Evaluating F(x) = 5x² - 45x + 101: A Step-by-Step Guide

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In this article, we will delve into the quadratic function f(x) = 5x² - 45x + 101. Our primary goal is to evaluate this function for specific input values and to determine the input values that yield a particular output. We will address the following tasks:

  • (a) Calculate f(4)
  • (b) Calculate f(-1/2)
  • (c) Calculate f(a)
  • (d) Calculate f(2/m)
  • (e) Find the values of x such that f(x) = 1

This exploration will provide a comprehensive understanding of how the quadratic function behaves and how to manipulate it effectively. Understanding quadratic functions is crucial in various fields, including physics, engineering, and economics, as they model many real-world phenomena. This article serves as a detailed guide, providing step-by-step solutions and explanations, making it an invaluable resource for students and anyone seeking to enhance their mathematical skills.

(a) Evaluating f(4)

To begin, let's calculate f(4). This involves substituting x = 4 into the function f(x) = 5x² - 45x + 101. This is a fundamental operation in algebra and is crucial for understanding how the function behaves at a specific point. The process involves careful substitution and arithmetic calculation, which reinforces basic mathematical skills. The result will provide us with the function's value when the input is 4, giving us a concrete point on the parabola that the quadratic function represents. This calculation is not just a mathematical exercise; it provides a tangible understanding of the function's output for a specific input. It helps in visualizing the graph of the function and understanding its behavior. By meticulously performing this calculation, we gain confidence in our ability to evaluate functions and interpret their results. This skill is essential for more advanced mathematical concepts and applications in various fields. Understanding how to evaluate functions at specific points is also crucial for solving real-world problems modeled by mathematical functions. For example, in physics, we might use a function to describe the trajectory of a projectile, and evaluating the function at a specific time would tell us the projectile's position at that time. Similarly, in economics, we might use a function to model the cost of production, and evaluating the function at a specific quantity would tell us the cost of producing that quantity. Therefore, mastering the skill of function evaluation is not just about performing calculations; it's about gaining a deeper understanding of the relationship between inputs and outputs, which is fundamental to mathematical modeling and problem-solving.

Let's perform the substitution:

  • f(4) = 5(4)² - 45(4) + 101
  • f(4) = 5(16) - 180 + 101
  • f(4) = 80 - 180 + 101
  • f(4) = 1

Therefore, f(4) = 1.

(b) Evaluating f(-1/2)

Next, we will evaluate f(-1/2). This involves substituting x = -1/2 into the function f(x) = 5x² - 45x + 101. This step is important as it involves working with fractions and negative numbers, which can be a common area of difficulty for students. By carefully substituting the value and performing the arithmetic operations, we reinforce our understanding of how to handle fractions and negative numbers in algebraic expressions. This calculation also demonstrates the function's behavior for a negative input value, which can be different from its behavior for positive values. The process of evaluating f(-1/2) involves squaring a fraction, multiplying by a constant, and then adding and subtracting the resulting terms. Each of these operations requires a solid understanding of basic arithmetic principles. By working through this example, we not only find the value of the function at x = -1/2, but we also strengthen our foundational mathematical skills. This skill is essential for various mathematical applications, including calculus and linear algebra, where fractions and negative numbers are frequently encountered. Furthermore, evaluating functions with fractional inputs is crucial in many real-world scenarios, such as calculating growth rates or determining the concentration of a substance in a solution. Therefore, mastering this skill is essential for both academic success and practical problem-solving. Understanding how the function behaves for both positive and negative inputs is crucial for sketching the graph of the function and interpreting its properties, such as its symmetry and intercepts. This detailed evaluation process helps in building a comprehensive understanding of the function's characteristics.

Let's perform the substitution:

  • f(-1/2) = 5(-1/2)² - 45(-1/2) + 101
  • f(-1/2) = 5(1/4) + 45/2 + 101
  • f(-1/2) = 5/4 + 90/4 + 404/4
  • f(-1/2) = 499/4

Therefore, f(-1/2) = 499/4 or 124.75.

(c) Evaluating f(a)

Now, let's calculate f(a). This involves substituting x = a into the function f(x) = 5x² - 45x + 101. This step is crucial because it introduces the concept of evaluating a function with a variable, rather than a specific number. This is a fundamental step towards understanding function composition and more advanced algebraic manipulations. Substituting a variable allows us to express the function's output in terms of the input variable, which is essential for understanding the function's general behavior. The result, f(a), will be an expression involving a, representing the function's value for any input a. This abstraction is a cornerstone of algebra, allowing us to work with functions in a more general and powerful way. Understanding how to evaluate f(a) is essential for various applications, including finding the roots of the equation, determining the vertex of the parabola, and analyzing the function's transformations. It also forms the basis for more advanced concepts like function composition and inverse functions. By performing this substitution, we are essentially creating a template for the function's output, where we can plug in any value for a to find the corresponding output. This is a powerful tool for analyzing functions and understanding their properties. Furthermore, evaluating f(a) helps in visualizing the function's graph and understanding how the output changes as the input a varies. This provides a deeper insight into the function's behavior and its relationship with the input variable. Therefore, this step is not just a simple substitution; it's a crucial step towards developing a deeper understanding of function manipulation and analysis.

Substituting x = a, we get:

  • f(a) = 5a² - 45a + 101

This expression represents the value of the function f at the input a.

(d) Evaluating f(2/m)

Next, we'll evaluate f(2/m). This involves substituting x = 2/m into the function f(x) = 5x² - 45x + 101. This step is particularly important as it introduces a variable in the denominator, which can sometimes be a point of confusion for students. By carefully substituting and simplifying the expression, we reinforce our understanding of how to work with algebraic fractions and expressions. This calculation demonstrates the function's behavior when the input is a ratio, which is common in many mathematical and scientific applications. The process of evaluating f(2/m) involves squaring the fraction, multiplying by constants, and then combining the resulting terms. This requires a solid understanding of algebraic manipulation and simplification. By working through this example, we not only find the expression for f(2/m), but we also strengthen our skills in dealing with complex algebraic expressions. This skill is essential for more advanced mathematical concepts, such as calculus and differential equations, where functions with variable denominators are frequently encountered. Furthermore, evaluating functions with fractional expressions is crucial in many real-world scenarios, such as calculating rates of change or determining the relationship between variables in a physical system. Therefore, mastering this skill is essential for both academic success and practical problem-solving. This step also highlights the importance of considering the domain of the function. Since m is in the denominator, we must ensure that m ≠ 0 to avoid division by zero, which is undefined. Understanding these domain restrictions is crucial for a complete understanding of the function's behavior.

Substituting x = 2/m, we get:

  • f(2/m) = 5(2/m)² - 45(2/m) + 101
  • f(2/m) = 5(4/m²) - 90/m + 101
  • f(2/m) = 20/m² - 90/m + 101

This expression represents the value of the function f at the input 2/m.

(e) Finding Values of x such that f(x) = 1

Finally, we need to find the values of x for which f(x) = 1. This involves setting the function equal to 1 and solving the resulting quadratic equation. This is a crucial skill in algebra as it allows us to find the roots or zeros of a function, which are the points where the function intersects the x-axis. Solving the equation 5x² - 45x + 101 = 1 requires us to rearrange the equation into the standard quadratic form and then use either the quadratic formula, factoring, or completing the square to find the solutions. This process reinforces our understanding of quadratic equations and the different methods for solving them. The solutions we find will be the x-values for which the function's output is 1, which corresponds to the points where the graph of the function intersects the horizontal line y = 1. This visual interpretation of the solutions is important for understanding the function's behavior and its relationship with the x-axis. Furthermore, finding the values of x such that f(x) = 1 has applications in various fields, such as finding the time when a projectile reaches a certain height or determining the input value that yields a specific output in a production process. Therefore, mastering this skill is essential for both academic success and practical problem-solving. This step also highlights the importance of algebraic manipulation and the ability to apply different techniques to solve equations. The quadratic formula, factoring, and completing the square are all powerful tools for solving quadratic equations, and choosing the appropriate method depends on the specific equation and the solver's preference.

Setting f(x) = 1, we get:

  • 5x² - 45x + 101 = 1
  • 5x² - 45x + 100 = 0

We can divide the equation by 5 to simplify it:

  • x² - 9x + 20 = 0

Now, we can factor the quadratic equation:

  • (x - 4)(x - 5) = 0

This gives us two possible solutions:

  • x - 4 = 0 => x = 4
  • x - 5 = 0 => x = 5

Therefore, the values of x such that f(x) = 1 are x = 4 and x = 5.

Conclusion

In this article, we have thoroughly evaluated the quadratic function f(x) = 5x² - 45x + 101 for various inputs. We calculated f(4), f(-1/2), f(a), and f(2/m), and we determined the values of x for which f(x) = 1. This comprehensive exploration demonstrates the importance of function evaluation and algebraic manipulation in understanding the behavior of quadratic functions. The ability to evaluate functions for specific inputs, including numbers, variables, and algebraic expressions, is a fundamental skill in mathematics. Furthermore, solving equations like f(x) = 1 allows us to find the roots or zeros of a function, which are crucial for understanding its graph and its applications in various fields. By working through these examples, we have reinforced our understanding of quadratic functions and the techniques for analyzing them. This knowledge is essential for more advanced mathematical concepts and for solving real-world problems modeled by quadratic functions. The skills developed in this article, such as function evaluation, algebraic manipulation, and equation solving, are not only valuable in mathematics but also in various other disciplines, including physics, engineering, economics, and computer science. Therefore, mastering these skills is essential for academic success and for practical problem-solving in a wide range of fields. This article serves as a valuable resource for students and anyone seeking to enhance their mathematical skills and understanding of quadratic functions.