1. **State the problem:** We need to graph a parabola with x-intercepts at $x = -3$ and $x = 5$, and a minimum value of $y = -4$. 2. **Formula and important rules:** 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 parabola's opening and vertical stretch. 3. **Apply the roots:** Here, $r_1 = -3$ and $r_2 = 5$, so $$y = a(x + 3)(x - 5)$$ 4. **Find the vertex:** The vertex's x-coordinate is the midpoint of the roots: $$x_v = \frac{-3 + 5}{2} = 1$$ 5. **Use the minimum value:** The parabola's minimum value is $y = -4$ at $x = 1$, so substitute into the equation: $$-4 = a(1 + 3)(1 - 5) = a(4)(-4) = -16a$$ 6. **Solve for $a$:** $$-4 = -16a \implies a = \frac{-4}{-16} = \frac{1}{4}$$ 7. **Final equation:** $$y = \frac{1}{4}(x + 3)(x - 5)$$ 8. **Expand for clarity:** $$y = \frac{1}{4}(x^2 - 2x - 15) = \frac{1}{4}x^2 - \frac{1}{2}x - \frac{15}{4}$$ This is the parabola with the given properties.