<em>Note: You may have unintentionally missed to add the diagram, so after a little research I was able to find the diagram, which anyways would help you clear your concept.</em>
Answer:
The function shown in the graph is
.
Step-by-step explanation:
From the graph, it is clear that
when x = 0, y = -5
also when x = 2, y = 1
so
![y=3(x-2)+1](https://tex.z-dn.net/?f=y%3D3%28x-2%29%2B1)
as
putting (0, -5)
![y=3\left(0-2\right)+1](https://tex.z-dn.net/?f=y%3D3%5Cleft%280-2%5Cright%29%2B1)
![y=-6+1](https://tex.z-dn.net/?f=y%3D-6%2B1)
![y=-5](https://tex.z-dn.net/?f=y%3D-5)
also putting (2, 1)
![y=3(x-2)+1](https://tex.z-dn.net/?f=y%3D3%28x-2%29%2B1)
![y=3\left(2-2\right)+1](https://tex.z-dn.net/?f=y%3D3%5Cleft%282-2%5Cright%29%2B1)
![y=1](https://tex.z-dn.net/?f=y%3D1)
So, the points (0, -5) and (2, 1) lies on the line
.
Also
![y=3(x-2)+1](https://tex.z-dn.net/?f=y%3D3%28x-2%29%2B1)
![y=3x-6+1](https://tex.z-dn.net/?f=y%3D3x-6%2B1)
![y=3x-5](https://tex.z-dn.net/?f=y%3D3x-5)
![\mathrm{For\:a\:line\:equation\:for\:the\:form\:of\:}\mathbf{y=mx+b}\mathrm{,\:the\:slope\:is\:}\mathbf{m}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3Aa%5C%3Aline%5C%3Aequation%5C%3Afor%5C%3Athe%5C%3Aform%5C%3Aof%5C%3A%7D%5Cmathbf%7By%3Dmx%2Bb%7D%5Cmathrm%7B%2C%5C%3Athe%5C%3Aslope%5C%3Ais%5C%3A%7D%5Cmathbf%7Bm%7D)
![m=3](https://tex.z-dn.net/?f=m%3D3)
Therefore, the function shown in the graph is
.