Hi,

solving algebraically:  Parabola with V(8,3) and passing thru (5,0) 

 y = a(x -8)^2 + 3

 0 = a(5-8)^2 + 3

 -3/9 = a

 y = (-1/3)(x-8)^2 + 3

 0 = (-1/3)(x-8)^2 + 3

      -3 = (-1/3)(x-8)^2

       (x-8)^2 = 9

        (x-8) = ± 3

          x = 5 and x = 11 

    (11,0) are the coordinates of the other x- intercept.   

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