Problem 25 - Algebra for College Students

Question

The rational function \(f(x)=\frac{(x-2)(x+3)}{x-2}\) has a domain of all real numbers except 2 and can be simplified to \(f(x)=x+3\). Thus its graph is a straight line with a hole at \((2,5)\). Graph each of the following functions. See answer section. (a) \(f(x)=\frac{(x+4)(x-1)}{x+4}\) (b) \(f(x)=\frac{x^{2}-5 x+6}{x-2}\) (c) \(f(x)=\frac{x-1}{x^{2}-1}\) (d) \(f(x)=\frac{x+2}{x^{2}+6 x+8}\)

Step-by-Step Solution

Answer

(a) Line with hole at (-4, -5); (b) Line with hole at (2, -1); (c) \(f(x)=\frac{1}{x+1}\) with hole at (1, 0.5) and asymptote at \(x=-1\); (d) \(f(x)=\frac{1}{x+4}\) with hole at (-2, 0.25) and asymptote at \(x=-4\).

Step 1: Understand the problem

We have four rational functions, and we need to determine their graphs and find any discontinuities, like holes or asymptotes, similar to the example provided.

Step 2: Simplifying the function in (a)

The function is given by \(f(x)=\frac{(x+4)(x-1)}{x+4}\). Since the factor \((x+4)\) appears both in the numerator and denominator, it can be cancelled, provided \(x eq -4\). Thus, the simplified function is \(f(x) = x - 1\). However, there is a hole at \(x = -4\), located at the point \((-4, -5)\).

Step 3: Simplifying the function in (b)

The function is \(f(x) = \frac{x^{2} - 5x + 6}{x-2}\). Factor the numerator: \(x^2 - 5x + 6 = (x-2)(x-3)\). Thus, \(f(x) = \frac{(x-2)(x-3)}{x-2}\). After cancelling \((x-2)\), simplify to \(f(x) = x - 3\), with a hole at \(x = 2\), located at the point \((2, -1)\).

Step 4: Analyzing the function in (c)

For \(f(x)=\frac{x-1}{x^{2}-1}\), recognize that \(x^2 - 1\) can be factored as \((x-1)(x+1)\). Then the function becomes \(f(x) = \frac{x-1}{(x-1)(x+1)}\). Cancel the \((x-1)\) factors, giving \(f(x) = \frac{1}{x+1}\), with a hole at \(x=1\), located at the point \((1, 0.5)\). An additional vertical asymptote exists at \(x = -1\).

Step 5: Analyzing the function in (d)

For \(f(x)=\frac{x+2}{x^{2}+6x+8}\), factor the denominator: \(x^2 + 6x + 8 = (x+2)(x+4)\). The function becomes \(f(x) = \frac{x+2}{(x+2)(x+4)}\). After cancelling \((x+2)\), simplify to \(f(x) = \frac{1}{x+4}\), with a hole at \(x = -2\), located at the point \((-2, 0.25)\), and a vertical asymptote at \(x = -4\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (typically x-values) for which the function is defined. In the context of rational functions, there are often restrictions within the domain due to values that can make the denominator zero.

A zero in the denominator creates undefined points in the function because division by zero is not allowed in mathematics. For instance, in the given function example, \( f(x) = \frac{(x-2)(x+3)}{x-2} \), the domain is all real numbers except \( x = 2 \). That's because plugging 2 into the denominator makes it zero, which is mathematically not permissible.

• To find the domain, look at the denominator: any value of \( x \) that zeroes it out must be excluded from the domain.
• A quick check is to solve the equation where the denominator equals zero to find restricted x-values.

Without these restrictions, the function would be undefined at certain points, breaking its continuity.

Holes in Graphs

In rational functions, a hole occurs when a factor is canceled from both the numerator and denominator. This indicates a removable discontinuity and represents a point where the function is not defined on its graph.

Imagine you're simplifying a rational expression like \( f(x) = \frac{(x+4)(x-1)}{x+4} \). Here, the \( (x+4) \) factor cancels from the numerator and denominator, suggesting a hole at \( x = -4 \). Even though the expression simplifies to \( f(x) = x - 1 \), there's a specific point on the graph, namely \( (-4, -5) \), that needs to be marked as undefined.

• A hole is not a break in the line but rather a single undefined point.
• Holes can be found by setting the canceled factor equal to zero and solving for x.
• Evaluate the simplified expression at this x-value to find the y-coordinate of the hole.

This ensures there is no ambiguity in the graph's behavior at these points.

Vertical Asymptotes

Vertical asymptotes in rational functions occur where the function grows indefinitely towards positive or negative infinity as x approaches a specific value. These are distinct from holes because they are not removable.

In short, where the denominator is non-cancelled and zero, a vertical asymptote arises. For instance, when analyzing \( f(x)=\frac{x-1}{x^2-1} \), you factor the denominator to get \( (x-1)(x+1) \). After canceling \( (x-1) \), you're left with \( f(x) = \frac{1}{x+1} \) and a vertical asymptote at \( x = -1 \).

• Vertical asymptotes occur when the factors in the denominator do not cancel out with those in the numerator.
• The x-values that make these remaining factors zero identify the vertical asymptotes.
• You can represent them graphically as dashed lines approaching which the function graph zooms towards infinity.

Understanding and identifying vertical asymptotes helps clarify the comprehensive shape and behavior of a graph.

Simplifying Rational Expressions

Simplifying rational expressions involves reducing them by canceling common factors from the numerator and denominator. This simplifies the function to its most basic form for easier analysis and graphing.

The key is to factor both the numerator and denominator first. For example, take the rational expression \( f(x) = \frac{x^2-5x+6}{x-2} \). Factor the numerator as \( (x-2)(x-3) \). You can then cancel the \( (x-2) \) terms, yielding a simplified form \( f(x) = x-3 \).

• Simplify by canceling terms where possible, but remember the original restrictions imposed by those terms.
• Note that simplification may sometimes reveal additional features of the function not visible in the unsimplified form.
• The original function may have graph features, such as holes, while the simplified function makes it easier to determine the general shape of the graph.

Always ensure that any canceled factors don't create overlooked holes or asymptotes in the graph.