You could use
In your image, you'd roughly have something like $z_1=1/2$, $z_2=1$, $z_3=2$. Here's a plot.
[Edit in response to the comment:]
Yes, you can make the curves more like quarter-circles by using
instead. Here's a plot for that. Then the curve for $z=1/2$ is exactly a quarter-circle (it's distorted by different scaling on the axes on the W|A plot), but the curve for $z=2$ isn't exactly the corresponding reflected quarter-circle. I think to get that to be the case with a single formula while fulfilling the other constraints would be tricky, but you can always use the above expression for $z\le1$ and then for $z\gt1$ just use the one for $1/z$ reflected; that would be
Here's a plot for that.
I played around with this some more. I didn't like the fact that $x$ and $y$ were playing different roles whereas the quarter-circles and the straight line seemed to indicate that the curves should ideally be symmetric with respect to interchange of $x$ and $y$. I came up with something symmetric, which also has the advantage of requiring only a square root and not arbitrary exponentiation.
I'll ignore the factor $n$ for now to keep things simple; as above, we can simply divide $x$ by $n$ in the end to get that scaling right. Now consider the coordinate transform
which rotates the coordinate system by $\pi/4$ and maps the straight blue line to the line $t=0$, $s\in[-1,1]$. A curve for which $t$ is an even function of $s$ is symmetric with respect to interchange of $x$ an $y$. In these coordinates, the circle equation $x^2+y^2=1$ becomes
On the other hand, to get the extreme case $y=1$ for all $x$, we need $t=s+1$, or, to make it look more like the other equation,
Now we have two equations that differ only in the coefficient $\pm1$ in one place, so we can interpolate between them by introducing a variable $\alpha$ for that coefficient:
For $\alpha\to\infty$ we recover $t=0$, which is the straight blue line. Thus, the range $\alpha\in[-1,\infty]$ yields an interpolation between the straight blue line ($\alpha=\infty$), the red quarter-circle ($\alpha=1$) and the extreme case $y=1$ ($\alpha=-1$). To get a nice range and make it match up with a corresponding range for the curves below the straight line, we can transform to $z=1/(\alpha+1)$ to get the range $z\in[0,\infty]$, with $z=0$ corresponding to the straight blue line.
Putting everything together by transforming from $s,t$ back to $x,y$ and from $\alpha$ to $z$ and solving for $y$ yields
Taking into account the factor of $n$ and obtaining the curves below the the straight line for $z<0$ by reflection as above, we get the final result
1 - f(1 - x/n,-z)&z<0\;,
with $f$ defined as above. Here are some plots.