Finding The Vertex Of F(x) = -x² + 6x - 8 A Step-by-Step Guide

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In the realm of mathematics, quadratic functions hold a prominent position, shaping parabolas and influencing a myriad of real-world phenomena. Understanding the key features of these functions, such as the vertex, is crucial for comprehending their behavior and applications. In this comprehensive guide, we embark on a journey to unravel the secrets of the quadratic function f(x) = -x² + 6x - 8, with a specific focus on pinpointing its vertex. We will delve into the fundamental concepts, explore different methods for vertex determination, and illustrate the significance of this pivotal point in the function's overall characteristics.

Grasping the Essence of Quadratic Functions and the Significance of the Vertex

Before we embark on our quest to find the vertex of f(x) = -x² + 6x - 8, let's take a moment to solidify our understanding of quadratic functions and the profound importance of the vertex. A quadratic function, in its most general form, is expressed as f(x) = ax² + bx + c, where a, b, and c are constants, and a is non-zero. The graph of a quadratic function is a parabola, a symmetrical U-shaped curve that opens either upwards or downwards, depending on the sign of the coefficient a. If a is positive, the parabola opens upwards, while a negative a signifies a downward-opening parabola.

The vertex, the focal point of our exploration, holds a position of paramount importance within the parabola. It represents the point where the parabola changes direction, marking either the minimum value (for upward-opening parabolas) or the maximum value (for downward-opening parabolas) of the function. The vertex serves as a critical landmark, providing valuable insights into the function's behavior, range, and symmetry. Its coordinates, typically denoted as (h, k), encapsulate the essence of the quadratic function's extreme point and its location on the coordinate plane.

Understanding the vertex allows us to solve a variety of real-world problems, from optimizing projectile motion to designing efficient structures. For instance, if we were modeling the trajectory of a ball thrown into the air, the vertex would represent the ball's highest point. Similarly, in engineering, the vertex can help determine the optimal shape for an arch or a bridge to maximize its strength and stability. Therefore, mastering the techniques for finding the vertex is not just an academic exercise but a practical skill with far-reaching implications.

Unveiling the Vertex: Method 1 - Completing the Square

One of the most elegant and insightful methods for determining the vertex of a quadratic function is the technique of completing the square. This method involves transforming the quadratic expression into a perfect square trinomial, which readily reveals the vertex coordinates. Let's apply this technique to our function, f(x) = -x² + 6x - 8.

Step-by-Step Transformation

  1. Factor out the coefficient of x²: In our case, the coefficient of x² is -1. Factoring this out, we get:

    f(x) = -1(x² - 6x + 8)

  2. Focus on the expression inside the parentheses: We now concentrate on the quadratic expression x² - 6x + 8. To complete the square, we need to add and subtract a specific value that will transform this expression into a perfect square trinomial. This value is determined by taking half of the coefficient of the x term (-6), squaring it ((-3)² = 9), and adding and subtracting it inside the parentheses:

    f(x) = -1(x² - 6x + 9 - 9 + 8)

  3. Rearrange and group terms: Now, we rearrange the terms to group the perfect square trinomial together:

    f(x) = -1((x² - 6x + 9) - 9 + 8)

  4. Express the perfect square trinomial as a squared binomial: The expression x² - 6x + 9 is a perfect square trinomial, which can be written as (x - 3)²:

    f(x) = -1((x - 3)² - 1)

  5. Distribute the -1 and simplify: Distribute the -1 back into the parentheses:

    f(x) = -(x - 3)² + 1

Vertex Revelation

Now, we have successfully transformed the quadratic function into vertex form: f(x) = a(x - h)² + k, where (h, k) represents the vertex coordinates. Comparing our transformed function, f(x) = -(x - 3)² + 1, with the vertex form, we can readily identify the vertex coordinates:

  • h = 3
  • k = 1

Therefore, the vertex of the quadratic function f(x) = -x² + 6x - 8 is located at the point (3, 1). This method not only reveals the vertex but also provides valuable insights into the function's transformations and symmetry.

Unveiling the Vertex: Method 2 - The Vertex Formula

Another powerful method for finding the vertex of a quadratic function is through the application of the vertex formula. This formula provides a direct and efficient way to calculate the vertex coordinates, given the coefficients of the quadratic expression. Let's explore this method and apply it to our function, f(x) = -x² + 6x - 8.

The Vertex Formula Unveiled

The vertex formula states that for a quadratic function in the standard form f(x) = ax² + bx + c, the x-coordinate of the vertex, denoted as h, is given by:

h = -b / 2a

Once we have the x-coordinate, we can find the y-coordinate, denoted as k, by substituting h back into the original function:

k = f(h)

Applying the Formula to f(x) = -x² + 6x - 8

Let's apply this formula to our function, f(x) = -x² + 6x - 8. First, we identify the coefficients:

  • a = -1
  • b = 6
  • c = -8

Now, we can calculate the x-coordinate of the vertex:

h = -b / 2a = -6 / (2 * -1) = 3

Next, we substitute h = 3 back into the original function to find the y-coordinate:

k = f(3) = -(3)² + 6(3) - 8 = -9 + 18 - 8 = 1

Vertex Confirmation

Therefore, using the vertex formula, we have determined that the vertex of the quadratic function f(x) = -x² + 6x - 8 is located at the point (3, 1). This result aligns perfectly with our findings from the completing the square method, further solidifying our understanding of the vertex's location.

The vertex formula offers a straightforward and efficient approach to finding the vertex, particularly when dealing with quadratic functions in standard form. Its direct application eliminates the need for algebraic manipulation, making it a valuable tool in our mathematical arsenal.

Significance of the Vertex: Unveiling the Function's Behavior

The vertex, as we've discovered, is not merely a point on the parabola; it's a pivotal landmark that unveils the function's behavior and characteristics. Its coordinates provide crucial information about the function's maximum or minimum value, its axis of symmetry, and its overall shape. Let's delve deeper into the significance of the vertex in understanding the function f(x) = -x² + 6x - 8.

Maximum Value and Range

The vertex, (3, 1), represents the maximum point of the parabola. This is because the coefficient of the x² term is negative (a = -1), indicating that the parabola opens downwards. The y-coordinate of the vertex, k = 1, represents the maximum value of the function. This means that no value of x will produce a function value greater than 1. Consequently, the range of the function is (-∞, 1], encompassing all real numbers less than or equal to 1.

Axis of Symmetry

The parabola possesses a line of symmetry that passes through the vertex, known as the axis of symmetry. This line divides the parabola into two mirror-image halves. The equation of the axis of symmetry is given by x = h, where h is the x-coordinate of the vertex. For our function, f(x) = -x² + 6x - 8, the axis of symmetry is the vertical line x = 3. This symmetry highlights the balanced nature of the quadratic function around its vertex.

Parabola's Shape and Orientation

The vertex, in conjunction with the coefficient a, provides insights into the parabola's shape and orientation. Since a is negative (a = -1), we know that the parabola opens downwards. The magnitude of a also influences the parabola's width. A larger absolute value of a results in a narrower parabola, while a smaller absolute value leads to a wider parabola. In our case, a = -1, indicating a moderately wide, downward-opening parabola.

Vertex as a Critical Point

In calculus, the vertex corresponds to a critical point of the function, where the derivative is equal to zero. This reinforces the vertex's significance as a point where the function's behavior changes direction. The vertex serves as a cornerstone for analyzing the function's increasing and decreasing intervals, concavity, and other key characteristics.

Conclusion: The Vertex - A Gateway to Understanding Quadratic Functions

In conclusion, the vertex of a quadratic function is far more than just a point on a graph. It's a gateway to understanding the function's maximum or minimum value, axis of symmetry, shape, and overall behavior. By mastering the techniques for finding the vertex, such as completing the square and applying the vertex formula, we equip ourselves with valuable tools for analyzing and interpreting quadratic functions in a wide range of applications. The vertex, in essence, encapsulates the essence of the quadratic function, providing a focal point for our exploration and comprehension.

In the case of f(x) = -x² + 6x - 8, the vertex at (3, 1) reveals that the function reaches a maximum value of 1, is symmetrical around the line x = 3, and forms a moderately wide, downward-opening parabola. This comprehensive understanding empowers us to effectively utilize this quadratic function in various mathematical and real-world scenarios.