SOLUTION: the vertex of 2x^2 + 8x + 1 is at what point

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Question 249751: the vertex of
2x^2 + 8x + 1 is at what point
Found 2 solutions by jsmallt9, tutorcecilia:
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
I assume the equation is
y+=+2x%5E2+%2B+8x+%2B+1 When the equation of a parabola is in the form y+=+ax%5E2+%2B+bx+%2Bc, then the x coordinate of the vertex is %28-b%29%2F2a. In your equation, the "a" is 2 and the "b" is 8, so the x coordinate of the vertex is
%28-%288%29%29%2F2%282%29+=+%28-8%29%2F4+=+-2

Now that we know the x coordinate of the vertex we can use the equation to figure out its y coordinate:
y+=+2%28-2%29%5E2+%2B+8%28-2%29+%2B+1
y+=+2%284%29+%2B+8%28-2%29+%2B+1
y+=+8+%2B+%28-16%29+%2B+1
y+=+-7

So the coordinates of the vertex are: (2, -7)

Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
2x^2 + 8x + 1 is at what point
%28-b%2F2a%29 [use the formula for finding the x-value of the vertex]
%28-%288%29%2F2%282%29%29 [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)