Question 384116

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Hi,         
1) Let x and (x+ 2yd) represent the width and length respectively
  x(x+2) = 80 yd^2
  x^2 + 2x  - 80 = 0
  factoring
(x+10)(x-8) = 0 Note:SUM of the inner product(10*x) and the outer product(-8x) = 2x
(x+10)=0
  x = -10  Tossing out negative solution
(x-8) = 0
  x = 8yd , the width, the length 10yd     (8yd + 2yd)

2)find the coordinates of the vertex:
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y=2x^2+4x-1  Completing the square
 y = 2[(x +1)^2 -1] - 1
 y = 2(x+1)^2 - 2 - 1
 y = 2(x+1)^2 - 3
  Vertex Pt(-1,-3)
{{{graph( 300, 300,-6,6,-6,6,2x^2+4x-1) }}}