Graph the parabola. \[ y=\frac{1}{2} x^{2}-6 \] Plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. Then click on the graph-a-function button.
Solution
Step 1 ：The given equation is a quadratic equation in the form of \(y = ax^2 + bx + c\), where \(a = \frac{1}{2}\), \(b = 0\), and \(c = -6\).
Step 2 ：The vertex of the parabola is given by the point \(-\frac{b}{2a}, f(-\frac{b}{2a})\). In this case, since \(b = 0\), the x-coordinate of the vertex is 0. Substituting \(x = 0\) in the equation gives the y-coordinate of the vertex as -6. So, the vertex of the parabola is \((0, -6)\).
Step 3 ：To find the other four points, we can choose two x-values to the left and right of the vertex and substitute them into the equation to get the corresponding y-values. Let's choose \(x = -1\) and \(x = -2\) for the points to the left of the vertex, and \(x = 1\) and \(x = 2\) for the points to the right of the vertex.
Step 4 ：Substituting these x-values into the equation gives the points \((-1, -5.5)\), \((-2, -4)\), \((1, -5.5)\), and \((2, -4)\).
Step 5 ：\(\boxed{\text{Final Answer: The five points on the parabola are the vertex (0, -6), two points to the left of the vertex (-1, -5.5) and (-2, -4), and two points to the right of the vertex (1, -5.5) and (2, -4). The graph of the function } y=\frac{1}{2} x^{2}-6 \text{ is a parabola opening upwards with its vertex at (0, -6).}}\)
