Plotting Parabolas: Finding Points For Y = -6x^2 + 3x + 8

Hey guys! Let's dive into the fascinating world of quadratic equations and parabolas. Today, we're tackling the equation y = -6x² + 3x + 8 and figuring out how to plot its parabola. This might sound intimidating, but trust me, it's totally doable. We’ll break it down step-by-step, so you'll be sketching parabolas like a pro in no time. Understanding how to graph these equations is super useful in math and even real-world applications, from physics to engineering. So, grab your pencils and let's get started!

Understanding Quadratic Equations and Parabolas

Before we jump into plotting specific points, let's make sure we're all on the same page about quadratic equations and parabolas. Quadratic equations are polynomial equations of the second degree, meaning the highest power of the variable (usually x) is 2. 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. In our case, for the equation y = -6x² + 3x + 8, we have a = -6, b = 3, and c = 8. Remember these values, because they're gonna be important in our calculations.

The graph of a quadratic equation is a parabola, which is a U-shaped curve. Parabolas can open upwards or downwards, depending on the sign of the coefficient a. If a is positive, the parabola opens upwards, and if a is negative, like in our equation (a = -6), the parabola opens downwards. This is a crucial piece of information because it gives us a general idea of what our graph will look like. Also, every parabola has a vertex, which is the point where the parabola changes direction. The vertex is either the minimum point (if the parabola opens upwards) or the maximum point (if it opens downwards). Finding the vertex is one of the key steps in plotting a parabola accurately. Once you understand the relationship between the equation and the shape of the parabola, you are already halfway there. The next step is to determine key points that help us sketch the curve, which we will explore in detail below.

Finding the Vertex: The Turning Point

The vertex is the most important point on a parabola. It's where the parabola changes direction, and it helps us understand the overall shape and position of the graph. To find the vertex of the parabola for the equation y = -6x² + 3x + 8, we use a simple formula. The x-coordinate of the vertex, often denoted as h, is given by h = -b / (2a). Remember those coefficients we identified earlier? Here's where they come in handy!

In our equation, a = -6 and b = 3, so we plug these values into the formula: h = -3 / (2 * -6) = -3 / -12 = 1/4. So, the x-coordinate of the vertex is 1/4 or 0.25. Now, to find the y-coordinate of the vertex, often denoted as k, we substitute this x-value back into the original equation. This will give us the y-value corresponding to the vertex. So, we calculate k = -6(1/4)² + 3(1/4) + 8 = -6(1/16) + 3/4 + 8 = -3/8 + 6/8 + 64/8 = 67/8, which is approximately 8.375. Therefore, the vertex of our parabola is at the point (1/4, 67/8) or (0.25, 8.375). This is a crucial point to plot, as it gives us a reference for the rest of the parabola. Knowing the vertex helps us visualize the parabola's peak and how the curve will extend on either side. We will use this as a centerpiece for our graph and then find additional points to accurately draw the curve.

Finding the Y-intercept: Where the Parabola Crosses the Y-axis

The y-intercept is another essential point to plot when sketching a parabola. It’s the point where the parabola intersects the y-axis. To find the y-intercept, we set x = 0 in our quadratic equation and solve for y. This works because any point on the y-axis has an x-coordinate of 0.

For our equation, y = -6x² + 3x + 8, we substitute x = 0: y = -6(0)² + 3(0) + 8 = 0 + 0 + 8 = 8. So, the y-intercept is at the point (0, 8). This point is relatively easy to find and provides a convenient reference on the graph. It gives us another fixed point that helps in determining the shape and position of the parabola. Think of the y-intercept as another anchor point on our graph; we know the parabola will pass through this point, allowing us to sketch the curve more precisely. Along with the vertex, the y-intercept provides a solid foundation for drawing an accurate representation of the quadratic equation. After finding the y-intercept, we can move on to finding additional points to ensure the parabola's shape is correctly represented on the graph.

Finding Additional Points: Symmetry to the Rescue

Now that we have the vertex and the y-intercept, we need a few more points to get a good sense of the parabola's shape. One of the beautiful things about parabolas is their symmetry. They are symmetrical around the vertical line that passes through the vertex. This line is called the axis of symmetry, and it simplifies our task of finding additional points. The equation for the axis of symmetry is x = h, where h is the x-coordinate of the vertex. In our case, the axis of symmetry is the vertical line x = 1/4 or x = 0.25.

To use this symmetry, we can pick a few x-values on one side of the vertex, calculate the corresponding y-values, and then use the symmetry to find the corresponding points on the other side. For example, let's choose x = 1. Substituting this into our equation, we get y = -6(1)² + 3(1) + 8 = -6 + 3 + 8 = 5. So, we have the point (1, 5). Now, because of symmetry, there is a corresponding point on the other side of the axis of symmetry. The x-value of this point will be the same distance from the axis of symmetry as x = 1, but on the opposite side. The distance between x = 1/4 and x = 1 is 3/4. So, we go 3/4 to the left of x = 1/4, which gives us x = 1/4 - 3/4 = -1/2 or x = -0.5. The corresponding point will have the same y-value, so we have the point (-1/2, 5). This trick effectively gives us two points for the price of one! Now let's pick another value to give us even more accuracy when plotting our parabola.

Let’s pick x = 2. Substituting this into our equation gives us y = -6(2)² + 3(2) + 8 = -6(4) + 6 + 8 = -24 + 6 + 8 = -10. So, we have the point (2, -10). To find the symmetrical point, we look at the distance from x = 2 to the axis of symmetry, x = 1/4. The distance is 2 - 1/4 = 7/4. Going the same distance to the left of the axis of symmetry, we have x = 1/4 - 7/4 = -6/4 = -3/2 or x = -1.5. Therefore, the symmetrical point is (-3/2, -10). By using symmetry, we've efficiently doubled the number of points we have, making our parabola sketch even more accurate. These additional points help us to see how steeply the parabola curves and ensure we capture its shape correctly when we graph it.

Plotting the Parabola: Connecting the Dots

Now that we've found several key points – the vertex (0.25, 8.375), the y-intercept (0, 8), and additional points like (1, 5), (-0.5, 5), (2, -10), and (-1.5, -10) – we're ready to plot our parabola. Start by drawing a coordinate plane with the x and y-axes. Then, carefully plot each of these points on the plane. The more accurate your plotting, the better your parabola will look.

Once the points are plotted, the final step is to connect them with a smooth, curved line. Remember that a parabola is a U-shaped curve, not a V-shaped one, so aim for a smooth, continuous curve. The vertex should be the turning point of the parabola, and the curve should be symmetrical around the axis of symmetry. If your points are plotted correctly, the parabola should pass through all of them smoothly. If some points seem off, double-check your calculations or plotting. The key is to ensure the curve gracefully connects the points, reflecting the nature of the quadratic equation. As you sketch, consider the overall shape: since the coefficient of our x² term is negative (a = -6), the parabola should open downwards, confirming our plotted points' trend. The final result should be a clear, symmetrical parabola that represents the equation y = -6x² + 3x + 8 accurately.

Conclusion: Mastering the Parabola

So, there you have it! We've successfully navigated the process of plotting the parabola for the quadratic equation y = -6x² + 3x + 8. We started by understanding the basic form of a quadratic equation and recognizing that its graph is a parabola. We then identified the key steps: finding the vertex using the formula h = -b / (2a) and substituting to find the y-coordinate, determining the y-intercept by setting x = 0, and using the symmetry of the parabola to find additional points efficiently.

By plotting these points and connecting them with a smooth curve, we created an accurate representation of the parabola. Remember, the vertex is the turning point, the y-intercept is where the parabola crosses the y-axis, and symmetry is your friend for finding additional points. These skills are invaluable not just in mathematics but also in various fields that use quadratic equations to model real-world phenomena. Keep practicing, and you'll become a parabola-plotting pro in no time! Now you have the tools and the know-how to tackle similar problems. Go forth and conquer those quadratic equations!