The given equation is $y = -x^2$, which represents a downward-facing parabola because of the negative sign in front of $x^2$.

Steps to plot the points:

1. Vertex: The vertex of this parabola is at $(0, 0)$, as the equation is in the standard form $y = ax^2 + bx + c$, where $a = -1$, $b = 0$, and $c = 0$.

2. Choose points to the left and right of the vertex: To the left of the vertex (where $x = -1$ and $x = -2$):

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

   To the right of the vertex (where $x = 1$ and $x = 2$):

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

Points to plot:

1. Vertex: $(0, 0)$
2. To the left: $(-1, -1)$ and $(-2, -4)$
3. To the right: $(1, -1)$ and $(2, -4)$

If you were graphing, these points form a symmetric shape around the vertex $(0, 0)$, characteristic of a parabola.

Would you like more details or any further questions?

Here are 5 related questions for practice:

1. What is the general form of a quadratic equation?
2. How does the sign of the coefficient $a$ in $y = ax^2$ affect the orientation of the parabola?
3. How can you find the vertex of a parabola if it is not at the origin?
4. How do you determine the axis of symmetry for a parabola?
5. How do you calculate the y-intercept and x-intercepts of a parabola?

Tip: The axis of symmetry of a parabola $y = ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$.