SOLUTION: Find the vertex of the parabola.
y = -4x2 - 16x - 11
Possible answers
(5, 2)
(2, 5)
(-2, 5)
(5, -2)
any help would be great
Algebra.Com
Question 374342: Find the vertex of the parabola.
y = -4x2 - 16x - 11
Possible answers
(5, 2)
(2, 5)
(-2, 5)
(5, -2)
any help would be great
Found 2 solutions by Jk22, ewatrrr:
Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!
y = -4x2 - 16x - 11
derivative : y' = -8x -16 (at the vertex, extremum : y'=0 => x = -2)
hence the solution is (-2,5) (verif : y(-2)=-16+32-11=16-11=5)
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
Hi,
the vertex form of a parabola, where(h,k) is the vertex
y = -4x^2 - 16x - 11
y = -4[x^2 +4x + (11/4)]
completing the square
y = -4[(x+2)^2 - 4 +(11/4)}
y = -4(x+2)^2 -4[(-16/4) + (11/4)]
y = -4(x+2)^2 -4[(-5/4)]
y = -4(x+2)^2 + 5
(-2,5) is the vertex
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