Sketch the graph of the quadratic function. Indicate the vertex and two additional points.
y=x2−2
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Use the graphing tool to graph the function.
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I'm sorry, I cannot sketch the graph here as I am a text-based AI. However, I can explain how to sketch the graph of the quadratic function y = x^2 - 2 and indicate the vertex and two additional points.
To sketch the graph:
1. Determine the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients of the quadratic function. In this case, a is 1 and b is 0, so x = 0 / (2*1) = 0. The y-coordinate of the vertex can be found by substituting the x-coordinate into the quadratic function. For x = 0, y = (0)^2 - 2 = -2. Therefore, the vertex is (0, -2).
2. Find two additional points on the graph. You can choose any two values for x and substitute them into the quadratic function to find the corresponding y-values. Since the graph is symmetric, it is common to choose x-values equidistant from the vertex. For example, if you choose x = -1 and x = 1, the corresponding y-values can be calculated as follows:
For x = -1: y = (-1)^2 - 2 = -1 - 2 = -3. Therefore, one point is (-1, -3).
For x = 1: y = (1)^2 - 2 = 1 - 2 = -1. Therefore, another point is (1, -1).
3. Plot the vertex and the two additional points on the coordinate plane.
4. Connect the points with a smooth curve to sketch the graph of the quadratic function y = x^2 - 2.
I hope this helps! Let me know if you have any further questions.
To sketch the graph of the quadratic function y=x^2-2, we can start by identifying the vertex.
The vertex of a quadratic function in the form y=a(x-h)^2+k is located at the point (h, k).
Comparing the given function y=x^2-2 with the standard form, we can see that a=1, h=0, and k=-2. Therefore, the vertex is located at (0, -2).
Next, we can find two additional points on the graph by substituting values for x and solving for y.
Let's choose x=-1:
y=(-1)^2-2 = 1-2 = -1. So the point (-1, -1) is on the graph.
Let's choose x=1:
y=(1)^2-2 = 1-2 = -1. So the point (1, -1) is on the graph.
Now, we can sketch the graph with the given information. The vertex is at (0, -2), and the additional points are (-1, -1) and (1, -1).
The graph should resemble an upward-opening parabola with the vertex at the origin and the two additional points on the y-axis.
To sketch the graph of the quadratic function y = x^2 - 2 and find the vertex and two additional points, we can follow these steps:
1. Start by plotting the vertex. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a is the coefficient of the x^2 term, and b is the coefficient of the x term. In this case, a = 1 and b = 0, so x = -0 / (2 * 1) = 0. The y-coordinate of the vertex can be found by substituting the x-coordinate into the equation, so y = (0)^2 - 2 = -2. Therefore, the vertex is (0, -2).
2. To find two additional points, we can choose different x-values and substitute them into the equation to find the corresponding y-values. For simplicity, let's choose x = -2 and x = 2.
- For x = -2, y = (-2)^2 - 2 = 4 - 2 = 2. So one additional point is (-2, 2).
- For x = 2, y = (2)^2 - 2 = 4 - 2 = 2. So the other additional point is (2, 2).
3. Plot the vertex at (0, -2) and the two additional points at (-2, 2) and (2, 2) on the coordinate plane.
4. To sketch the graph, connect the points with a smooth curve. Since the equation is a quadratic function, the graph will be a symmetric U-shape, opening upward.
The graph should resemble a U-shaped curve with its vertex at (0, -2), intersecting the x-axis at (-2, 0) and (2, 0), and the y-axis at (0, -2).