1. **State the problem:** Find the equation of a parabola whose axis is parallel to the x-axis and passes through points (3, 1), (0, 0), and (8, -4). 2. **Recall the form of the parabola:** Since the axis is parallel to the x-axis, the parabola can be expressed as $$ (y - k)^2 = 4p(x - h) $$ where $(h, k)$ is the vertex. 3. **Use the general form:** We can rewrite the equation as $$ y^2 + Ay + Bx + C = 0 $$ or better, express $x$ in terms of $y$ as $$ x = a y^2 + b y + c $$ because the axis is horizontal. 4. **Set up equations using the points:** Substitute each point $(x, y)$ into $$ x = a y^2 + b y + c $$ For (3, 1): $$ 3 = a(1)^2 + b(1) + c = a + b + c $$ For (0, 0): $$ 0 = a(0)^2 + b(0) + c = c $$ For (8, -4): $$ 8 = a(-4)^2 + b(-4) + c = 16a - 4b + c $$ 5. **From the second equation:** $$ c = 0 $$ 6. **Rewrite the other equations:** From (3,1): $$ 3 = a + b + 0 = a + b $$ From (8,-4): $$ 8 = 16a - 4b + 0 = 16a - 4b $$ 7. **Solve the system:** From $$ a + b = 3 $$, we get $$ b = 3 - a $$ Substitute into the second: $$ 8 = 16a - 4(3 - a) = 16a - 12 + 4a = 20a - 12 $$ Add 12 to both sides: $$ 8 + 12 = 20a $$ $$ 20 = 20a $$ Divide both sides by 20: $$ \cancel{20} = \cancel{20}a $$ $$ 1 = a $$ 8. **Find b:** $$ b = 3 - a = 3 - 1 = 2 $$ 9. **Write the equation:** $$ x = y^2 + 2y $$ or equivalently, $$ x = y^2 + 2y + 0 $$ 10. **Final answer:** The equation of the parabola is $$ \boxed{x = y^2 + 2y} $$