Question: Of the quadratic equations are given below, which equation produces a parabola with vertex $(-1, 5)$?
y = -2(x - 1)^2 - 5
y = -2(x + 1)^2 - 5
y = -2(x + 1)^2 + 5
y = -2(x - 1)^2 + 5
1. **State the problem:** We need to find which quadratic equation produces a parabola with vertex $(-1, 5)$.
2. **Recall the vertex form of a quadratic equation:**
$$y = a(x - h)^2 + k$$
where $(h, k)$ is the vertex of the parabola.
3. **Identify the vertex in each equation:**
- For $y = -2(x - 1)^2 - 5$, vertex is $(1, -5)$.
- For $y = -2(x + 1)^2 - 5$, vertex is $(-1, -5)$.
- For $y = -2(x + 1)^2 + 5$, vertex is $(-1, 5)$.
- For $y = -2(x - 1)^2 + 5$, vertex is $(1, 5)$.
4. **Compare vertices to the target vertex $(-1, 5)$:**
Only the equation $y = -2(x + 1)^2 + 5$ has the vertex $(-1, 5)$.
5. **Conclusion:** The equation that produces a parabola with vertex $(-1, 5)$ is:
$$y = -2(x + 1)^2 + 5$$