Looking at the graph, we have a periodic function, and it's period is equal to 2 units.
The graph also goes from negative infinity to positive infinity in each cycle.
So this is a tangent function. The tangent function is given by:
![\begin{gathered} f(x)=c\tan (a(x+b))+d \\ \text{Period}=(\pi)/(a) \\ horizontal\text{ shift}=b \\ \text{vertical scaling}=c \\ \text{vertical shift}=d \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/ggtgh5qa2nfflkccv9gqg539d9etq5trs4.png)
The curvature of the curve in each cycle changes when y = -1 (and the first curvature after zero occurs at x = 1), so the vertical shift is d = -1 and the horizontal shift is b = -1.
The period is approximately 2 units, so we have:
![\begin{gathered} (\pi)/(a)=2 \\ a=(\pi)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/xpzwzc0w2k57f08ir5d3rx124e0r7d5zfb.png)
Now, to find the vertical scaling, let's use the approximate point (1.5, 2):
![\begin{gathered} f(x)=c\cdot\tan ((\pi)/(2)(x-1))-1 \\ 2=c\cdot\tan ((\pi)/(2)(1.5-1))-1 \\ 2=c\cdot\tan ((\pi)/(4))-1 \\ 2=c\cdot1-1 \\ c-1=2 \\ c=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2h2pbuzimdikagu6anttsn2pbgf5qrc1l0.png)
So our function is:
![f(x)=3\cdot\tan ((\pi)/(2)(x-1))-1](https://img.qammunity.org/2023/formulas/mathematics/high-school/vqvfr7ltp8n51pplt9e7sksz3nxgqe6u0b.png)