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A parabola's vertex is a crucial point as it represents the highest or lowest point on the graph, depending on the direction the parabola opens. For the quadratic function given in vertex form \( f(x) = a(x-h)^2 + k \), the vertex is the point (h, k). In the provided exercise, the function is \( f(x) = -\frac{1}{2}(x+1)^2 + 2 \). By comparing this with the vertex form, we see that h = -1 and k = 2. Therefore, the vertex of this parabola is at (-1, 2). The vertex tells us where the peak or trough is located. For this function, because the coefficient of \( (x+1)^2 \) is negative (-\frac{1}{2}), the vertex represents the maximum point on the parabola. This means the parabola opens downwards from this vertex.

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves, passing through the vertex. This line helps in plotting the rest of the parabola as any point on one side of the axis will have a corresponding point on the other side. For a quadratic function in vertex form \( f(x) = a(x-h)^2 + k \), the axis of symmetry is given by the equation x = h. For the exercise function \( f(x) = -\frac{1}{2}(x+1)^2 + 2 \), the vertex is (-1, 2), so the axis of symmetry is x = -1. This means every point on the parabola has a symmetric counterpart across the line x = -1.

The domain and range of a quadratic function give us valuable information about the scope of x and y values, respectively.
Domain:
The domain of any quadratic function is all real numbers, symbolized as \( (-\infty, \infty) \). This is because there's no restriction on the x-values; we can input any real number for x and get an output.
Range:
The range varies depending on whether the parabola opens upwards or downwards. Given that the parabola in the exercise opens downwards (since the coefficient of \( (x+1)^2 \) is negative), the vertex represents the highest point. The y-coordinate of the vertex determines the maximum value of the function. In this case, the vertex is (-1, 2), so the maximum y-value is 2. Thus, the range includes all real numbers less than or equal to 2, written as \( (-\infty, 2] \).

Graphing a quadratic function involves a few steps to ensure accuracy and a clear representation. Below are the general steps to follow:

• Identify the vertex and plot it on the coordinate plane.
• Draw the axis of symmetry as a dashed vertical line through the vertex.
• Plot additional points on either side of the axis of symmetry. Use symmetric points to make the process easier.
• Connect these points with a smooth curve to complete the parabola.
• Make sure to indicate the direction the parabola opens (upwards or downwards).

For the given function \( f(x) = -\frac{1}{2}(x+1)^2 + 2 \), start by plotting the vertex at (-1, 2) and drawing the axis of symmetry at x = -1. Calculate additional points by plugging values into the quadratic function and plotting these points symmetrically around the axis. This will give a clearer understanding of the parabola's shape and trajectory.