If you change your second equation to be
y2 = -y+c
Then
1/exp(x) = -1/exp(x) + c
thus
c = 2/exp(x)
so just choose what the x-value of the intersection should be, plug it in that formula and that's your c
. So for a crossing at x==4
, c=2/exp(x
) which is 0.036631
so
y2 = -y + 0.036631
So (assuming you can alter c
) all you need to choose the x-vlaue of the point of intersection is to put that desired x-value into the equation
c = 2/exp(x) %//note this is a scalar value of x, not the whole vector