Question: Write an equation for the graph shown.
$(x - 1)^2 - 2$
$(0, 0)$
$(1, -2)$
Graph shape: a downward-opening parabola with vertex at $(0, 0)$ and a labeled point at $(1, -2)$; position_hint = center
1. **State the problem:**
We need to find the equation of a parabola that opens downward, has a vertex at $(0,0)$, and passes through the point $(1,-2)$.
2. **Recall the vertex form of a parabola:**
The vertex form is $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex.
3. **Plug in the vertex:**
Since the vertex is at $(0,0)$, the equation simplifies to $$y = a(x - 0)^2 + 0 = a x^2$$.
4. **Use the point $(1,-2)$ to find $a$:**
Substitute $x=1$ and $y=-2$:
$$-2 = a(1)^2 = a$$
So, $$a = -2$$.
5. **Write the final equation:**
$$y = -2x^2$$.
6. **Check the shape:**
Since $a = -2 < 0$, the parabola opens downward, matching the problem description.
**Final answer:**
$$y = -2x^2$$