How can I scale a 0-5v signal down to both a 0.5-4.5v signal and a 0.5-2.0 signal simultaneously?
Link to simulation.


Left graph = The triangle wave source that goes from 0 V to 5 V
Middle graph = upper Y network's output
Right graph = bottom Y network's output


With two Y networks you can solve your scaling issues. 

The formula for calculating the node at the middle of the Y network is the following equation:

\$V_{middle}=\dfrac{V_aR_bR_c+V_bR_aR_c+V_cR_aR_b}{R_aR_b+R_bR_c+R_aR_b}\$

As you can see, it is linear, it's namely an averaging function, like \$\frac{V_a+V_b+V_c}{3}\$ but with weights. 



You can solve this mathematically, and plug equations into each other and end up with some value for the resistors. Or you can just enter your favorite simulator, mine is CircuitJS. Set the input to the maximum of the limits and change the resistors until you get your wanted output value. And then set your input to the minimum value and change the resistors again, go back to top value and repeat. 

I had calculated the 125 Ω from the equation above by turning \$R_c\$ into \$R_b\$ because 0.5 and 4.5 is balanced (as you can see by the two 1 kΩ resistors).

The bottom Y network that gives 0.5 V and 2 V was made by simply changing the resistors in the simulator until I saw the output that you sought.



If you are going to attach a load to the output of these Y networks, then you will definitely want to buffer the output. A load in this sense would be some resistor between the output of the Y network and ground that is less than... say 10 kΩ. Above 10 kΩ you will still distort the signal.

If your input has high impedance, meaning that it will have issues with giving the current into the Y network, then you might want to buffer the input. 



Edit

Op is right, brute-forcing your way to a solution is against the engineer code.

So here's the proper mathematical expression for \$R_b\$ and \$R_c\$: 

\$
\scriptsize{
\begin{array}{l l}
V_a^+ &amp; \text{= Maximum input voltage}\\
V_o^+ &amp; \text{= Output at that voltage}\\
V_a^- &amp; \text{= Minimum input voltage}\\
V_o^- &amp; \text{= Output at that voltage}\\
V_b   &amp; \text{= Positive rail voltage}\\
V_c   &amp; \text{= Negative rail voltage}\\
R_a   &amp; \text{= Resistor from output to input}\\
R_b   &amp; \text{= Resistor from output to positive rail}\\
R_c   &amp; \text{= Resistor from output to negative rail}\\
\end{array}
}
\$


\$
R_b = \dfrac{V_b-V_o^+}{\frac{V_a^+-V_o^+}{R_a}+\frac{V_c-V_o^+}{R_c}}\\
R_c = Ra\dfrac{(V_b-V_o^+)(V_c-V_o^-)-(V_c-V_o^+)(V_b-V_o^-)}{(V_b-V_o^-)(V_a^+-V_o^+)-(V_b-V_o^+)(V_a^--V_o^-)}
\$

So the idea here is that you choose \$R_a\$, then solve for \$R_c\$ and use that value to get \$R_b\$.

Or you just use this equation for \$R_b\$:

\$
R_b = -R_a\dfrac{(V_b-V_o^+)(V_c-V_o^-)-(V_c-V_o^+)(V_b-V_o^-)}{(V_c-V_o^-)(V_a^+-V_o^+)-(V_c-V_o^+)(V_a^--V_o^-)}
\$

Example 1:

\$
\scriptsize{
\begin{array}{l l l}
V_a^+ &amp; 5\text{ V} &amp;\text{= Maximum input voltage}\\
V_o^+ &amp; 4.5\text{ V} &amp;\text{= Output at that voltage}\\
V_a^- &amp; 0\text{ V} &amp;\text{= Minimum input voltage}\\
V_o^- &amp; 0.5\text{ V} &amp;\text{= Output at that voltage}\\
V_b   &amp; 5\text{ V} &amp;\text{= Positive rail voltage}\\
V_c   &amp; 0\text{ V} &amp;\text{= Negative rail voltage}\\
R_a   &amp; 125\text{ Ω} &amp;\text{= Resistor from output to input}\\
R_b   &amp; ?\text{ Ω} &amp;\text{= Resistor from output to positive rail}\\
R_c   &amp; ?\text{ Ω} &amp;\text{= Resistor from output to negative rail}\\
\end{array}
}
\$


$$
\begin{align}
R_c &amp;= Ra\dfrac{(V_b-V_o^+)(V_c-V_o^-)-(V_c-V_o^+)(V_b-V_o^-)}{(V_b-V_o^-)(V_a^+-V_o^+)-(V_b-V_o^+)(V_a^--V_o^-)}\\
\\
&amp;= 125\dfrac{(5-4.5)(0-0.5)-(0-4.5)(5-0.5)}{(5-0.5)(5-4.5)-(5-4.5)(0-0.5)}\\
\\
&amp;= 1000Ω
\end{align}
$$

And we don't need to calculate \$R_b\$ because we know it is also 1 kΩ in this case, so let's calculate \$R_b\$ in another example.

Example 2:

\$
\scriptsize{
\begin{array}{l l l}
V_a^+ &amp; 5\text{ V} &amp;\text{= Maximum input voltage}\\
V_o^+ &amp; 2\text{ V} &amp;\text{= Output at that voltage}\\
V_a^- &amp; 0\text{ V} &amp;\text{= Minimum input voltage}\\
V_o^- &amp; 0.5\text{ V} &amp;\text{= Output at that voltage}\\
V_b   &amp; 5\text{ V} &amp;\text{= Positive rail voltage}\\
V_c   &amp; 0\text{ V} &amp;\text{= Negative rail voltage}\\
R_a   &amp; 2000\text{ Ω} &amp;\text{= Resistor from output to input}\\
R_b   &amp; ?\text{ Ω} &amp;\text{= Resistor from output to positive rail}\\
R_c   &amp; ?\text{ Ω} &amp;\text{= Resistor from output to negative rail}\\
\end{array}
}
\$


$$
\begin{align}
R_b &amp;= -R_a\dfrac{(V_b-V_o^+)(V_c-V_o^-)-(V_c-V_o^+)(V_b-V_o^-)}{(V_c-V_o^-)(V_a^+-V_o^+)-(V_c-V_o^+)(V_a^--V_o^-)}\\
\\
&amp;= -2000\dfrac{(5-2)(0-0.5)-(0-2)(5-0.5)}{(0-0.5)(5-2)-(0-2)(0-0.5)}\\
\\
&amp;= 6000Ω
\end{align}
$$

And if you would've used the previous expression for \$R_c\$, then you would've gotten 1 kΩ.

So both of my equations agree with my schematic at the top of this answer.

Tada!