Question 249751
2x^2 + 8x + 1 is at what point 
{{{(-b/2a)}}}  [use the formula for finding the x-value of the vertex]
{{{(-(8)/2(2))}}}  [plug-in the appropriate values from the equation]
x=-2
.
y=2x^2 + 8x + 1
y=2(-2)^2 + 8(-2) + 1  [plug-in the x-value (-2) and solve for the y-value]
y=8-16+1
y=-7
.
So, the vertex point is (-2, -7)