Finding The Line Of Symmetry Parabola Equation Y=(1/5)(x-6)^2-5

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In the realm of quadratic equations and their graphical representations, parabolas hold a position of prominence. Understanding the properties of parabolas, such as their vertex and line of symmetry, is crucial for both theoretical comprehension and practical applications. This article delves into the process of determining the equation of the line of symmetry for a parabola, using the example equation y = (1/5)(x - 6)^2 - 5 as a case study. We'll break down the concepts, explore different methods, and provide clear explanations to help you master this essential skill.

Understanding Parabolas and Symmetry

At its core, a parabola is a symmetrical U-shaped curve. This symmetry stems from the quadratic nature of its equation. The standard form of a quadratic equation is y = ax^2 + bx + c, where 'a', 'b', and 'c' are constants. When graphed, this equation produces a parabola. The line of symmetry is a vertical line that divides the parabola into two mirror-image halves. The vertex of the parabola, which is the point where the curve changes direction (either the minimum or maximum point), lies precisely on this line of symmetry.

The vertex form of a quadratic equation, y = a(x - h)^2 + k, provides a more direct way to identify the vertex and the line of symmetry. In this form, the vertex is located at the point (h, k), and the line of symmetry is the vertical line x = h. Recognizing and utilizing the vertex form is a key strategy for solving problems related to parabolas.

The line of symmetry is a crucial feature of any parabola. It's the vertical line that cuts the parabola perfectly in half, creating two mirror images. This line always passes through the vertex, which is either the lowest point (minimum) or the highest point (maximum) on the parabola. Understanding the line of symmetry helps us visualize the parabola's shape and location on the coordinate plane.

Identifying the Vertex Form

Our given equation, y = (1/5)(x - 6)^2 - 5, is already conveniently presented in vertex form. This makes our task significantly easier. Let's compare it to the general vertex form y = a(x - h)^2 + k:

  • a = 1/5: This value determines the parabola's width and direction. Since 'a' is positive, the parabola opens upwards.
  • h = 6: This value represents the x-coordinate of the vertex.
  • k = -5: This value represents the y-coordinate of the vertex.

Therefore, we can immediately identify the vertex of the parabola as the point (6, -5). The vertex plays a central role in determining the line of symmetry, as it lies directly on it. By recognizing the vertex form of the equation, we bypass the need for complex calculations or transformations, allowing us to quickly pinpoint the key parameters of the parabola.

Determining the Line of Symmetry

As mentioned earlier, the line of symmetry for a parabola in vertex form y = a(x - h)^2 + k is the vertical line x = h. Since we've identified h = 6 in our equation, the line of symmetry is simply x = 6.

The line of symmetry is always a vertical line expressed in the form x = constant. This constant represents the x-coordinate of every point on the line, and in the case of a parabola, it's the same as the x-coordinate of the vertex. The line of symmetry acts as a mirror, reflecting one half of the parabola onto the other. Its equation provides a concise and precise description of this fundamental property.

Alternative Methods (Optional)

While the vertex form provides the most straightforward method, let's briefly explore alternative approaches to find the line of symmetry. These methods are particularly useful if the equation is not initially given in vertex form.

1. Converting to Standard Form and Using the Formula

We can expand the given equation to convert it to the standard form y = ax^2 + bx + c:

y = (1/5)(x - 6)^2 - 5 y = (1/5)(x^2 - 12x + 36) - 5 y = (1/5)x^2 - (12/5)x + (36/5) - 5 y = (1/5)x^2 - (12/5)x + 11/5

In standard form, the line of symmetry is given by the formula x = -b / 2a. In our case, a = 1/5 and b = -12/5. Plugging these values into the formula:

x = -(-12/5) / (2 * 1/5) x = (12/5) / (2/5) x = 6

This method confirms our previous result, although it involves more algebraic manipulation. The standard form of a quadratic equation provides a general representation that allows us to apply formulas and techniques to extract key information, including the line of symmetry. While not as direct as using the vertex form, this method demonstrates the versatility of algebraic approaches in problem-solving.

2. Finding the x-intercepts and Averaging

If the parabola has two distinct x-intercepts, the line of symmetry will pass through the midpoint of these intercepts. To find the x-intercepts, we set y = 0 and solve for x:

0 = (1/5)(x - 6)^2 - 5 5 = (1/5)(x - 6)^2 25 = (x - 6)^2 ±5 = x - 6 x = 6 ± 5

This gives us two x-intercepts: x = 1 and x = 11. The average of these x-intercepts is:

(1 + 11) / 2 = 6

Again, we arrive at the line of symmetry x = 6. This method highlights the relationship between the x-intercepts and the line of symmetry. When a parabola intersects the x-axis at two distinct points, the line of symmetry always bisects the segment connecting these points. This geometric property provides an alternative way to visualize and calculate the line of symmetry.

Conclusion

Determining the equation of the line of symmetry for a parabola is a fundamental skill in algebra and calculus. By recognizing the vertex form of the quadratic equation, we can quickly identify the line of symmetry as x = h, where 'h' is the x-coordinate of the vertex. Alternative methods, such as converting to standard form or averaging the x-intercepts, provide additional approaches to solve this problem. Understanding these techniques enhances our ability to analyze and interpret quadratic functions and their graphical representations.

In the specific case of the equation y = (1/5)(x - 6)^2 - 5, the line of symmetry is x = 6. This result aligns with our understanding of parabolas and their symmetrical nature. Mastering these concepts provides a solid foundation for further exploration of quadratic functions and their applications in various fields.

Answer

The correct answer is C. x = 6.