1. **State the problem:** We are given a quadratic function $y = ax^2 + bx + c$ with a minimum point at $(5, -3)$ and it passes through the point $(4, 0)$. We need to find the values of $a$, $b$, and $c$. 2. **Recall the vertex form and properties:** The vertex of a parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Since the vertex is a minimum, $a > 0$. 3. **Use the vertex coordinates:** The vertex is at $(5, -3)$, so: $$5 = -\frac{b}{2a} \implies b = -10a$$ 4. **Use the vertex value for $y$:** At $x=5$, $y=-3$: $$-3 = a(5)^2 + b(5) + c = 25a + 5b + c$$ Substitute $b = -10a$: $$-3 = 25a + 5(-10a) + c = 25a - 50a + c = -25a + c$$ So, $$c = -3 + 25a$$ 5. **Use the point $(4,0)$:** The parabola passes through $(4,0)$: $$0 = a(4)^2 + b(4) + c = 16a + 4b + c$$ Substitute $b = -10a$ and $c = -3 + 25a$: $$0 = 16a + 4(-10a) + (-3 + 25a) = 16a - 40a - 3 + 25a = (16a - 40a + 25a) - 3 = 1a - 3$$ So, $$a - 3 = 0 \implies a = 3$$ 6. **Find $b$ and $c$:** $$b = -10a = -10(3) = -30$$ $$c = -3 + 25a = -3 + 25(3) = -3 + 75 = 72$$ **Final answer:** $$a = 3, \quad b = -30, \quad c = 72$$