1. The problem asks us to find which quadratic equation has a vertex at $(-3,4)$. 2. Recall the vertex form of a quadratic equation is $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. 3. Here, the vertex is $(-3,4)$, so $h = -3$ and $k = 4$. 4. Substitute $h$ and $k$ into the vertex form: $$y = a(x - (-3))^2 + 4 = a(x+3)^2 + 4$$ 5. Now check each option: - $y = 2(x+3)^2 + 4$ matches the form with $a=2$, vertex $(-3,4)$. - $y = (x+3)^2 - 4$ has vertex $(-3,-4)$, so no. - $y = (x-3)^2 + 4$ has vertex $(3,4)$, so no. - $y = 2(x-3)^2 + 4$ has vertex $(3,4)$, so no. 6. Therefore, the correct equation is $$y = 2(x+3)^2 + 4$$ which has vertex $(-3,4)$.