Verified answer

The easiest way is let x = t and y = f(t)

So for your example, x = t, y = e^(2t-1)

some suggestions... in case you finished the sq., you get (x-2)=(y-3)^2 which shows its a parabola (not a hyperbola) dealing with appropriate with its vertex at (2,3) as you propose. In parametric style you ought to write y = t +3 and x = t^2 + 2 the place t is the parameter.