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

1 Answer
Nov 25, 2017

Answer:

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#.

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