Finding The Vertex Of Y=-1/2(x+2)(x-6) A Step-by-Step Guide

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Determining the vertex of a quadratic function is a fundamental concept in algebra, with broad applications across various fields, including physics, engineering, and economics. The vertex represents the maximum or minimum point of the parabola, providing crucial information about the function's behavior. In this article, we'll delve deep into understanding how to find the vertex, focusing on the quadratic function expressed in factored form: $y=-\frac{1}{2}(x+2)(x-6)$. We'll explore the underlying principles, step-by-step methods, and practical implications of this essential concept.

Understanding Quadratic Functions and Their Graphs

Before we jump into the specifics of finding the vertex, it's essential to establish a solid understanding of quadratic functions and their graphical representations. A quadratic function is a polynomial function of degree two, generally expressed in the standard form as: $f(x) = ax^2 + bx + c$, 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, depending on the sign of the coefficient a.

  • If a > 0, the parabola opens upwards, and the vertex represents the minimum point of the function.
  • If a < 0, the parabola opens downwards, and the vertex represents the maximum point of the function.

The vertex is a critical point on the parabola, serving as the axis of symmetry. This means that the parabola is symmetrical about the vertical line passing through the vertex. The x-coordinate of the vertex is often referred to as the axis of symmetry.

Quadratic functions can also be expressed in other forms, such as the vertex form $f(x) = a(x - h)^2 + k$, where (h, k) are the coordinates of the vertex. This form makes it incredibly easy to identify the vertex directly. Another important form is the factored form, which is the form presented in our question: $y = a(x - r_1)(x - r_2)$, where r1 and r2 are the roots or x-intercepts of the function. Understanding the relationships between these forms is key to efficiently solving quadratic problems.

Finding the Vertex from Factored Form

Our given function, $y=-\frac{1}{2}(x+2)(x-6)$, is in factored form. This form provides us with a unique advantage in finding the vertex. The factored form directly reveals the x-intercepts (or roots) of the quadratic function. The x-intercepts are the points where the parabola intersects the x-axis, meaning y = 0. In the factored form $y = a(x - r_1)(x - r_2)$, the x-intercepts are r1 and r2. To find the x-intercepts for our function, we set y = 0:

0=−12(x+2)(x−6)0 = -\frac{1}{2}(x + 2)(x - 6)

This equation is satisfied when either (x + 2) = 0 or (x - 6) = 0. Solving these equations gives us the x-intercepts:

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

So, the x-intercepts are -2 and 6. Now, here's the crucial step: the x-coordinate of the vertex lies exactly in the middle of the two x-intercepts due to the symmetry of the parabola. We can find the x-coordinate of the vertex by calculating the average of the x-intercepts:

xvertex=r1+r22=−2+62=42=2x_\text{vertex} = \frac{r_1 + r_2}{2} = \frac{-2 + 6}{2} = \frac{4}{2} = 2

Therefore, the x-coordinate of the vertex is 2. To find the y-coordinate of the vertex, we substitute this x-value back into the original equation:

yvertex=−12(2+2)(2−6)=−12(4)(−4)=−12(−16)=8y_\text{vertex} = -\frac{1}{2}(2 + 2)(2 - 6) = -\frac{1}{2}(4)(-4) = -\frac{1}{2}(-16) = 8

Thus, the y-coordinate of the vertex is 8. Combining the x and y coordinates, we find that the vertex of the parabola is (2, 8).

Step-by-Step Method for Finding the Vertex from Factored Form

To summarize, here's a step-by-step method for finding the vertex of a quadratic function given in factored form:

  1. Identify the x-intercepts (roots): Set the function equal to zero and solve for x. These are your r1 and r2.
  2. Calculate the x-coordinate of the vertex: Use the formula $x_\text{vertex} = \frac{r_1 + r_2}{2}$. This finds the midpoint between the roots.
  3. Calculate the y-coordinate of the vertex: Substitute the x-coordinate of the vertex back into the original equation and solve for y.
  4. Write the vertex as a coordinate pair: The vertex is (x_vertex, y_vertex).

Applying this method to our example, $y=-\frac{1}{2}(x+2)(x-6)$, we found the vertex to be (2, 8).

Alternative Methods for Finding the Vertex

While finding the vertex from the factored form is efficient, it's worth noting other methods that can be used, especially when the quadratic function is in standard form ($f(x) = ax^2 + bx + c$).

1. Using the Vertex Formula

When the quadratic function is in standard form, the x-coordinate of the vertex can be directly calculated using the vertex formula:

xvertex=−b2ax_\text{vertex} = -\frac{b}{2a}

After finding the x-coordinate, substitute it back into the equation to find the y-coordinate, just as we did with the factored form. To use this method with our example, we would first need to expand the factored form into standard form:

y=−12(x+2)(x−6)=−12(x2−6x+2x−12)=−12(x2−4x−12)=−12x2+2x+6y = -\frac{1}{2}(x + 2)(x - 6) = -\frac{1}{2}(x^2 - 6x + 2x - 12) = -\frac{1}{2}(x^2 - 4x - 12) = -\frac{1}{2}x^2 + 2x + 6

Now, we can identify a = -1/2, b = 2, and c = 6. Applying the vertex formula:

xvertex=−22(−12)=−2−1=2x_\text{vertex} = -\frac{2}{2(-\frac{1}{2})} = -\frac{2}{-1} = 2

As before, substituting x = 2 back into the standard form equation gives us y = 8, confirming the vertex (2, 8).

2. Completing the Square

Completing the square is another method to transform the standard form into the vertex form, $f(x) = a(x - h)^2 + k$, which directly reveals the vertex (h, k). To complete the square, we manipulate the quadratic expression to create a perfect square trinomial. Let's apply this to our example, starting from the standard form: $y = -\frac{1}{2}x^2 + 2x + 6$

  1. Factor out the coefficient of the x² term from the first two terms:

    y=−12(x2−4x)+6y = -\frac{1}{2}(x^2 - 4x) + 6

  2. Take half of the coefficient of the x term (-4), square it ((-2)² = 4), and add and subtract it inside the parentheses:

    y=−12(x2−4x+4−4)+6y = -\frac{1}{2}(x^2 - 4x + 4 - 4) + 6

  3. Rewrite the perfect square trinomial:

    y=−12((x−2)2−4)+6y = -\frac{1}{2}((x - 2)^2 - 4) + 6

  4. Distribute the -1/2 and simplify:

    y=−12(x−2)2+2+6y = -\frac{1}{2}(x - 2)^2 + 2 + 6

    y=−12(x−2)2+8y = -\frac{1}{2}(x - 2)^2 + 8

Now, the equation is in vertex form, and we can directly identify the vertex as (2, 8).

Practical Applications of the Vertex

The vertex of a parabola is not just a theoretical concept; it has significant practical applications in various real-world scenarios. Here are a few examples:

  1. Optimization Problems: In optimization problems, we often seek to maximize or minimize a certain quantity. If the quantity can be modeled by a quadratic function, the vertex will give us the optimal value. For example, if a company's profit is modeled by a quadratic function, the vertex will tell us the maximum profit and the level of production needed to achieve it.
  2. Projectile Motion: The path of a projectile (like a ball thrown in the air) can be modeled by a parabola. The vertex represents the highest point the projectile reaches. Knowing the vertex allows us to calculate the maximum height and the time it takes to reach that height.
  3. Engineering Design: Engineers use parabolas in designing various structures, such as bridges and antennas. The vertex plays a crucial role in ensuring the structural integrity and optimal performance of these designs.
  4. Economics: Quadratic functions are used to model cost, revenue, and profit in economics. The vertex can help determine the break-even points, maximum profit, and minimum cost.

Common Mistakes to Avoid

When finding the vertex of a quadratic function, it's essential to be aware of common mistakes that can lead to incorrect answers. Here are a few to watch out for:

  1. Incorrectly identifying x-intercepts: Ensure you correctly solve for the roots in the factored form. A sign error can lead to incorrect x-intercepts and, consequently, an incorrect vertex.
  2. Forgetting the sign in the vertex formula: The vertex formula is $x_\text{vertex} = -\frac{b}{2a}$. Forgetting the negative sign can lead to an incorrect x-coordinate of the vertex.
  3. Substituting the x-coordinate incorrectly: When finding the y-coordinate of the vertex, make sure to substitute the x-coordinate back into the original equation (or the correct form of the equation). Substituting into an intermediate step can lead to errors.
  4. Misinterpreting the vertex form: When using the vertex form $f(x) = a(x - h)^2 + k$, remember that the vertex is (h, k), not (-h, k). Pay attention to the signs.

Conclusion

Finding the vertex of a quadratic function is a critical skill in algebra with far-reaching applications. Whether you're working with factored form, standard form, or vertex form, understanding the different methods and their underlying principles is key to success. In the case of our example, $y=-\frac{1}{2}(x+2)(x-6)$, we successfully identified the vertex as (2, 8) using the factored form method. By mastering these techniques and avoiding common pitfalls, you'll be well-equipped to tackle a wide range of quadratic function problems and appreciate their significance in various real-world contexts. Remember to practice regularly and apply these concepts to different scenarios to solidify your understanding.

Answer

The correct answer is B. (2, 8).