Question

已知抛物线y=2x²+bx+c在x轴上截得的线段长为4，且顶点坐标是在直线y=-2x-1上，求该抛物线表达式

Answer 1

顶点坐标(t,-2t-1) => y=2(x-t)^2 -2t-1 =2x^2-4tx +2t^2-2t-1 => b=-4t ; c=2t^2-2t-1
=> 设交X轴於(α,0);(β,0) ;α<β=> α+ β=2t ; αβ=t^2-t-0.5 ; β-α=4
=> (β-α)^2 = (α+ β)^2- 4αβ => 16=4t^2 - 4t^2+4t +2 =>t=3.5
=> y=2(x-3.5)^2 -8 =2x^2 -14x +16.5......ans

Answer 2

顶点坐标(t,-2t-1) => y=2(x-t)^2 -2t-1 =2x^2-4tx +2t^2-2t-1 => b=-4t ; c=2t^2-2t-1
=> 设交X轴於(α,0);(β,0) ;α<β=> α+ β=2t ; αβ=t^2-t-0.5 ; β-α=4
=> (β-α)^2 = (α+ β)^2- 4αβ => 16=4t^2 - 4t^2+4t +2 =>t=3.5
=> y=2(x-3.5)^2 -8 =2x^2 -14x +16.5......ans 望楼主采纳 我会很开心的哟