it.

Holes in Graphs

Holes in graphs occur when a factor is canceled from both the numerator and the denominator, indicating a common factor that does not contribute to a vertical asymptote. For the function \(f(x) = \frac{x+1}{x(x^2+4x+3)}\), the factor \(x+1\) appears both in the numerator and the denominator. This results in a hole at \(x = -1\). Although this point is not noticeable directly on the graph as a break, mathematically, the function is undefined there. Instead of a single point on the graph, there is a removable discontinuity. Always check for such factors as they influence the graph's domain privately.

X-Intercepts

The \(x\)-intercepts are the points where the graph crosses the x-axis, or where \(f(x) = 0\). Therefore, these intercepts are found by setting the numerator equal to zero. Here, \(x+1 = 0\) suggests an \(x\)-intercept at \(x = -1\). However, because this factor corresponds to a hole rather than a vertical crossing, the actual graph does not cross the x-axis at this point. Thus, the graph of this function has no visible \(x\)-intercepts, underlining the importance of understanding the impact of canceled factors.

Y-Intercepts

The \(y\)-intercept is the point where the graph of a function meets the \(y\)-axis and can be found by evaluating \(f(x)\) at \(x = 0\). However, for this function, trying to calculate \(f(0)\) leads to division by zero due to the denominator term \(0 \times (0^2 + 4 \times 0 + 3)\). This results in an undefined expression, indicating that there is no \(y\)-intercept. This is a special case where the function does not have a defined output at \(x = 0\), often tied to a vertical asymptote or hole at this x-value, as seen with this function.