To sketch the graph of the quadratic function $g(x) = -x^2 + 4$ and compare it with $y = x^2$, we can analyze their features:

1. Shape:

   • $y = x^2$ is a parabola that opens upwards.
   • $g(x) = -x^2 + 4$ is a parabola that opens downwards due to the negative coefficient of $x^2$.

2. Vertex:

   • The vertex of $y = x^2$ is at (0, 0).
   • The vertex of $g(x)$ is at (0, 4).

3. Intercepts:

   • For $y = x^2$:
     • $y$-intercept: (0, 0)
     • $x$-intercepts: (0, 0)
   • For $g(x) = -x^2 + 4$:
     • $y$-intercept: (0, 4)
     • $x$-intercepts: Solve $-x^2 + 4 = 0$ → $x^2 = 4$ → $x = ±2$ (intercepts at (2, 0) and (-2, 0)).

4. Direction and Width:

   • The graph of $g(x)$ is wider than that of $y = x^2$ due to the coefficient of $x^2$ being less than 1 in magnitude.

Sketch

1. Graph of $y = x^2$: A standard upward-opening parabola with vertex at (0, 0).
2. Graph of $g(x) = -x^2 + 4$: A downward-opening parabola with vertex at (0, 4) and intercepts at (2, 0) and (-2, 0).

Comparison

• $g(x) = -x^2 + 4$ is a reflection of $y = x^2$ across the horizontal line $y = 2$ (the line through its vertex).

If you need a visual graph or further analysis, please let me know!

Questions for Further Exploration

1. How do transformations affect the graph of a quadratic function?
2. What are the implications of the vertex form of a quadratic function?
3. How can we determine the maximum or minimum value of a quadratic function?
4. What role do the coefficients play in determining the shape of a parabola?
5. Can you find the axis of symmetry for both quadratic functions?

Tip: Always identify the vertex and intercepts when sketching quadratic functions, as they provide key points for the graph.