1. **State the problem:** We need to find the equation of a parabola with x-intercepts at $x = -3$ and $x = 5$, and a minimum value (vertex) at $y = -4$. 2. **Recall the form of a parabola:** A parabola with roots $r_1$ and $r_2$ can be written as $$y = a(x - r_1)(x - r_2)$$ where $a$ determines the opening direction and width. 3. **Substitute the roots:** Here, $r_1 = -3$ and $r_2 = 5$, so $$y = a(x + 3)(x - 5)$$ 4. **Find the vertex:** The vertex $x$-coordinate is the midpoint of the roots, $$x_v = \frac{-3 + 5}{2} = 1$$ 5. **Use the vertex to find $a$:** The vertex is at $(1, -4)$, so substitute $x=1$, $y=-4$: $$-4 = a(1 + 3)(1 - 5) = a(4)(-4) = -16a$$ 6. **Solve for $a$:** $$a = \frac{-4}{-16} = \frac{1}{4}$$ 7. **Write the final equation:** $$y = \frac{1}{4}(x + 3)(x - 5)$$ 8. **Expand for clarity:** $$y = \frac{1}{4}(x^2 - 5x + 3x - 15) = \frac{1}{4}(x^2 - 2x - 15) = \frac{1}{4}x^2 - \frac{1}{2}x - \frac{15}{4}$$ This parabola opens upwards (since $a > 0$), has roots at $x = -3$ and $x = 5$, and a minimum value of $-4$ at $x=1$.