Evaluating Polynomial Functions A Step-by-Step Guide
Introduction: Understanding Polynomial Functions
Polynomial functions are fundamental building blocks in mathematics, appearing in various fields, from algebra and calculus to engineering and computer science. Understanding how to evaluate these functions for specific values is a crucial skill. In this comprehensive guide, we will walk you through the process of evaluating the polynomial function f(x) = x³ + 5x² - 15x + 1 at x = -2. This involves substituting -2 for x in the function and simplifying the expression. Mastering this skill allows you to analyze the behavior of polynomial functions, find their roots, and graph them accurately. This article will provide a step-by-step solution, ensuring you grasp every detail of the process. So, let's embark on this mathematical journey and solve the problem together!
Problem Statement: Evaluating f(-2)
Our primary task is to evaluate the polynomial function f(x) = x³ + 5x² - 15x + 1 at x = -2. This means we need to substitute -2 for every instance of x in the function and then simplify the resulting expression using the order of operations. Polynomial functions are expressions consisting of variables raised to non-negative integer powers, combined with coefficients and constants. Evaluating a polynomial function at a specific value involves replacing the variable with that value and performing the necessary arithmetic operations. This process helps us understand the function's behavior at that particular point. In the given function, we have terms involving x raised to the powers of 3, 2, and 1, along with a constant term. The coefficients are the numerical values multiplying the variable terms, and the constant term is the value that doesn't depend on x. Let's move on to the detailed steps for solving this problem.
Step 1: Substitute -2 for x
The first crucial step in evaluating f(-2) is to substitute -2 for every x in the polynomial function. This process replaces the variable x with the specific value we are interested in, allowing us to compute the function's output at that point. The function f(x) = x³ + 5x² - 15x + 1 becomes f(-2) = (-2)³ + 5(-2)² - 15(-2) + 1. It is essential to perform this substitution carefully, ensuring that the negative signs and exponents are correctly placed. This step sets the stage for the subsequent arithmetic operations. Substituting values into functions is a fundamental technique in algebra and calculus, and it's the basis for evaluating functions at particular points, finding roots, and graphing functions. The next step will involve simplifying the expression by performing the exponentiations and multiplications according to the order of operations. Let's proceed to the simplification stage to arrive at the final answer.
Step 2: Simplify the Expression
Now that we have substituted -2 for x, we need to simplify the expression f(-2) = (-2)³ + 5(-2)² - 15(-2) + 1. We'll follow the order of operations (PEMDAS/BODMAS), which dictates that we first handle exponents, then multiplication and division, and finally addition and subtraction. First, we evaluate the exponents: (-2)³ = -8 and (-2)² = 4. Next, we perform the multiplications: 5(-2)² = 5(4) = 20 and -15(-2) = 30. The expression now becomes f(-2) = -8 + 20 + 30 + 1. Finally, we add the numbers together: f(-2) = -8 + 20 + 30 + 1 = 12 + 30 + 1 = 42 + 1 = 43. Therefore, the simplified value of the expression is 43. This step-by-step simplification ensures accuracy and clarity in the evaluation process. Simplifying expressions is a crucial skill in algebra, and it's used extensively in solving equations, graphing functions, and more. With the expression simplified, we have arrived at the final value of f(-2). Let's state our conclusion.
Step 3: Conclusion and Final Answer
After carefully substituting x = -2 into the polynomial function f(x) = x³ + 5x² - 15x + 1 and simplifying the resulting expression, we have arrived at the solution. The evaluation of f(-2) yields the value 43. That is, f(-2) = 43. This result represents the y-coordinate of the point on the graph of the function where x = -2. Evaluating polynomial functions is a fundamental skill in mathematics, and it has numerous applications in various fields. Understanding how to substitute values into functions and simplify the expressions is essential for solving problems in algebra, calculus, and other areas of mathematics. In this case, our systematic approach, starting with substitution and following the order of operations, has led us to the correct answer. Therefore, the final answer to the problem is 43, which corresponds to option D in the given choices. This detailed walkthrough demonstrates the importance of methodical problem-solving and careful execution of each step.
Detailed Explanation of the Solution: f(-2) = 43
To provide a comprehensive understanding, let's recap the step-by-step solution and delve deeper into the reasoning behind each operation. We started with the polynomial function f(x) = x³ + 5x² - 15x + 1 and aimed to find the value of f(-2). The first step involved substituting -2 for every instance of x in the function, which transformed the function into f(-2) = (-2)³ + 5(-2)² - 15(-2) + 1. This substitution is the cornerstone of evaluating functions at specific points. Next, we simplified the expression by following the order of operations. We first evaluated the exponents: (-2)³ equals -8 because -2 multiplied by itself three times is -2 * -2 * -2 = -8. Similarly, (-2)² equals 4 because -2 multiplied by itself is -2 * -2 = 4. After handling the exponents, we moved on to the multiplications. We multiplied 5 by (-2)², which is 4, resulting in 5 * 4 = 20. We also multiplied -15 by -2, which gives us 30. The expression now looked like f(-2) = -8 + 20 + 30 + 1. Finally, we performed the additions from left to right. -8 plus 20 equals 12. Then, 12 plus 30 equals 42. And finally, 42 plus 1 equals 43. Thus, we concluded that f(-2) = 43. This detailed breakdown clarifies the arithmetic and logical steps involved, reinforcing the method for evaluating polynomial functions.
Common Mistakes to Avoid When Evaluating Polynomial Functions
When evaluating polynomial functions, several common mistakes can lead to incorrect answers. Being aware of these pitfalls can help you avoid them and ensure accurate results. One frequent mistake is misapplying the order of operations. It’s crucial to remember PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division, Addition and Subtraction) to perform operations in the correct sequence. For example, failing to evaluate exponents before multiplication can lead to significant errors. Another common mistake involves mishandling negative signs. When substituting negative values for variables, especially when raised to powers, it’s essential to pay close attention to the signs. For instance, (-2)³ is -8, while (-2)² is 4. Forgetting to apply the negative sign correctly or squaring a negative number and incorrectly keeping the result negative are frequent errors. Additionally, errors can occur during arithmetic calculations, especially when dealing with multiple terms. It's advisable to double-check each step of the simplification process to minimize these errors. Another mistake is not substituting the value into all instances of the variable. Ensure that every x in the function is replaced with the given value. By being mindful of these common pitfalls and practicing careful, methodical problem-solving, you can improve your accuracy when evaluating polynomial functions.
Practice Problems: Evaluate Polynomial Functions
To reinforce your understanding of evaluating polynomial functions, let’s work through some practice problems. These examples will help you apply the step-by-step process we’ve discussed and build confidence in your skills. Problem 1: Evaluate g(x) = 2x³ - 4x² + 6x - 1 at x = 3. Solution: Substitute 3 for x: g(3) = 2(3)³ - 4(3)² + 6(3) - 1. Simplify: g(3) = 2(27) - 4(9) + 18 - 1 = 54 - 36 + 18 - 1 = 18 + 18 - 1 = 36 - 1 = 35. Therefore, g(3) = 35. Problem 2: Evaluate h(x) = -x⁴ + 3x² - 2x + 5 at x = -1. Solution: Substitute -1 for x: h(-1) = -(-1)⁴ + 3(-1)² - 2(-1) + 5. Simplify: h(-1) = -(1) + 3(1) + 2 + 5 = -1 + 3 + 2 + 5 = 2 + 2 + 5 = 4 + 5 = 9. Therefore, h(-1) = 9. Problem 3: Evaluate p(x) = x⁵ - 2x³ + x - 4 at x = 0. Solution: Substitute 0 for x: p(0) = (0)⁵ - 2(0)³ + (0) - 4. Simplify: p(0) = 0 - 0 + 0 - 4 = -4. Therefore, p(0) = -4. These practice problems illustrate the consistent approach to evaluating polynomial functions, emphasizing substitution and simplification. By working through various examples, you can enhance your proficiency in this essential mathematical skill.
Real-World Applications of Polynomial Functions
Polynomial functions are not just abstract mathematical concepts; they have numerous practical applications in various real-world scenarios. Understanding and evaluating these functions is crucial in fields ranging from engineering and physics to economics and computer graphics. In physics, polynomial functions can model the trajectory of a projectile. For example, the height of a ball thrown into the air can be described by a quadratic polynomial function, where the function's evaluation at a specific time gives the ball's height at that moment. In engineering, polynomials are used to design curves and surfaces, such as those found in bridges and roads. The shape of these structures can be precisely determined using polynomial equations, ensuring stability and functionality. In economics, polynomial functions can model cost and revenue curves, helping businesses analyze their financial performance. For instance, a cubic polynomial might represent the total cost of production, and evaluating this function at a given quantity of output provides the total cost. In computer graphics, polynomials are used to create smooth curves and surfaces for 3D models and animations. Bézier curves, which are based on polynomials, are widely used in computer-aided design (CAD) and animation software. Furthermore, polynomial regression is a statistical technique that uses polynomials to model relationships between variables, which is applied in various fields, including data analysis and forecasting. These diverse applications underscore the importance of mastering polynomial functions for practical problem-solving and real-world analysis.
