To graph the parabola $y = -x^2$, we need to:

1. Identify the vertex.
2. Calculate two points to the left and two points to the right of the vertex.
3. Plot these points on the graph.

Step-by-Step Solution:

1. Vertex: The equation $y = -x^2$ is in the form $y = ax^2 + bx + c$, where $a = -1$, $b = 0$, and $c = 0$.

   • The vertex of a parabola $y = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.
   • Since $b = 0$, the vertex is at $x = 0$.
   • Substituting $x = 0$ into $y = -x^2$: $y = -(0)^2 = 0$.
   • Vertex: $(0, 0)$.

2. Points to the Left of the Vertex:

   • Let $x = -1$: $y = -(-1)^2 = -1$. So, the point is $(-1, -1)$.
   • Let $x = -2$: $y = -(-2)^2 = -4$. So, the point is $(-2, -4)$.

3. Points to the Right of the Vertex:

   • Let $x = 1$: $y = -(1)^2 = -1$. So, the point is $(1, -1)$.
   • Let $x = 2$: $y = -(2)^2 = -4$. So, the point is $(2, -4)$.

4. Points for Plotting:

   • Vertex: $(0, 0)$
   • Points to the left: $(-1, -1)$ and $(-2, -4)$
   • Points to the right: $(1, -1)$ and $(2, -4)$

These points can be plotted on a coordinate plane to graph the parabola. The graph will open downwards because the coefficient of $x^2$ is negative.

Would you like more details or have any questions?

Additional Questions:

1. How do you determine the direction of a parabola's opening based on its equation?
2. What effect does the coefficient $a$ have on the width of a parabola?
3. How can you determine the axis of symmetry for a given parabola?
4. What are some characteristics of a parabola with a vertex at the origin?
5. How would the graph change if the equation were $y = x^2$ instead of $y = -x^2$?

Tip:

The vertex form of a parabola, $y = a(x-h)^2 + k$, makes it easier to identify the vertex and direction of opening directly.