Question

How do you graph `y=log_5(2x+2)+5`?

Answer 1

It is the graph of `y=log_ 5x` with a horizontal translation of 1 unit left, horizontal compression of `1/2`, and a vertical translation of 5 units up!

Explanation:

To graph `y=log_5x`, you can change it to an exponential equation, which would be `5^y=x` and pick some values of y to find x values.
This would give you the 'original' graph.

`y=\log_5\(2x+2)+5` could be changed to `y=\log_5\2(x+1)+5`

From that graph, the transformed values are:
--> K = `-1`, which means that the graph of `y=log_5x` is horizontally translated 1 unit left.
--> D = 2, which means that `y=log_5x` is horizontally compressed by a factor of `1/2`.
--> H = 5 which means that `y=log_5x` is vertically translated 5 units up.

Also note that due to these transformations, the vertical asymptote is translated 1 unit left, to `x=-1`.