Solving Quadratic Functions Find F(1/2) Step-by-Step
This article dives into the world of quadratic functions, exploring how to determine the value of a quadratic function at a specific point given certain conditions. We'll break down the process step-by-step, making it easy to understand even if you're not a math whiz. Our specific problem involves finding the value of f(1/2) for a quadratic function f, given that f(0) = 6, f(1) = 2, and f(-2) = 20. This is a classic problem that demonstrates the power of using given information to construct and solve for unknown parameters in a function. Mastering these techniques is essential for anyone studying algebra and calculus, as quadratic functions form the foundation for many advanced mathematical concepts. So, let's embark on this mathematical journey and unlock the secrets hidden within these functions.
What is a Quadratic Function?
At its core, a quadratic function is a polynomial function of degree two. This means that the highest power of the variable (usually 'x') is 2. The general form of a quadratic function is expressed as:
f(x) = ax² + bx + c
where 'a', 'b', and 'c' are constants, and 'a' cannot be zero (otherwise, it would be a linear function). These constants play a crucial role in determining the shape and position of the parabola, the U-shaped curve that graphically represents a quadratic function. The coefficient 'a' dictates the direction of the parabola's opening (upward if 'a' is positive, downward if 'a' is negative) and its width. The coefficient 'b' influences the parabola's horizontal position, and the constant 'c' represents the y-intercept, the point where the parabola intersects the y-axis. Understanding the interplay between these coefficients and the resulting graph is key to effectively working with quadratic functions. The versatility of quadratic functions stems from their ability to model a wide range of real-world phenomena, from the trajectory of a projectile to the shape of a satellite dish. This makes them indispensable tools in fields like physics, engineering, and economics.
Problem Statement: Finding f(1/2)
Our main goal is to find the value of the quadratic function f(x) when x is equal to 1/2. We are given three crucial pieces of information: f(0) = 6, f(1) = 2, and f(-2) = 20. These are essentially three points that lie on the parabola represented by the quadratic function. By strategically using these points, we can create a system of equations that will allow us to determine the unique values of the coefficients a, b, and c in the general form of the quadratic function. Once we have these coefficients, we can simply substitute x = 1/2 into the equation to find f(1/2). The process may seem like a puzzle at first, but by carefully applying algebraic principles, we can systematically solve for the unknowns and arrive at the solution. The beauty of this approach lies in its applicability to a wide range of problems involving quadratic functions. By mastering this method, you'll be well-equipped to tackle similar challenges in various mathematical contexts. The ability to extract information from given conditions and translate it into a solvable system of equations is a fundamental skill in mathematics and problem-solving in general.
Setting up the Equations
To determine the specific quadratic function, we need to find the values of a, b, and c in the general form f(x) = ax² + bx + c. We can use the given information – f(0) = 6, f(1) = 2, and f(-2) = 20 – to create a system of three equations. Each piece of information will provide us with one equation. Let's start by substituting x = 0 into the general form:
f(0) = a(0)² + b(0) + c = 6
This simplifies to:
c = 6
Now, substitute x = 1:
f(1) = a(1)² + b(1) + c = 2
Which gives us:
a + b + c = 2
Finally, substitute x = -2:
f(-2) = a(-2)² + b(-2) + c = 20
This simplifies to:
4a - 2b + c = 20
We now have a system of three equations:
- c = 6
- a + b + c = 2
- 4a - 2b + c = 20
This system of equations is the key to unlocking the values of a, b, and c. By systematically solving this system, we can determine the unique quadratic function that satisfies the given conditions. This process highlights the power of translating function evaluations into algebraic equations, allowing us to apply techniques for solving systems of equations to find the unknown parameters of the function. This approach is a cornerstone of mathematical problem-solving and has broad applications beyond quadratic functions.
Solving the System of Equations
Now that we have our system of equations, let's solve for a, b, and c. We already know that c = 6. We can substitute this value into the other two equations to simplify them:
Equation 2 becomes:
a + b + 6 = 2
Which simplifies to:
a + b = -4
Equation 3 becomes:
4a - 2b + 6 = 20
Which simplifies to:
4a - 2b = 14
We now have a system of two equations with two unknowns:
- a + b = -4
- 4a - 2b = 14
We can use several methods to solve this system, such as substitution or elimination. Let's use elimination. Multiply the first equation by 2:
2(a + b) = 2(-4)
2a + 2b = -8
Now add this modified equation to the second equation:
(2a + 2b) + (4a - 2b) = -8 + 14
6a = 6
Divide by 6:
a = 1
Now that we have a, we can substitute it back into the equation a + b = -4:
1 + b = -4
Subtract 1:
b = -5
So, we have found a = 1, b = -5, and c = 6. This systematic approach demonstrates the power of algebraic manipulation in solving systems of equations. The elimination method, in particular, is a versatile technique that can be applied to a wide range of systems, making it an essential tool in any mathematician's arsenal. By carefully organizing the equations and applying the rules of algebra, we can efficiently solve for the unknowns and gain valuable insights into the relationships between variables.
Constructing the Quadratic Function
With the values of a, b, and c determined, we can now construct the quadratic function. Recall the general form: f(x) = ax² + bx + c. We found that a = 1, b = -5, and c = 6. Substituting these values, we get:
f(x) = 1x² - 5x + 6
Which simplifies to:
f(x) = x² - 5x + 6
This is the specific quadratic function that satisfies the given conditions: f(0) = 6, f(1) = 2, and f(-2) = 20. We have successfully navigated the process of starting with a general form, using given information to create a system of equations, solving the system, and finally, constructing the unique quadratic function. This function now serves as a mathematical model that encapsulates the relationships described by the initial conditions. The ability to construct such models is a fundamental skill in mathematics and its applications. It allows us to represent real-world phenomena with mathematical expressions, enabling us to analyze, predict, and understand complex systems.
Evaluating f(1/2)
Our final step is to evaluate the quadratic function f(x) at x = 1/2. We have the function:
f(x) = x² - 5x + 6
Substitute x = 1/2:
f(1/2) = (1/2)² - 5(1/2) + 6
Simplify:
f(1/2) = 1/4 - 5/2 + 6
To add these terms, we need a common denominator. The least common denominator for 4 and 2 is 4. So, we rewrite the terms:
f(1/2) = 1/4 - 10/4 + 24/4
Now, combine the numerators:
f(1/2) = (1 - 10 + 24) / 4
f(1/2) = 15 / 4
Therefore, f(1/2) = 15/4. This final calculation demonstrates the power of having a constructed function. By simply substituting a value for x, we can determine the corresponding value of the function. This ability to evaluate functions at specific points is a cornerstone of mathematical analysis and has countless applications in fields ranging from physics and engineering to economics and computer science. The value 15/4 represents a specific point on the parabola defined by our quadratic function, and it provides valuable information about the function's behavior at that particular location.
Conclusion
In this article, we successfully determined the value of f(1/2) for a quadratic function given the conditions f(0) = 6, f(1) = 2, and f(-2) = 20. We achieved this by first understanding the general form of a quadratic function, f(x) = ax² + bx + c, and then using the given information to set up a system of three equations. We solved this system to find the values of a, b, and c, which allowed us to construct the specific quadratic function. Finally, we evaluated the function at x = 1/2 to find f(1/2) = 15/4. This problem illustrates a fundamental approach to working with functions: using given information to determine unknown parameters and then using the resulting function to make predictions or calculations. The techniques we employed, such as setting up and solving systems of equations, are essential tools in mathematics and have broad applicability across various disciplines. The journey from the initial problem statement to the final solution highlights the power of mathematical reasoning and problem-solving skills. The ability to break down complex problems into smaller, manageable steps, apply relevant concepts and techniques, and arrive at a logical solution is a valuable skill that extends far beyond the realm of mathematics.
Keywords
quadratic function, f(1/2), system of equations, parabola, coefficients, algebraic manipulation, function evaluation, problem-solving, mathematical reasoning
