Question: User: y 8 (-2, 6) (-1, 3) 4 x Graph: two downward-opening parabolas/curves are shown, with one peak near (-2, 6) and a marked point near (-1, 3); position_hint: center Can you make an equation in vertex form

1. **State the problem:** We are given a parabola with vertex at $(-2, 6)$ and a point on the parabola at $(-1, 3)$. We want to find the equation of the parabola in vertex form. 2. **Recall the vertex form of a parabola:** $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. **Substitute the vertex:** Here, $h = -2$ and $k = 6$, so $$y = a(x + 2)^2 + 6$$ 4. **Use the given point $(-1, 3)$ to find $a$:** Substitute $x = -1$ and $y = 3$: $$3 = a(-1 + 2)^2 + 6$$ $$3 = a(1)^2 + 6$$ $$3 = a + 6$$ 5. **Solve for $a$:** $$a = 3 - 6 = -3$$ 6. **Write the final equation:** $$y = -3(x + 2)^2 + 6$$ This is the equation of the parabola in vertex form that passes through the given points.