How To Graph Quadratic Equations Y = X² + 6 And Others

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This article serves as a comprehensive guide to graphing quadratic equations, focusing on four specific examples. We will delve into the process of identifying key features of a parabola, such as the vertex, axis of symmetry, and intercepts, and how to use these features to accurately plot the graph. Understanding these concepts is crucial for anyone studying algebra or calculus, as quadratic equations appear in various applications, from physics to engineering to economics.

1. Graphing y = x² + 6

Let's start by graphing the first equation: y = x² + 6. This is a basic quadratic equation in the form y = ax² + c, where a = 1 and c = 6. Recognizing this form immediately tells us that the graph will be a parabola that opens upwards (since a is positive) and its vertex is shifted vertically by 6 units compared to the parent function y = x².

To graph this equation, we first need to identify the vertex. For a quadratic in the form y = ax² + c, the vertex is simply (0, c). In this case, the vertex is (0, 6). This point is the minimum point on the parabola, as the parabola opens upwards. Next, we can find some additional points to plot. We can choose some x-values and substitute them into the equation to find the corresponding y-values. A symmetrical approach is often helpful, choosing x-values that are equidistant from the x-coordinate of the vertex (which is 0 in this case). For instance, let's try x = -2, -1, 1, and 2.

  • When x = -2, y = (-2)² + 6 = 4 + 6 = 10. So, we have the point (-2, 10).
  • When x = -1, y = (-1)² + 6 = 1 + 6 = 7. So, we have the point (-1, 7).
  • When x = 1, y = (1)² + 6 = 1 + 6 = 7. So, we have the point (1, 7).
  • When x = 2, y = (2)² + 6 = 4 + 6 = 10. So, we have the point (2, 10).

Now we have a set of points: (0, 6), (-2, 10), (-1, 7), (1, 7), and (2, 10). Plotting these points on a coordinate plane and connecting them with a smooth curve will give us the graph of y = x² + 6. We can observe that the parabola is symmetrical about the y-axis, which is the axis of symmetry for this equation. This symmetry is a characteristic feature of parabolas and is always present about the vertical line passing through the vertex. By understanding the basic form of the equation and plotting a few key points, we can accurately graph this quadratic function.

2. Graphing y = (1/2)x² - 2x + 6

The second equation we'll tackle is y = (1/2)x² - 2x + 6. This equation is in the general quadratic form y = ax² + bx + c, where a = 1/2, b = -2, and c = 6. This form is slightly more complex than the previous one, but we can still find the key features of the parabola. The most important feature to start with is the vertex. To find the x-coordinate of the vertex, we use the formula x = -b / 2a.

In this case, a = 1/2 and b = -2, so x = -(-2) / (2 * 1/2) = 2 / 1 = 2. To find the y-coordinate of the vertex, we substitute x = 2 back into the equation: y = (1/2)(2)² - 2(2) + 6 = (1/2)(4) - 4 + 6 = 2 - 4 + 6 = 4. Therefore, the vertex is (2, 4). Since a = 1/2 is positive, the parabola opens upwards, and the vertex represents the minimum point.

The axis of symmetry is a vertical line that passes through the vertex. Its equation is simply x = the x-coordinate of the vertex, which in this case is x = 2. The axis of symmetry helps us to understand the symmetrical nature of the parabola and plot points more efficiently.

Next, let's find some additional points. We can choose x-values on either side of the vertex. Let's try x = 0, 1, 3, and 4:

  • When x = 0, y = (1/2)(0)² - 2(0) + 6 = 6. So, we have the point (0, 6).
  • When x = 1, y = (1/2)(1)² - 2(1) + 6 = 1/2 - 2 + 6 = 4.5. So, we have the point (1, 4.5).
  • When x = 3, y = (1/2)(3)² - 2(3) + 6 = (1/2)(9) - 6 + 6 = 4.5. So, we have the point (3, 4.5).
  • When x = 4, y = (1/2)(4)² - 2(4) + 6 = (1/2)(16) - 8 + 6 = 8 - 8 + 6 = 6. So, we have the point (4, 6).

Now we have the points (2, 4), (0, 6), (1, 4.5), (3, 4.5), and (4, 6). Plotting these points and connecting them with a smooth curve gives us the graph of y = (1/2)x² - 2x + 6. Notice how the parabola is symmetrical about the line x = 2. The coefficient of x² (1/2) affects the width of the parabola; a smaller coefficient makes the parabola wider compared to y = x².

3. Graphing y = x² + 4x - 12

Now, let's graph the equation y = x² + 4x - 12. This is another quadratic equation in the general form y = ax² + bx + c, where a = 1, b = 4, and c = -12. We'll follow a similar process to the previous example. First, we find the vertex using the formula x = -b / 2a.

In this case, a = 1 and b = 4, so x = -4 / (2 * 1) = -2. To find the y-coordinate of the vertex, we substitute x = -2 back into the equation: y = (-2)² + 4(-2) - 12 = 4 - 8 - 12 = -16. Thus, the vertex is (-2, -16). Since a = 1 is positive, the parabola opens upwards, and the vertex is the minimum point.

The axis of symmetry is the vertical line passing through the vertex, which is x = -2.

This time, let's also find the x-intercepts and the y-intercept to help us graph the parabola. The x-intercepts are the points where the parabola crosses the x-axis, which means y = 0. So, we need to solve the equation x² + 4x - 12 = 0. This can be factored as (x + 6)(x - 2) = 0. Therefore, the x-intercepts are x = -6 and x = 2. The points are (-6, 0) and (2, 0).

The y-intercept is the point where the parabola crosses the y-axis, which means x = 0. Substituting x = 0 into the equation, we get y = (0)² + 4(0) - 12 = -12. So, the y-intercept is (0, -12).

Now, let's find a few additional points. We can choose x-values around the vertex. Let's try x = -4 and x = 0 (we already know the y-intercept at x = 0):

  • When x = -4, y = (-4)² + 4(-4) - 12 = 16 - 16 - 12 = -12. So, we have the point (-4, -12).

We have the vertex (-2, -16), the x-intercepts (-6, 0) and (2, 0), the y-intercept (0, -12), and the additional point (-4, -12). Plotting these points and connecting them with a smooth curve gives us the graph of y = x² + 4x - 12. The x-intercepts give us important information about where the parabola intersects the x-axis, and they help us to accurately sketch the curve.

4. Graphing y = (1/2)x² + 2x - 6

Finally, let's graph the equation y = (1/2)x² + 2x - 6. This is another quadratic equation in the general form y = ax² + bx + c, where a = 1/2, b = 2, and c = -6. Again, we'll start by finding the vertex using the formula x = -b / 2a.

In this case, a = 1/2 and b = 2, so x = -2 / (2 * 1/2) = -2 / 1 = -2. To find the y-coordinate of the vertex, we substitute x = -2 back into the equation: y = (1/2)(-2)² + 2(-2) - 6 = (1/2)(4) - 4 - 6 = 2 - 4 - 6 = -8. So, the vertex is (-2, -8). Since a = 1/2 is positive, the parabola opens upwards, and the vertex is the minimum point.

The axis of symmetry is the vertical line passing through the vertex, which is x = -2.

To enhance our graph, we will also identify the x-intercepts and the y-intercept. The x-intercepts occur where y = 0, so we need to solve the equation (1/2)x² + 2x - 6 = 0. To make it easier, we can multiply the entire equation by 2 to eliminate the fraction: x² + 4x - 12 = 0. This is the same quadratic equation we solved in the previous example! So, the x-intercepts are x = -6 and x = 2. The points are (-6, 0) and (2, 0).

The y-intercept is the point where the parabola crosses the y-axis (x = 0). Substituting x = 0 into the equation, we get y = (1/2)(0)² + 2(0) - 6 = -6. So, the y-intercept is (0, -6).

Now, let's find a few additional points. Since we already have the intercepts, we can choose x-values near the vertex. Let's try x = -4:

  • When x = -4, y = (1/2)(-4)² + 2(-4) - 6 = (1/2)(16) - 8 - 6 = 8 - 8 - 6 = -6. So, we have the point (-4, -6).

We have the vertex (-2, -8), the x-intercepts (-6, 0) and (2, 0), the y-intercept (0, -6), and the additional point (-4, -6). Plotting these points and connecting them with a smooth curve gives us the graph of y = (1/2)x² + 2x - 6. Comparing this graph with the previous one, we see that the coefficient of x² (1/2) makes the parabola wider, just as we observed before.

Conclusion

Graphing quadratic equations involves understanding the key features of a parabola, such as the vertex, axis of symmetry, and intercepts. By finding these features and plotting a few additional points, we can accurately graph the equation. Remember to use the formula x = -b / 2a to find the x-coordinate of the vertex, and then substitute this value back into the equation to find the y-coordinate. Finding the x-intercepts by setting y = 0 and the y-intercept by setting x = 0 can provide valuable information about the shape and position of the parabola. Through practice and a systematic approach, graphing quadratic equations becomes a straightforward and valuable skill in mathematics.