Finding The Vertex Of Quadratic Function F(x) = (x-6)(x+2)

Unlocking the secrets of quadratic functions often starts with identifying key features, and the vertex stands out as one of the most crucial. The vertex represents the highest or lowest point on the parabola, dictating the function's minimum or maximum value. In this article, we'll embark on a comprehensive exploration of the quadratic function f(x) = (x-6)(x+2), focusing on how to determine its vertex and what that vertex tells us about the function's behavior. We will delve into the different forms of quadratic equations, the methods for finding the vertex, and the significance of the vertex in real-world applications. Understanding the vertex is not just a mathematical exercise; it's a gateway to grasping the nature and behavior of quadratic functions, which have widespread applications in physics, engineering, economics, and computer science.

Quadratic Functions: A Quick Recap

Before we dive into the specifics of f(x) = (x-6)(x+2), let's refresh our understanding of quadratic functions in general. A quadratic function is a polynomial function of degree two, meaning the highest power of the variable x is 2. The standard form of a quadratic function is given by:

f(x) = ax² + bx + c

where a, b, and c are constants, and a ≠ 0. The graph of a quadratic function is a parabola, a U-shaped curve that opens either upwards or downwards. The direction in which the parabola opens is determined by the coefficient a: if a > 0, the parabola opens upwards, and if a < 0, it opens downwards. The vertex is the turning point of the parabola; it's the minimum point if the parabola opens upwards and the maximum point if the parabola opens downwards.

Another important form of a quadratic function is the vertex form:

f(x) = a(x - h)² + k

where (h, k) represents the coordinates of the vertex. This form is particularly useful because the vertex is immediately apparent. The value of h gives the x-coordinate of the vertex, and the value of k gives the y-coordinate. Recognizing the vertex form makes it incredibly easy to identify the vertex without any further calculations.

Finally, the factored form of a quadratic function, which is the form our function f(x) = (x-6)(x+2) is presented in, is given by:

f(x) = a(x - r₁)(x - r₂)

where r₁ and r₂ are the roots or zeros of the function, i.e., the values of x for which f(x) = 0. While the factored form doesn't directly reveal the vertex, it provides crucial information about the x-intercepts of the parabola, which can be used to determine the vertex. Understanding these different forms is essential for analyzing quadratic functions and extracting meaningful information.

Finding the Vertex of f(x) = (x-6)(x+2)

Now, let's focus on our specific quadratic function, f(x) = (x-6)(x+2). This function is presented in factored form, which, as we mentioned, provides the roots directly. To find the vertex, we can employ a couple of different methods. The first method involves converting the factored form to standard form and then using a formula to find the vertex. The second method leverages the symmetry of the parabola and the roots to directly calculate the x-coordinate of the vertex.

Method 1: Converting to Standard Form

To convert f(x) = (x-6)(x+2) to standard form, we simply need to expand the product:

f(x) = (x - 6)(x + 2) f(x) = x² + 2x - 6x - 12 f(x) = x² - 4x - 12

Now we have the function in the standard form f(x) = ax² + bx + c, where a = 1, b = -4, and c = -12. The x-coordinate of the vertex, denoted as h, can be found using the formula:

h = -b / 2a

Plugging in our values for a and b:

h = -(-4) / (2 * 1) h = 4 / 2 h = 2

So, the x-coordinate of the vertex is 2. To find the y-coordinate, denoted as k, we substitute h = 2 back into the standard form of the function:

k = f(2) = (2)² - 4(2) - 12 k = 4 - 8 - 12 k = -16

Therefore, the vertex of the quadratic function f(x) = (x-6)(x+2) is (2, -16). This method is a straightforward approach that utilizes the standard form of the quadratic equation and a well-established formula for the vertex.

Method 2: Using Symmetry and Roots

The second method takes advantage of the parabola's inherent symmetry. The vertex lies exactly midway between the roots of the quadratic function. From the factored form f(x) = (x-6)(x+2), we can easily identify the roots by setting each factor equal to zero:

x - 6 = 0 => x = 6 x + 2 = 0 => x = -2

Thus, the roots are 6 and -2. The x-coordinate of the vertex, h, is the average of these roots:

h = (r₁ + r₂) / 2 h = (6 + (-2)) / 2 h = 4 / 2 h = 2

This gives us the same x-coordinate, h = 2, as we found using the standard form method. To find the y-coordinate, k, we substitute h = 2 back into the original factored form of the function:

k = f(2) = (2 - 6)(2 + 2) k = (-4)(4) k = -16

Again, we find that the y-coordinate of the vertex is -16. Therefore, using this method, we also arrive at the vertex (2, -16). This approach is often quicker and more intuitive, especially when the function is already given in factored form. It beautifully illustrates the relationship between the roots and the vertex of a parabola.

Significance of the Vertex

The vertex of the quadratic function f(x) = (x-6)(x+2), which we've determined to be (2, -16), holds significant information about the function's behavior and characteristics. Since the coefficient a in the standard form (x² - 4x - 12) is positive (a = 1), the parabola opens upwards. This means that the vertex represents the minimum point of the function. The y-coordinate of the vertex, -16, is the minimum value that the function can attain. In practical terms, if this function were modeling a physical scenario, such as the trajectory of a projectile or the cost of production, the vertex would represent the point of minimum height or minimum cost.

Furthermore, the vertex provides the axis of symmetry for the parabola. The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. For our function, the axis of symmetry is the line x = 2. This symmetry is a fundamental property of parabolas and is crucial in understanding their graphical representation and behavior. Knowing the axis of symmetry allows us to easily sketch the parabola and predict its values for different inputs.

In addition, the vertex helps us determine the range of the quadratic function. Since the parabola opens upwards and the minimum value is -16, the range of f(x) = (x-6)(x+2) is y ≥ -16. This means that the function's output values will always be greater than or equal to -16. This information is vital in various applications, such as optimization problems, where we need to find the minimum or maximum value of a function within a certain range.

Real-World Applications of the Vertex

The concept of the vertex extends far beyond theoretical mathematics and finds practical applications in numerous real-world scenarios. One common example is in projectile motion. When an object is thrown or launched into the air, its trajectory often follows a parabolic path, which can be modeled by a quadratic function. The vertex of this parabola represents the maximum height the object reaches. Engineers and physicists use this information to calculate the optimal launch angle and initial velocity to achieve a desired range or height.

Another application is in optimization problems. Many real-world scenarios involve finding the optimal value of a quantity, such as minimizing costs or maximizing profits. Quadratic functions are frequently used to model these scenarios, and the vertex provides the point of minimum or maximum value. For instance, a business might use a quadratic function to model the relationship between the price of a product and the profit generated. The vertex would then indicate the price that maximizes profit.

In engineering and architecture, parabolas are used in the design of bridges, arches, and satellite dishes. The vertex plays a crucial role in ensuring structural stability and optimal performance. For example, the cables of a suspension bridge often form a parabolic shape, and the vertex represents the lowest point of the cable, which is critical for load distribution and overall stability.

Moreover, quadratic functions and their vertices are used in computer graphics and game development to create realistic trajectories and movements. Artists and programmers use parabolas to simulate the motion of objects, such as projectiles, characters jumping, or balls bouncing. The vertex helps control the apex of the trajectory and ensures smooth and realistic animations.

Conclusion

In summary, finding the vertex of a quadratic function is a fundamental skill with far-reaching applications. For the specific function f(x) = (x-6)(x+2), we've demonstrated two methods to determine the vertex: converting to standard form and using the symmetry and roots. Both methods lead us to the same vertex, (2, -16), which represents the minimum point of the parabola. The vertex provides crucial information about the function's behavior, including its minimum value, axis of symmetry, and range.

Understanding the vertex empowers us to analyze and interpret quadratic functions effectively. Whether it's modeling projectile motion, optimizing business profits, or designing structures, the vertex serves as a key element in solving real-world problems. By mastering the techniques for finding the vertex and recognizing its significance, we unlock a deeper understanding of quadratic functions and their widespread applications in various fields. The journey to understanding quadratic functions is ongoing, and the vertex is just one of the many fascinating aspects that make this topic so rich and rewarding.