Graphing The Parabola Y=x^2-6x+11: Vertex And Points Method

This article offers a detailed walkthrough on graphing the parabola defined by the equation y = x² - 6x + 11. We'll cover how to find the vertex, which is the parabola's turning point, and how to identify two points to the left and two points to the right of the vertex. By plotting these key points, we can accurately sketch the parabola and understand its characteristics.

Understanding Parabolas

Before we dive into graphing the specific equation, let's establish some foundational knowledge about parabolas. A parabola is a U-shaped curve defined by a quadratic equation, which is a polynomial equation of degree two. The general form of a quadratic equation is y = ax² + bx + c, where a, b, and c are constants, and a is not equal to zero. The sign of a determines whether the parabola opens upwards (a > 0) or downwards (a < 0). In our case, y = x² - 6x + 11, a = 1, b = -6, and c = 11. Since a is positive, this parabola will open upwards.

The vertex is a crucial point on the parabola. It represents either the minimum value of y (if the parabola opens upwards) or the maximum value of y (if the parabola opens downwards). The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves. Understanding these key components helps in accurately graphing any parabola.

Finding the Vertex: Completing the Square

To graph the parabola y = x² - 6x + 11, our first crucial step is to find the vertex. There are two common methods for determining the vertex: completing the square and using the vertex formula. In this section, we'll focus on the method of completing the square because it provides valuable insights into the structure of the equation and the position of the vertex. Completing the square transforms the quadratic equation into vertex form, which directly reveals the coordinates of the vertex.

The vertex form of a quadratic equation is y = a(x - h)² + k, where (h, k) represents the coordinates of the vertex. This form makes it easy to identify the vertex, as h is the x-coordinate and k is the y-coordinate. Transforming the given equation into this form involves a series of algebraic manipulations. We start by isolating the x² and x terms: y = (x² - 6x) + 11. Next, we need to add and subtract a value inside the parentheses to create a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. To find this value, we take half of the coefficient of the x term (-6), which is -3, and square it, resulting in 9. So, we add and subtract 9 inside the parentheses: y = (x² - 6x + 9 - 9) + 11.

Now, we can rewrite the perfect square trinomial as a squared binomial: y = ((x - 3)²) - 9) + 11. Distribute any necessary terms and combine the constants outside the parenthesis: y = (x - 3)² - 9 + 11. Finally, we simplify the equation to vertex form: y = (x - 3)² + 2. From this form, we can easily identify the vertex. The vertex is at the point (3, 2). This means the parabola's turning point is at x = 3 and y = 2. Completing the square not only helps us find the vertex but also provides a clear understanding of how the parabola is shifted from the basic y = x² graph. The (x - 3)² term indicates a horizontal shift of 3 units to the right, and the + 2 term indicates a vertical shift of 2 units upwards. This visual understanding is crucial for accurately graphing the parabola.

Finding the Vertex: Using the Vertex Formula

Another effective method for determining the vertex of the parabola y = x² - 6x + 11 is by using the vertex formula. This method is particularly useful when you need a quick and direct way to find the vertex without going through the process of completing the square. The vertex formula is derived from the standard form of a quadratic equation, y = ax² + bx + c, and it provides a straightforward calculation for the x-coordinate of the vertex. Once we have the x-coordinate, we can substitute it back into the original equation to find the corresponding y-coordinate.

The formula for the x-coordinate of the vertex, often denoted as h, is given by h = -b / 2a. In our equation, y = x² - 6x + 11, we have a = 1 and b = -6. Plugging these values into the formula, we get h = -(-6) / (2 * 1) = 6 / 2 = 3. This tells us that the x-coordinate of the vertex is 3.

Now that we have the x-coordinate, we can find the y-coordinate, often denoted as k, by substituting x = 3 into the original equation: y = (3)² - 6(3) + 11 = 9 - 18 + 11 = 2. Therefore, the y-coordinate of the vertex is 2. Combining these results, we find that the vertex is at the point (3, 2). This result aligns with what we found by completing the square, confirming the accuracy of both methods. The vertex formula is a powerful tool for quickly locating the vertex, especially when dealing with more complex quadratic equations. It provides a reliable and efficient way to determine the turning point of the parabola, which is essential for graphing and analyzing the function.

Choosing Points for Graphing: Left and Right of the Vertex

Now that we've determined the vertex of the parabola y = x² - 6x + 11 to be at (3, 2), the next step is to choose points to the left and right of the vertex to accurately sketch the curve. Since parabolas are symmetrical around their axis of symmetry, selecting points equidistant from the vertex will yield symmetrical results, making the graphing process more efficient. We aim to find at least two points on each side of the vertex to ensure we capture the shape of the parabola accurately. This approach leverages the symmetry of the parabola to minimize calculations while maximizing the precision of the graph.

To select points to the left of the vertex, we'll choose x-values less than 3. Let's pick x = 1 and x = 2. These values are symmetrically positioned with respect to the axis of symmetry, which passes through the vertex at x = 3. For x = 1, we substitute it into the equation: y = (1)² - 6(1) + 11 = 1 - 6 + 11 = 6. This gives us the point (1, 6). For x = 2, we substitute it into the equation: y = (2)² - 6(2) + 11 = 4 - 12 + 11 = 3. This gives us the point (2, 3). These two points on the left side of the vertex provide a good representation of the curve's behavior as it approaches the vertex.

Similarly, to select points to the right of the vertex, we'll choose x-values greater than 3. Let's pick x = 4 and x = 5. These values maintain the symmetrical balance with our choices on the left side. For x = 4, we substitute it into the equation: y = (4)² - 6(4) + 11 = 16 - 24 + 11 = 3. This gives us the point (4, 3). Notice that the y-value is the same as when x = 2, which confirms the symmetry of the parabola. For x = 5, we substitute it into the equation: y = (5)² - 6(5) + 11 = 25 - 30 + 11 = 6. This gives us the point (5, 6), which has the same y-value as when x = 1, further illustrating the symmetrical nature of the parabola. By selecting these five points – the vertex and two points on each side – we have a solid foundation for accurately graphing the parabola.

Plotting the Points and Sketching the Parabola

With the vertex and four additional points calculated, we are now ready to plot these points on a coordinate plane and sketch the parabola y = x² - 6x + 11. Plotting these key points accurately is essential for drawing the curve of the parabola and visualizing its characteristics. The points we have are the vertex (3, 2), two points to the left (1, 6) and (2, 3), and two points to the right (4, 3) and (5, 6).

Start by drawing a coordinate plane with the x-axis and y-axis clearly marked. The scale should be appropriate to accommodate the points we are plotting. In this case, the x-values range from 1 to 5, and the y-values range from 2 to 6, so a standard scale will work well. First, plot the vertex (3, 2). This point serves as the central reference for the parabola and is the turning point of the curve. Next, plot the points to the left of the vertex, (1, 6) and (2, 3). These points show how the parabola rises as we move away from the vertex on the left side. Then, plot the points to the right of the vertex, (4, 3) and (5, 6). These points mirror the behavior on the left side, demonstrating the symmetry of the parabola.

Once the points are plotted, carefully sketch the parabola by drawing a smooth, U-shaped curve that passes through all five points. The curve should be symmetrical around the vertical line that passes through the vertex (the axis of symmetry). The vertex should be the lowest point on the parabola since a is positive, indicating that the parabola opens upwards. Ensure the curve is smooth and continuous, without any sharp angles or breaks. The resulting graph provides a visual representation of the quadratic function y = x² - 6x + 11, showing its shape, vertex, and symmetry. This visual representation is invaluable for understanding the function's behavior and its relationship to the equation.

Key Takeaways and Conclusion

In this comprehensive guide, we've walked through the process of graphing the parabola y = x² - 6x + 11. We began by understanding the general characteristics of parabolas, including the importance of the vertex and the axis of symmetry. We then explored two methods for finding the vertex: completing the square and using the vertex formula. Completing the square allowed us to rewrite the equation in vertex form, directly revealing the vertex coordinates. The vertex formula provided a quicker, more direct method for finding the vertex.

Next, we discussed how to choose points to the left and right of the vertex to accurately sketch the parabola. By selecting points equidistant from the vertex, we leveraged the parabola's symmetry to simplify the graphing process. We calculated the y-values for these chosen x-values and obtained five key points: the vertex and two points on each side. These points formed the foundation for sketching the parabola.

Finally, we plotted these points on a coordinate plane and drew a smooth, U-shaped curve that passed through all the points. The resulting graph visually represents the quadratic function, illustrating its shape, turning point, and symmetrical nature. This process not only allows us to graph the parabola accurately but also deepens our understanding of quadratic functions and their graphical representations. Mastering these techniques is crucial for further studies in mathematics and related fields, providing a solid foundation for analyzing and visualizing quadratic relationships.

This method can be applied to any quadratic equation to graph the corresponding parabola. By finding the vertex and additional points, you can accurately represent the curve and understand the behavior of the quadratic function.