Binet's (or Euler's or de Moivre's) formula for the nth Fibonacci number is (((sqrt(5)+1)/2)^n-(-(sqrt(5)+1)/2)^-n)/sqrt(5). Additions are O(n). Multiplications vary depending on the algorithm, the best known (Harvey-Hoeven) is O(n log(n)) but comes with some nasty constants that probably mean classic Karatsuba or Toom-Cook multiplication (O(n^1.X) for some X will be faster for the range to be graphed, but I'll stick with O(n log(n)) for this. The exponentials will be O(n log(n)^2) in the best known case IIRC. That factor dominates, so I'd say it's probably O(n log(n)^2) but that a graph probably won't start cooperating quickly due to the algorithms picked having factors that are ignored by big-O notation for the small inputs given.

However you do it it probably can't be linear, since multiplication is probably at best O(n log(n)), though that lower bound hasn't been proven. A naive recursive calculation will be even worse since that has exponential complexity.