1. **Problem a:** Write the equation of a parabola in the form $y = ax^2 + bx + c$ passing through points $(0,0)$, $(5,5)$, and $(6,0)$, opening downward. 2. Since the parabola passes through $(0,0)$, substituting $x=0$, $y=0$ gives: $$0 = a\cdot0^2 + b\cdot0 + c \implies c = 0$$ 3. Using points $(5,5)$ and $(6,0)$, substitute into $y = ax^2 + bx + 0$: $$5 = 25a + 5b$$ $$0 = 36a + 6b$$ 4. Solve the system: From second equation: $$6b = -36a \implies b = -6a$$ Substitute into first: $$5 = 25a + 5(-6a) = 25a - 30a = -5a \implies a = -1$$ Then: $$b = -6(-1) = 6$$ 5. Equation for part a: $$y = -1x^2 + 6x + 0 = -x^2 + 6x$$ 6. **Problem b:** Write the equation of a parabola in the form $y = ax^2 + bx + c$ passing through points $(-4,0)$, $(6,0)$, and $(1,-5)$, opening upward. 7. Since $(-4,0)$ and $(6,0)$ are roots, the parabola can be written as: $$y = a(x + 4)(x - 6) = a(x^2 - 2x - 24) = a x^2 - 2a x - 24a$$ 8. Use point $(1,-5)$ to find $a$: $$-5 = a(1)^2 - 2a(1) - 24a = a - 2a - 24a = -25a \implies a = \frac{-5}{-25} = \frac{1}{5}$$ 9. Final equation for part b: $$y = \frac{1}{5}x^2 - \frac{2}{5}x - \frac{24}{5}$$