To graph the quadratic function f(x) = -\frac{1}{2}(x + 1)^2 + 2, we start by identifying key features of the parabola. The function is in vertex form, which is expressed as f(x) = a(x - h)^2 + k, where (h, k) is the vertex. In this case, we can rewrite the function to find the vertex:

Here, x + 1 can be interpreted as x - (-1), indicating that h = -1 and k = 2. Thus, the vertex is at the point (-1, 2). Since the coefficient a = -\frac{1}{2} is negative, the parabola opens downward, confirming that the vertex is a maximum point.

The axis of symmetry for the parabola is given by the line x = h, which in this case is x = -1.

Next, we find the x-intercepts by setting f(x) = 0:

- \frac{1}{2}(x + 1)^2 + 2 = 0

Rearranging gives:

- \frac{1}{2}(x + 1)^2 = -2

Multiplying both sides by -2 results in:

(x + 1)^2 = 4

Taking the square root of both sides yields:

x + 1 = \pm 2

Solving for x gives two x-intercepts: x = 1 and x = -3.

To find the y-intercept, we evaluate f(0):

f(0) = -\frac{1}{2}(0 + 1)^2 + 2 = -\frac{1}{2}(1) + 2 = -\frac{1}{2} + 2 = \frac{3}{2}

Thus, the y-intercept is (0, \frac{3}{2}).

Now we can summarize the key points for graphing:

• Vertex: (-1, 2) (maximum point)
• Axis of symmetry: x = -1
• X-intercepts: (1, 0) and (-3, 0)
• Y-intercept: (0, \frac{3}{2})

When graphing, plot the vertex, axis of symmetry, x-intercepts, and y-intercept. Connect these points with a smooth curve to form the parabola, which opens downward.

For the domain of the quadratic function, it is always (-\infty, \infty), indicating that all real numbers are included. The range, however, is limited by the maximum point at the vertex, so it is (-\infty, 2], where 2 is included in the range.

Lastly, the intervals of increase and decrease can be determined from the graph. The function is increasing on the interval (-\infty, -1) and decreasing on the interval (-1, \infty). The vertex at x = -1 is not included in these intervals as it represents the transition point.