Introduction
In this article, we will delve into the process of evaluating a function for specific input values. Evaluating functions is a fundamental skill in mathematics, particularly in algebra and calculus. It allows us to determine the output of a function for a given input, providing insights into the function's behavior and properties. We will focus on the function v(x) = 12 - 2x - 5 and evaluate it for three distinct values of x: -2, 0, and 5. This exercise will demonstrate the straightforward yet crucial process of substituting values into a function and simplifying the expression to obtain the corresponding output. Understanding how to evaluate functions is essential for solving equations, graphing functions, and applying mathematical models to real-world scenarios. By the end of this article, you will have a clear understanding of how to evaluate the given function and will be equipped to handle similar problems with confidence. We will break down each step, providing clear explanations and justifications to ensure a comprehensive understanding of the process. This skill is not only vital for academic success but also for various practical applications in fields such as engineering, economics, and computer science, where functions are used to model and analyze different phenomena.
Understanding the Function v(x) = 12 - 2x - 5
Before we begin evaluating the function, it's crucial to understand its structure. The function v(x) = 12 - 2x - 5 is a linear function, which means it can be represented graphically as a straight line. The variable 'x' is the input, and 'v(x)' represents the output, which is the value of the function at that particular input. The expression 12 - 2x - 5 involves basic arithmetic operations: subtraction and multiplication. To evaluate the function, we will substitute the given values of 'x' into this expression and simplify. A linear function is characterized by its constant rate of change, which is represented by the coefficient of 'x'. In this case, the coefficient is -2, indicating that for every unit increase in 'x', the value of v(x) decreases by 2. The constant term, which can be found by simplifying 12 - 5, is 7. This constant term represents the y-intercept of the line when the function is graphed. Understanding these basic properties of the function helps us to predict its behavior and interpret the results of our evaluations. For example, we can expect that as 'x' increases, the value of v(x) will decrease due to the negative coefficient of 'x'. This initial understanding provides a valuable context for the numerical calculations that follow. By grasping the fundamental characteristics of the function, we can better appreciate the significance of the values we obtain when we evaluate it for specific inputs.
Evaluating v(x) for x = -2
Let's start by evaluating the function v(x) = 12 - 2x - 5 when x = -2. To do this, we substitute -2 for 'x' in the expression: v(-2) = 12 - 2(-2) - 5. The next step is to simplify the expression using the order of operations (PEMDAS/BODMAS), which dictates that we perform multiplication before addition and subtraction. So, -2 multiplied by -2 is 4, giving us: v(-2) = 12 + 4 - 5. Now we perform the addition and subtraction from left to right: 12 + 4 = 16, so we have v(-2) = 16 - 5. Finally, 16 - 5 = 11. Therefore, v(-2) = 11. This result tells us that when the input is -2, the output of the function is 11. This single evaluation provides a specific point on the line represented by the function. Understanding the process of substitution and simplification is crucial for accurately evaluating functions. By carefully following the order of operations, we can avoid errors and arrive at the correct result. This methodical approach is applicable to evaluating any function, regardless of its complexity. The result v(-2) = 11 gives us a valuable piece of information about the function's behavior around x = -2, and it serves as a foundation for further analysis and applications.
Evaluating v(x) for x = 0
Next, we will evaluate the function v(x) = 12 - 2x - 5 when x = 0. Substituting 0 for 'x' in the expression gives us: v(0) = 12 - 2(0) - 5. Again, we follow the order of operations. First, we perform the multiplication: 2 multiplied by 0 is 0, so we have v(0) = 12 - 0 - 5. Now we perform the subtraction from left to right: 12 - 0 = 12, which simplifies the expression to v(0) = 12 - 5. Finally, 12 - 5 = 7. Thus, v(0) = 7. This means that when the input is 0, the output of the function is 7. In the context of a graph, this result represents the y-intercept of the line, where the line crosses the y-axis. Evaluating a function at x = 0 often gives us a key piece of information about the function's behavior, particularly in linear functions. The value v(0) = 7 tells us the starting point of the function on the y-axis and provides a reference point for understanding how the function changes as 'x' changes. This evaluation is straightforward but highly significant in interpreting the function's properties. By substituting 0 for 'x', we isolate the constant term in the expression, which directly reveals the y-intercept. This understanding is crucial for graphing the function and analyzing its behavior in different contexts.
Evaluating v(x) for x = 5
Now, let's evaluate the function v(x) = 12 - 2x - 5 for x = 5. We substitute 5 for 'x' in the expression: v(5) = 12 - 2(5) - 5. Following the order of operations, we first perform the multiplication: 2 multiplied by 5 is 10, so we have v(5) = 12 - 10 - 5. Next, we perform the subtraction from left to right: 12 - 10 = 2, which gives us v(5) = 2 - 5. Finally, 2 - 5 = -3. Therefore, v(5) = -3. This result indicates that when the input is 5, the output of the function is -3. This point, along with the points we found earlier, helps us to visualize the line represented by the function. Evaluating the function at x = 5 provides further insight into its behavior as 'x' increases. We can observe that the value of v(x) decreases as 'x' increases, which is consistent with the negative coefficient of 'x' in the function's expression. The value v(5) = -3 gives us another specific point on the line and helps us to understand the rate at which the function is changing. This methodical evaluation, like the previous ones, demonstrates the importance of accurate substitution and simplification. By carefully following the order of operations, we can confidently determine the output of the function for any given input value. This skill is essential for solving equations, graphing functions, and applying mathematical models.
Conclusion
In conclusion, we have successfully evaluated the function v(x) = 12 - 2x - 5 for three different values of x: -2, 0, and 5. We found that v(-2) = 11, v(0) = 7, and v(5) = -3. These evaluations demonstrate the fundamental process of substituting values into a function and simplifying the expression to find the corresponding output. This skill is essential for a wide range of mathematical applications, including solving equations, graphing functions, and modeling real-world phenomena. Understanding how to evaluate functions accurately allows us to analyze and interpret their behavior, providing valuable insights into their properties. The methodical approach we used, following the order of operations, ensures that we arrive at the correct results. Each evaluation gives us a specific point on the line represented by the function, contributing to a comprehensive understanding of its characteristics. The y-intercept, found by evaluating at x = 0, is a particularly important feature, as it represents the starting point of the function on the y-axis. The changes in the function's value as 'x' changes, demonstrated by evaluating at different points, reflect the rate of change of the function. By mastering the process of evaluating functions, we gain a powerful tool for mathematical analysis and problem-solving.