Problem 64 - Precalculus Mathematics for Calculus

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

x-intercepts: 0, 1; y-intercept: 0; vertical asymptotes: x=-1, 2; horizontal asymptote: y=1; domain: x≠-1, 2; range: y≠1.

Step 1: Find the x-intercepts

To find the x-intercepts of the rational function, we set the numerator equal to zero and solve for \(x\). The numerator is \(x^3 - x^2\). Factoring out the greatest common factor, we have \(x^2(x - 1) = 0\). Solving for \(x\), we find \(x = 0\) and \(x = 1\). Thus, the x-intercepts are at \(x = 0\) and \(x = 1\).

Step 2: Find the y-intercept

To find the y-intercept, substitute \(x = 0\) into the function \(t(x)\). This gives us \(t(0) = \frac{0^3 - 0^2}{0^3 - 3 \cdot 0 - 2} = 0\). Therefore, the y-intercept is at \(y = 0\).

Step 3: Find the 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 \(x\). This cubic equation factors as \((x + 1)(x - 2)^2 = 0\), giving potential vertical asymptotes at \(x = -1\) and \(x = 2\). Confirm whether numerator is zero at these points: \(t(-1)\) and \(t(2)\) are undefined since the numerator is not zero here, thus these are vertical asymptotes.

Step 4: Identify horizontal asymptotes

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. Both the numerator and denominator are of degree 3, so the horizontal asymptote is \(y = \frac{1}{1} = 1\).

Step 5: State the domain

The domain of the function is all real numbers except where the denominator equals zero, leading to undefined values. Thus, the domain is \(x eq -1\) and \(x eq 2\).

Step 6: Determine the range

To determine the range, consider the behavior of the function as \(x\) approaches its asymptotes and beyond them. The function approaches the horizontal asymptote \(y=1\) as \(x\) approaches infinity. As \(x\) approaches the vertical asymptotes, the function can exceed, but cannot equal, \(y=1\). Therefore, the range is \((-fty, 1) \cup (1, fty)\).

Step 7: Sketch the graph

Plot the x-intercepts \((0, 0), (1, 0)\), the vertical asymptotes at \(x = -1\) and \(x = 2\), and the horizontal asymptote \(y = 1\). Sketch the curve respecting these features and approach asymptotes as expected. Verify using a graphing device.

Key Concepts

Understanding X-Intercepts in Rational Functions

To find the x-intercepts of a rational function like \(t(x) = \frac{x^{3}-x^{2}}{x^{3}-3x-2}\), focus on the numerator. X-intercepts, where the graph crosses the x-axis, occur when the output is zero, meaning the numerator must be zero as long as the denominator is not zero.
✔ Begin by setting the numerator \(x^3 - x^2\) to zero. This can be factored out to \(x^2(x - 1) = 0\).
✔ Solve the equation \(x^2(x - 1) = 0\) for \(x\), resulting in solutions \(x = 0\) and \(x = 1\). Therefore, the x-intercepts are at these points: \((0, 0)\) and \((1, 0)\).
Identify x-intercepts:

• Factoring simplifies finding solutions.
• Checking each factor for zero gives specific x-values.
• Make sure the denominator isn’t zero at these points.

Exploring Vertical Asymptotes

Vertical asymptotes in rational functions represent values of \(x\) where the function becomes undefined because the denominator is zero, but the numerator is not zero.
✔ For \(t(x)\), set the denominator \(x^3 - 3x - 2\) equal to zero and solve:

• Factor the equation to \((x + 1)(x - 2)^2 = 0\).
• Solutions are \(x = -1\) and \(x = 2\), indicating potential vertical asymptotes.

Verify if the numerator is zero at these points:

• If the numerator isn’t zero, as seen with \(t(-1)\) and \(t(2)\), these x-values become true asymptotes.

At \(x = -1\) and \(x = 2\), the function exhibits asymptotic behavior and the graph will appear to approach but never touch these vertical lines.

Domain and Range of Rational Functions

The domain and range describe the possible input and output values of the function, respectively.
✔ Domain: Consider where the function is undefined, typically when the denominator is zero.

• For \(t(x)\), exclude \(x = -1\) and \(x = 2\) because these are points where the denominator equals zero. Hence, domain is all real numbers except \(x = -1\) and \(x = 2\).

✔ Range:

• The horizontal asymptote \(y=1\) suggests that as \(x\) approaches infinity, \(y\) heads towards one but never equals it.
• Asymptote behavior implies that \(y\) can be every number except \(y = 1\). Thus, the range is \((-\infty, 1) \cup (1, \infty)\).

Understanding domain and range: By recognizing points of undefined conditions and behavior near infinity or asymptotes, you can effectively determine where the function operates.

Asymptotic Behavior in Rational Functions

Asymptotic behavior in rational functions details how the function acts around certain critical values.
✔ Vertical Asymptotes \((-1, 2)\): The function tends toward infinity or negative infinity as \(x\) approaches these x-values

• This indicates the graph nears—but never touches—vertical lines at \(x = -1\) and \(x = 2\).

✔ Horizontal Asymptote \((y = 1)\): Both the numerator and denominator have the same degree.

• The leading coefficients will determine the horizontal asymptote: In this case, it is \(y = \frac{1}{1} = 1\).

Understanding Asymptotes:

• They represent limits that the function approaches.
• Visualize asymptotes as invisible barriers the graph almost reaches but never crosses.