The function given is:

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

This is a quadratic function in vertex form:

$f(x) = a(x - h)^2 + k$

where $a = -\frac{1}{2}$, $h = -3$, and $k = 0$. The vertex of this parabola is at $(h, k) = (-3, 0)$.

Key Features:

1. The negative sign in front of the $a$ value indicates the parabola opens downward.
2. The vertex is at $(-3, 0)$.
3. The factor of $-\frac{1}{2}$ affects the "width" of the parabola, making it wider than a standard parabola.

Steps to Plot the Graph:

1. Vertex: Plot the vertex at $(-3, 0)$.

2. Another Point: To find another point, let's choose $x = -2$ and substitute into the equation:

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

   So, another point is $(-2, -\frac{1}{2})$.

Graph of the function:

• Plot the vertex at $(-3, 0)$.
• Plot the point $(-2, -\frac{1}{2})$.
• Reflect the point $(-2, -\frac{1}{2})$ across the axis of symmetry (which is $x = -3$) to find another point, which would be $(-4, -\frac{1}{2})$.

Symmetry:

Since this is a parabola, it is symmetric about the vertical line $x = -3$.

Would you like me to graph this function for you? If so, I can create and upload the plot.

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Follow-up questions:

1. How does the value of $a$ in the equation affect the width of the parabola?
2. What happens to the graph if the vertex is moved to a different point?
3. How does the negative sign in front of $a$ impact the direction of the parabola?
4. What is the axis of symmetry for this quadratic function?
5. How do you find the y-intercept of the parabola?

Tip: When graphing quadratic functions, always start by identifying the vertex and the direction in which the parabola opens (up or down).