Question 906321
{{{ y = 2x^2 + x  -  16 }}}
y = 2(x+1/4)^2 - 1/8 - 16
y = 2(x+.25)^2 - 129/8
0 = 2(x+.25)^2 - 129/8
129/16 = (x+.25)^2
-.25 ± sqrt(129/16) = x
x is 2.5895,  -3.0895

The coordinates of the Vertex are ( .25, -129/8). 
 This parabola has (2 ) x-intercepts. 
 These intercepts occur at (exact values): 
 x = (-.25 - √129/4) Smaller value
 x = (-.25 +√129/4  ) Bigger value
 This parabola attains a min value of -129/8