The function provided is:

$g(x) = (x + 2)^2$

This is a quadratic function and can be rewritten in the form of $g(x) = (x + h)^2 + k$ where $h = -2$ and $k = 0$. The graph of this function is a parabola that opens upwards.

Key points for graphing:

1. Vertex: The vertex of the parabola is at $(-2, 0)$.
2. Axis of Symmetry: The line of symmetry for the parabola is $x = -2$.
3. Y-intercept: When $x = 0$, $g(0) = (0 + 2)^2 = 4$. Therefore, the y-intercept is $(0, 4)$.
4. Shape: Since the coefficient of the squared term is positive, the parabola opens upwards.

Plotting the graph:

• Start at the vertex $(-2, 0)$.
• Draw the axis of symmetry $x = -2$.
• Identify another point like $(0, 4)$ and plot it symmetrically on the other side of the axis at $(-4, 4)$.

Would you like me to generate the actual graph for you?

Additional Questions

1. What is the domain and range of this function?
2. How does the function $g(x) = (x + 2)^2$ differ from $f(x) = x^2$?
3. What happens to the graph if we change the function to $g(x) = (x + 2)^2 + 3$?
4. Can you identify any transformations applied to the parent function $f(x) = x^2$?
5. How would you describe the concavity of the graph?

Tip: For any quadratic function in the form $y = (x + h)^2 + k$, the vertex will always be at $(-h, k)$.