Hi

y= 2x^2-4x -16

Note: the vertex form of a parabola,  where(h,k) is the vertex

y= 2x^2-4x -16   |Completing the square

y = 2[(x-1)^2 - 1] -16

y = 2(x-1)^2 - 2 -16

y = 2(x-1)^2 -18  Vertex(1,-18) a = 2 > 0 parabola opens upward

Line  of symmetry is x = 1

x-intercepts when y = 0 are also the solutions for the quadratic equation

 2(x-1)^2 = 18

    x-1 = ± 

      x = 1 ± 3     x = 4 or x = -2



 

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