Arectangular field has an area of 1,764 m^2. the width of the field is 13 m more than the length. what is the perimeter of the field?

Let the length of the field be x, then width = x + 13
x(x + 13) = 1764
x^2 + 13x - 1764 = 0
(x - 36)(x + 49) = 0
x = 36

Therefore, length is 36 m and width = 36 + 13 = 49
Perimeter = 2(length + width) = 2(36 + 49) = 2(85) = 170 m

The correct graph is the second one, with the points being at (-6,0), (2,0) and reaching down to (-2,-8)