Problem 91 - Algebra and Trigonometry

Question

Graphs with Holes In this chapter we adopted the convention that in rational functions, the numerator and denominator don't share a common factor. In this exercise we consider the graph of a rational function that does not satisfy this rule. (a) Show that the graph of $$ r(x)=\frac{3 x^{2}-3 x-6}{x-2} $$ is the line \(y=3 x+3\) with the point \((2,9)\) removed. [Hint. Factor. What is the domain of \(r ? ]\) (b) Graph the rational functions: $$ \begin{aligned} s(x) &=\frac{x^{2}+x-20}{x+5} \\ t(x) &=\frac{2 x^{2}-x-1}{x-1} \\ u(x) &=\frac{x-2}{x^{2}-2 x} \end{aligned} $$

Step-by-Step Solution

Answer

The function is the line \( y = 3x + 3 \) with a hole at \((2, 9)\).

Step 1: Factor the numerator of r(x)

To analyze where the graph of \( r(x) = \frac{3x^2 - 3x - 6}{x - 2} \) has a hole, we first need to see if the numerator can be factored such that a common factor with the denominator appears. The numerator can be factored as follows:\( 3x^2 - 3x - 6 = 3(x^2 - x - 2) = 3(x - 2)(x + 1) \)

Step 2: Simplify the function

We observe that the factor \( (x - 2) \) is present in both the numerator and the denominator of \( r(x) \). By canceling out \( (x - 2) \) from both, we get:\( r(x) = \frac{3(x - 2)(x + 1)}{x - 2} = 3(x + 1) \) for \( x eq 2 \).Thus, \( r(x) \) simplifies to \( y = 3x + 3 \) for \( x eq 2 \).

Step 3: Determine the point of discontinuity

The function \( r(x) \) is undefined at \( x = 2 \) because the original form had \( x - 2 \) in the denominator, which causes division by zero. The simplified form, \( y = 3x + 3 \), defines a line with a hole at the point \((2, 9)\) since:\[ y = 3(2) + 3 = 9 \].

Step 4: Conclusion for part (a)

The line \( y = 3x + 3 \) has a point removed at \( (2, 9) \) on the graph of the rational function \( r(x) \).

Step 5: Graph the function s(x) = (x^2 + x - 20)/(x + 5)

First, factor the numerator \( x^2 + x - 20 \) as \((x - 4)(x + 5)\). The function becomes:\( s(x) = \frac{(x - 4)(x + 5)}{x + 5} = x - 4 \) for \( x eq -5 \).The graph is a line \( y = x - 4 \) with a hole at \( x = -5 \) corresponding to the point \((-5, -9)\).

Step 6: Graph the function t(x) = (2x^2 - x - 1)/(x - 1)

Factor the numerator \( 2x^2 - x - 1 \) as \((2x + 1)(x - 1)\). The function becomes:\( t(x) = \frac{(2x + 1)(x - 1)}{x - 1} = 2x + 1 \) for \( x eq 1 \).The graph is a line \( y = 2x + 1 \) with a hole at \( x = 1 \) corresponding to the point \((1, 3)\).

Step 7: Graph the function u(x) = (x-2)/(x^2-2x)

Factor the denominator \( x^2 - 2x \) as \( x(x - 2) \). The function simplifies to:\( u(x) = \frac{x-2}{x(x - 2)} = \frac{1}{x} \) for \( x eq 2 \) and \( x eq 0 \).This is a hyperbola with vertical asymptote at \( x = 0 \) and a hole at \( x = 2 \). For \( x = 2 \), the point of discontinuity is \((2, 1/2)\).

Key Concepts

Factorization

Factorization is a crucial step in simplifying rational functions, which are fractions where both the numerator and the denominator are polynomials. The goal is to express these polynomials as products of simpler expressions. This process can uncover common factors, which are essential to fully simplify the function. For example, the function \( r(x) = \frac{3x^2 - 3x - 6}{x - 2} \) can be factored as \( 3(x - 2)(x + 1) \) in the numerator while the denominator remains \( x - 2 \). This factorization reveals a common factor of \( (x - 2) \) in both, allowing us to cancel it out, simplifying the function to \( y = 3(x + 1) \). Achieving this simplification hints at the potential for holes in the graph, as the cancellation of terms changes the function's behavior at specific points. Be mindful that factorization is possible only when the degree of the polynomial is suitable, which typically requires understanding quadratic, cubic, or higher-order polynomial structures.

Holes in Graphs

Holes in the graph of a rational function occur at points where both the numerator and denominator become zero, leading to an undefined value. When you simplify a rational function by canceling a common factor, the location where this factor equals zero becomes a hole, rather than a vertical asymptote. For instance, in the function \( r(x) = \frac{3(x - 2)(x + 1)}{x - 2} \), canceling the \( (x - 2) \) factor results in a hole at \( x = 2 \), rather than an asymptote. Thus, the graph is almost the same as \( y = 3x + 3 \) everywhere except at \( x = 2 \), where the point \((2, 9)\) is missing. To determine the hole’s coordinates, substitute the x-value that makes the cancelled factor equal to zero back into the simplified expression. This creates a tangible way to visualize what a hole represents: a point where a limit exists but the function itself isn't defined.

Graph Analysis

Graph analysis of rational functions often focuses on identifying the behavior of the function by examining features like intercepts, holes, and asymptotes. By factoring and simplifying, we can often transform complex expressions for easier graphing and insight. For instance, once the function is simplified, you can determine the y-intercept by evaluating the function at \( x = 0 \). The x-intercepts are found by setting the numerator equal to zero after simplification. Another key aspect is identifying holes, which, as previously discussed, occur where cancelled factors were originally present in the denominator. Simplifying \( r(x) = \frac{3x^2 - 3x - 6}{x - 2} \) into \( y = 3x + 3 \) shows a line, with a hole at \( x = 2 \). Graph analysis also involves understanding constraints of the domain (inputs for which the function is defined) by assessing where the original denominator equals zero, offering a complete picture of the function's behavior. Recognizing these nuances allows for accurate sketching of the graph, making analysis a handy tool in interpreting a rational function's characteristics.

Discontinuity

Discontinuity in rational functions arises from points where the function's behavior isn't smooth, usually caused by division by zero. This can manifest as either holes or vertical asymptotes. A hole, or removable discontinuity, occurs when a factor is common to both the numerator and denominator, and can be cancelled out. For instance, in \( r(x) = \frac{3(x - 2)(x + 1)}{x - 2} \), the factor \( x - 2 \) causes a removable discontinuity at \( x = 2 \), resulting in a hole. In contrast, vertical asymptotes arise when the denominator equals zero and the factor can't be cancelled. Rational functions like \( u(x) = \frac{x-2}{x(x - 2)} \) demonstrate a vertical asymptote at \( x = 0 \) because \( x \) in the denominator cannot be eliminated. Discontinuities are crucial in detailed graphing, as they alter the appearance and behavior of the graph. Knowing these concepts ensures a comprehensive understanding of function plotting, highlighting areas of undefined behavior and refining the overall graphical interpretation.