Problem 64 - Algebra and Trigonometry

Question

Find the intercepts and asymptotes, and then sketch a graph of the rational function and state the domain and range. Use a graphing device to confirm your answer. $$ t(x)=\frac{x^{3}-x^{2}}{x^{3}-3 x-2} $$

Step-by-Step Solution

Answer

The function has x-intercepts at \(x=0\) and \(x=1\), a y-intercept at \(y=0\), a vertical asymptote at \(x=-1\), and a horizontal asymptote at \(y=1\). Domain: \((-\infty,-1) \cup (-1,1) \cup (1, \infty)\); Range: \((-\infty,1) \cup (1, \infty)\).

Step 1: Find the x-intercepts

Set the numerator of the function equal to zero to find the x-intercepts. The numerator is \(x^3 - x^2\). Factor it out: \(x^2(x-1) = 0\). So, the x-intercepts are at \(x=0\) and \(x=1\).

Step 2: Find the y-intercept

To find the y-intercept, evaluate \(t(0)\). The function is \(\frac{0^3 - 0^2}{0^3 - 3 \cdot 0 - 2} = 0\). Thus, the y-intercept is \(y=0\).

Step 3: Determine vertical asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Set the denominator \(x^3 - 3x - 2 = 0\) and solve for \(x\). By factoring, \((x+1)(x-1)^2=0\). The solutions are \(x=-1\) and \(x=1\). Therefore, there is a vertical asymptote at \(x=-1\). Note that \(x=1\) is a removable discontinuity because \(\frac{x^3 - x^2}{x^3 - 3x - 2}\) simplifies to \(\frac{x^2}{x+1}\) for \(x eq 1\).

Step 4: Determine horizontal asymptotes

Consider the degrees of the numerator and denominator. Both are degree 3, so the horizontal asymptote is the ratio of the leading coefficients: \(\frac{1}{1}=1\). Therefore, the horizontal asymptote is \(y=1\).

Step 5: State the domain

The domain of \(t(x)\) is all real numbers except where the denominator is zero: \(x eq -1\) and \(x eq 1\). Therefore, the domain is \(( -\infty, -1) \cup (-1, 1) \cup (1, \infty)\).

Step 6: State the range

The range is determined by the horizontal asymptote and the behavior of the graph. Since \(y=1\) is a horizontal asymptote and does not intersect with any points of discontinuity or intercepts, the range of \(t(x)\) is all real numbers except \(y=1\): \(( -\infty, 1) \cup (1, \infty)\).

Step 7: Sketch the Graph

Sketch the x-intercepts \((0,0)\), \((1,0)\), the y-intercept \((0,0)\), the vertical asymptote at \(x=-1\), and the horizontal asymptote \(y=1\). Note the removable discontinuity at \(x=1\), where the graph is undefined but behaves as if the line \(y=x^2/(x+1)\) is followed closely.

Key Concepts

Intercepts

Rational functions, like the one given as \( t(x)=\frac{x^{3}-x^{2}}{x^{3}-3x-2} \), often have intercepts where the graph crosses the axes. These points are valuable as they provide insight into the function's behavior directly as it touches either the x-axis or the y-axis.

• X-intercepts: To find the x-intercepts of a rational function, set the numerator equal to zero. The numerator here is \( x^3 - x^2 \). By factoring, we get \( x^2(x-1) = 0 \), indicating the x-intercepts at \( x=0 \) and \( x=1 \). These points reveal where the graph crosses the x-axis.
• Y-intercept: For the y-intercept, substitute \( x=0 \) into the function. This evaluates the function at zero, yielding \( t(0) = 0 \), showing that the y-intercept is at \( y=0 \). The graph crosses the y-axis at this point.

Recognizing these intercepts helps form a basic sketch of the graph, indicating points where it will intersect the axes.

Asymptotes

Asymptotes are lines that the graph of a function approaches but never touches. They offer critical clues about a function's behavior as the inputs become quite large or approach certain values.

• Vertical Asymptotes: These occur where the denominator of a function equals zero, leading the function towards infinity. In \( t(x)=\frac{x^{3}-x^{2}}{x^{3}-3x-2} \), the denominator becomes zero at \( x^3 - 3x - 2 = 0 \). By factoring it, we find \( (x+1)(x-1)^2=0 \), giving us solutions \( x=-1 \) and \( x=1 \). However, since \( x=1 \) results in both the numerator and the denominator equaling zero, it is a removable discontinuity rather than a vertical asymptote. Thus, the only vertical asymptote is at \( x=-1 \).
• Horizontal Asymptotes: These are determined by comparing the degrees of the numerator and the denominator. Both are of degree 3, so the horizontal asymptote is the ratio of the leading coefficients: \( \frac{1}{1} = 1 \). This means the graph levels out towards \( y=1 \) as \( x \) approaches positive or negative infinity.

Understanding asymptotes is vital for knowing how the graph behaves at its extremes, providing more depth to the sketch of the graph.

Domain and Range

The domain and range of a rational function reveal all possible input and output values. They tell us where the function is defined and how the outputs are distributed.

• Domain: The domain consists of all real numbers except where the denominator is zero. For this function, the denominator \( x^3 - 3x - 2 \) equals zero at \( x=-1 \) and \( x=1 \). Thus, these values are excluded from the domain, leaving us with \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).
• Range: The range is all possible outputs, which are affected by horizontal asymptotes and discontinuities. Here, the horizontal asymptote at \( y=1 \) means the function will not reach \( y=1 \), given the graph never touches the asymptote. Hence, the range is \( (-\infty, 1) \cup (1, \infty) \).

Knowing the domain and range aids in understanding the complete behavior and limitations of the function, ensuring an accurate sketch and a proper understanding of where the function can and cannot "live."