Expression of output voltage
Let's redraw the schematic. (I'm sure you already know, but it may help others attempting to read your schematic.)







simulate this circuit &ndash; Schematic created using CircuitLab

This is done by following some very simple rules about schematic drafting, which work to aid understanding (and ignores the interests of those with a soldering iron in hand.) The top of the schematic sheet is the most positive, the bottom of the schematic sheet is the most negative, the left side is where signals of interest arrive, and the right side is where signals of interest leave the page. I've also avoided specific values, since they are incidental to the question.

Here, it's pretty easy to see that \$D_1\$ and \$D_2\$ are meant to remain inactive so long as \$NODE_1\$ remains remains between the \$+5\:\textrm{V}\$ and \$-5\:\textrm{V}\$ rails. As sstobbe mentioned, if the diodes are idealized (no voltage drop developed as current flows through them), then they are ON only when \$V_{IN}\$ either rises above \$+5\:\textrm{V}\$ or when \$V_{IN}\$ falls below \$-5\:\textrm{V}\$. In reality, of course, there will be a small voltage drop across any real diode. (But as the currents are so low [due to the high-valued resistors], they probably won't drop a lot more than about half a volt, anyway.)

You are interested in just one of the two interesting cases; namely, where \$V_{IN} &lt; -5\:\textrm{V}\$. In this case, \$D_2\$ is on and you have a simple resistor divider formed by \$R_2\$ and \$R_3\$, where the output voltage is:

$$\begin{align*}
V_{NODE_1}&amp;=\frac{\left(V_D-5\:\textrm{V}\right)\cdot R_3+V_{IN}\cdot R_2}{R_2+R_3}
\end{align*}$$

Assuming \$R=R_2=R_3\$ and \$V_D=0\:\textrm{V}\$ (idealized), then:

$$\begin{align*}
V_{NODE_1}&amp;=\frac{\left(V_D-5\:\textrm{V}\right)\cdot R_3+V_{IN}\cdot R_2}{R_2+R_3}\\\\
&amp;=\frac{-5\:\textrm{V}\cdot R+V_{IN}\cdot R}{2\: R}\\\\
&amp;=\frac{-5\:\textrm{V}\cdot R}{2\: R}+\frac{V_{IN}\cdot R}{2\: R}\\\\
&amp;=\frac{-5\:\textrm{V}}{2}+\frac{V_{IN}}{2}\\\\
&amp;=-2.5\:\textrm{V}+\frac{V_{IN}}{2}\\\\
&amp;=\frac{V_{IN}}{2}-2.5\:\textrm{V}\\\\
\end{align*}$$

Here, you can see that I arrived at the same equation you are supposed to get, not the one you say you got. Perhaps you can check the above work to see your error? It may be that you simply took the absolute magnitude of \$V_{IN}\$ rather than its signed magnitude?