Problem 44 - Precalculus: Functions and Graphs

Question

Simplify \(f(x),\) and sketch the graph of \(f\) $$f(x)=\frac{\left(x^{2}+x\right)(2 x-1)}{\left(x^{2}-3 x+2\right)(2 x-1)}$$

Step-by-Step Solution

Answer

Simplified: \( f(x) = \frac{x^2 + x}{(x - 1)(x - 2)} \), with discontinuities at \(x = \frac{1}{2}, 1,\) and \(2\).

Step 1: Factorize the Denominator

Identify factors of the quadratic equation in the denominator, \(x^2 - 3x + 2\). This can be factored as \((x - 1)(x - 2)\) because \( (x - 1)(x - 2) = x^2 - 2x - x + 2\).

Step 2: Simplify the Fraction

The function \(f(x)\) is given by \(\frac{(x^2 + x)(2x - 1)}{(x^2 - 3x + 2)(2x - 1)}\). Notice that \((2x - 1)\) is a common factor in both the numerator and the denominator. Cancel this common factor to simplify the function to \( \frac{x^2 + x}{x^2 - 3x + 2} \).

Step 3: Identify Restrictions

Even though we canceled \((2x - 1)\), it's important to remember that originally the denominator could not be zero. Hence, \(2x - 1 eq 0\), implying \(x eq \frac{1}{2}\). Additionally, from \(x^2 - 3x + 2 = (x - 1)(x - 2)\), we see \(x eq 1\) and \(x eq 2\). Therefore, \(x\) is undefined at \(x = \frac{1}{2}, 1,\) and \(2\).

Step 4: Sketch the Graph

Recognize that after simplifying, \(f(x) = \frac{x^2 + x}{(x - 1)(x - 2)}\). The graph should be sketched considering the undefined points and approaching negative and positive infinities at these points. Plot characteristic points and asymptotes at \(x = 1\) and \(x = 2\), and a removable discontinuity (hole) at \(x = \frac{1}{2}\).

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a critical skill in simplifying complex algebraic fractions. In the given function, the denominator is the quadratic expression \(x^2 - 3x + 2\). Factoring it involves finding two numbers that multiply to the constant term and add to the coefficient of the linear term. Here, these numbers are -1 and -2, because \((-1)\times(-2) = 2\) and \((-1)+(-2) = -3\).

Thus, we can write the expression as \((x - 1)(x - 2)\). To factor quadratics effectively, always check for basic patterns such as the difference of squares or perfect square trinomials, and remember to practice recognizing these patterns quickly.

Simplifying Rational Expressions

Simplifying rational expressions involves canceling out common factors between the numerator and the denominator. In the exercise, we start with the function \(f(x) = \frac{(x^2 + x)(2x - 1)}{(x^2 - 3x + 2)(2x - 1)}\).

Notice that \((2x - 1)\) appears in both the numerator and the denominator. By canceling \((2x - 1)\), we minimize the function to \(\frac{x^2 + x}{x^2 - 3x + 2}\).

• Ensure to only cancel factors, not terms, as this requires equal value in both numerator and denominator.
• Always identify common factors and make sure to account for all dividers, ensuring accurate and proper simplification.

Graphing Rational Functions

Graphing rational functions requires knowledge of characteristics such as intercepts, asymptotes, and discontinuities. After simplifying, you get \(f(x) = \frac{x^2 + x}{(x - 1)(x - 2)}\).

Step-by-step, begin by finding the intercepts. The y-intercept occurs where \(x = 0\), which means solving \(f(0)\). X-intercepts occur when the numerator equals zero, so you'll solve \(x^2 + x = 0\).

• Factoring reveals solutions \(x(x + 1) = 0\), thus, \(x = 0\) or \(x = -1\).

The next step is to determine the asymptotes. In this function, vertical asymptotes happen where the denominator is zero and the original function becomes undefined, namely \(x = 1\) and \(x = 2\).

• Find vertical asymptotes using the roots of the denominator's factored form.
• Horizontal asymptotes are identified by comparing degrees of the numerator and denominator, often leading to long division if necessary.

Identifying Asymptotes and Holes

Identifying asymptotes and holes is crucial for understanding the behavior of a rational function. In the simplified function \(f(x)\), asymptotes occur at \(x = 1\) and \(x = 2\), where the function approaches infinity.

Holes in a graph—also known as removable discontinuities—happen where factors were canceled. Since we canceled \((2x - 1)\), a hole exists at \(x = \frac{1}{2}\), where the original equation would become undefined.

• To determine holes, identify any canceled factors and solve to find their corresponding \(x\)-values.
• Asymptotes inform us of places the graph will never touch or intersect; vertical asymptotes indicate undefined regions within a function's graph.

The graph must reflect these aspects: approach vertical lines at asymptotes and have breaks at the hole without connecting across it.