Answer:
[tex]y=2x^2-4x-8[/tex]
Step-by-step explanation:
If the parabola has x-intecepts at [tex](1+\sqrt{5},0)[/tex] and [tex](1-\sqrt{5},0),[/tex] then its equation is
[tex]y=a(x-(1+\sqrt{5}))(x-(1-\sqrt{5}))\\ \\y=a(x-1-\sqrt{5})(x-1+\sqrt{5})[/tex]
This parabola passes through the point (4,8), then its coordinates satisfy the equation:
[tex]8=a(4-1-\sqrt{5})(4-1+\sqrt{5})\\ \\8=a(3-\sqrt{5})(3+\sqrt{5})\\ \\8=a(3^2-\sqrt{5}^2)\\ \\8=a(9-5)\\ \\8=4a\\ \\a=2[/tex]
Hence, parabola's equation is
[tex]y=2(x-1-\sqrt{5})(x-1+\sqrt{5})\\ \\y=2((x-1)^2-\sqrt{5}^2)\\ \\y=2((x-1)^2-5)\\ \\y=2(x^2-2x+1-5)\\ \\y=2(x^2-2x-4)\\ \\y=2x^2-4x-8[/tex]
