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The graph has a vertical asymptote at x=-2 so it is logarithmic. The graph also contains the points (0,-2) and (-1,-3).
The general form for a logarithm is `y=alog_b (x-h)+k` where h translates the base graph horizontally, k translates vertically, and a performs a vertical stretch.
Since the graph has a vertical asymptote at x=-2 the function is of the form `y=alog_b (x+2)+k` . From the point (-1,-3) we know that k=-3 since `log_b (1)=0` for any base. So we now have `y=alog_b (x+2)-3`
If the graph had (6,0) as a point this would be `y=log_2(x+2)-3` , but it appears to have (8,0).
A little experimenting with the natural log gives
`y=1.3ln(x+2)-3` as a good approximation.
Refer to following graph for answering question.
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