# Create a defining funcion (either logarthmic or expoential) for the following graph.

embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

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.

The graph:

jbrowner | Student, Undergraduate | (Level 1) eNoter

Posted on

Refer to following graph for answering question.

Thank you.

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