Solving For A In Quadratic Function F(x) = X^2 - 4x + 8 Where F(a) = 5

In the realm of mathematics, quadratic functions hold a special place, often encountered in various applications ranging from physics to economics. These functions, characterized by their parabolic curves, possess a unique charm and a set of properties that make them fascinating to explore. In this article, we embark on a journey to unravel the intricacies of a specific quadratic function, f(x) = x^2 - 4x + 8, and delve into the process of finding the value of 'a' when f(a) equals 5.

Understanding the Quadratic Function f(x) = x^2 - 4x + 8

The given function, f(x) = x^2 - 4x + 8, is a classic example of a quadratic function. Its defining characteristic is the presence of a term with x raised to the power of 2 (x^2), which dictates its parabolic shape when plotted on a graph. The coefficients of the terms, namely 1 for x^2, -4 for x, and 8 as the constant term, influence the parabola's orientation, width, and position on the coordinate plane. To truly grasp the essence of this function, let's delve into its key components:

• The x^2 Term: The presence of x^2 is the hallmark of a quadratic function. It dictates that the graph will be a parabola, a symmetrical U-shaped curve. The coefficient of x^2, which is 1 in this case, determines whether the parabola opens upwards (positive coefficient) or downwards (negative coefficient). Since our coefficient is positive, the parabola opens upwards.
• The -4x Term: This term introduces a linear component to the function, influencing the parabola's horizontal position. The coefficient -4 shifts the parabola horizontally, affecting its vertex (the lowest or highest point on the curve).
• The Constant Term 8: The constant term 8 determines the vertical position of the parabola. It represents the y-intercept, the point where the parabola intersects the y-axis. In our case, the parabola intersects the y-axis at the point (0, 8).

The Significance of f(a) = 5

The equation f(a) = 5 presents us with an intriguing challenge. It asks us to find the value(s) of 'a' that, when plugged into the function f(x), produce an output of 5. In essence, we are seeking the x-coordinate(s) on the parabola where the y-coordinate is 5. Graphically, this corresponds to finding the points where the horizontal line y = 5 intersects the parabola of f(x). To solve this, we'll substitute 'a' for 'x' in the function and set the expression equal to 5:

f(a) = a^2 - 4a + 8 = 5

This equation now sets the stage for our algebraic journey to uncover the value(s) of 'a'.

Embarking on the Algebraic Quest: Solving for 'a'

To solve the equation a^2 - 4a + 8 = 5, we'll employ the time-tested technique of rearranging it into a standard quadratic form and then applying either factoring or the quadratic formula. Let's begin by subtracting 5 from both sides to bring the equation into the familiar form of ax^2 + bx + c = 0:

a^2 - 4a + 8 - 5 = 0

Simplifying, we arrive at:

a^2 - 4a + 3 = 0

Now, we have a quadratic equation in standard form, poised for solution. Two primary methods stand before us: factoring and the quadratic formula. Let's explore the factoring approach first, as it often provides a more elegant and efficient solution when applicable.

The Elegance of Factoring

Factoring involves expressing the quadratic expression as a product of two linear expressions. To factor a^2 - 4a + 3, we seek two numbers that multiply to 3 (the constant term) and add up to -4 (the coefficient of the 'a' term). After a bit of contemplation, we find that -1 and -3 fit the bill perfectly (-1 * -3 = 3 and -1 + -3 = -4). Thus, we can rewrite the quadratic expression as:

(a - 1)(a - 3) = 0

The zero-product property, a cornerstone of algebra, dictates that if the product of two factors is zero, then at least one of the factors must be zero. Applying this principle, we set each factor equal to zero and solve for 'a':

• a - 1 = 0 => a = 1
• a - 3 = 0 => a = 3

Therefore, factoring gracefully unveils two possible values for 'a': 1 and 3.

The Versatile Quadratic Formula

In situations where factoring proves elusive, the quadratic formula stands as a reliable ally. This formula provides a universal solution for quadratic equations of the form ax^2 + bx + c = 0:

a = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 1, b = -4, and c = 3. Plugging these values into the quadratic formula, we get:

a = (4 ± √((-4)^2 - 4 * 1 * 3)) / (2 * 1)

Simplifying:

a = (4 ± √(16 - 12)) / 2

a = (4 ± √4) / 2

a = (4 ± 2) / 2

This yields two solutions:

• a = (4 + 2) / 2 = 3
• a = (4 - 2) / 2 = 1

As expected, the quadratic formula echoes the results obtained through factoring, confirming that the possible values for 'a' are indeed 1 and 3.

The Answer: One Possible Value of 'a'

The question posed seeks one possible value of 'a' for which f(a) = 5. Having solved the quadratic equation, we've unearthed two such values: 1 and 3. Therefore, we can confidently state that one possible value of 'a' is 1 or 3.

Visualizing the Solution: A Graphical Perspective

To solidify our understanding, let's cast a graphical eye upon the solution. The function f(x) = x^2 - 4x + 8, when plotted, manifests as an upward-opening parabola. The equation f(a) = 5 translates to finding the points on this parabola where the y-coordinate is 5. If we were to draw a horizontal line at y = 5 on the same graph, we'd observe that it intersects the parabola at two distinct points. The x-coordinates of these points of intersection correspond to the values of 'a' that satisfy f(a) = 5, which we've already determined to be 1 and 3.

This visual representation provides an intuitive grasp of why there are two solutions to this problem. The parabola's symmetrical shape allows for two distinct x-values to produce the same y-value.

Conclusion: Mastering the Art of Solving Quadratic Equations

In this exploration, we've successfully navigated the realm of quadratic functions, specifically f(x) = x^2 - 4x + 8, and adeptly determined the values of 'a' for which f(a) = 5. We've witnessed the power of algebraic techniques, such as factoring and the quadratic formula, in unraveling the solutions to quadratic equations. Furthermore, we've reinforced our understanding through a graphical interpretation, solidifying the connection between algebraic solutions and their visual counterparts.

This journey underscores the importance of mastering quadratic functions, as they serve as fundamental building blocks in various mathematical and scientific disciplines. By understanding their properties and solution methods, we equip ourselves to tackle a wide array of problems, from trajectory calculations in physics to optimization challenges in economics. The quest to solve f(a) = 5 has not only provided us with specific answers but has also enriched our understanding of the broader mathematical landscape.