nope, x is the position
How would you do this question from the same exam? (if my memory serves correctly)
A function y=f(x) has a tangent at point A(x, y) which crosses the x-axis at coordinate (x-1/2, 0)
The function has a y-intercept at coordinate (0, 1/e)
Find the equation of the curve.
Chavi can you confirm this question please 
Hey , I did this question - it is tricky, but once you understand what you're asking for it's pretty simple.
OK, so you have points A(x,y) and (x-1/2, 0) - for the
tangent to the curve - (don't confuse the y incpt of the function with the tangent to the curve)
then, find the gradient of the tangent by rise/run: dy/dx = y/(x-(x/1/2)) = 2y and thus, dx/dy = 1/2y
From this, x = (ln(y))/2 + c and at the y intercept, x=0 and y = 1/e
so you have 0 = 1/2(-1) + c, and c = 1/2
So, y = e^(2x-1)
Hint: Draw a diagram.