1. ## Graphing y=-ln|x+2| +3.

I'm having problems with this maths question:

Graph y=-ln|x+2| +3.

I don't understand why even tho +3 is at the back, the graph moved down 3 units instead of up? Does reflection in x-axis come AFTER translations?

2. Originally Posted by Lushorama
I'm having problems with this maths question:

Graph y=-ln|x+2| +3.

I don't understand why even tho +3 is at the back, the graph moved down 3 units instead of up? Does reflection in x-axis come AFTER translations?
y = -(ln|x + 2| - 3). Therefore do the translations and then reflect.

Alternatively, ln|x| --> ln|x + 2| --> -ln|x + 2| --> -ln|x + 2| + 3.

There's more than one set of ordered transformations that will achieve the same outcome ....

3. However, whether you do it as y= -(ln|x+2|- 3) or y= -ln|x+2|+ 3, the graph is 3 places higher than the graph of y= -3ln|x+2|, not lower!

For example, for x= -1, ln|-1+2|= ln|1|= 0 so y= -(0- 3)= 3 while -ln|x+2|+ 3= -ln|1|+ 3= 3. The graph y= -ln(x+2) passes through the point (-1, 0) while the graph of y= -ln|x+2|+ 3 passes through the point (-1, 3), three places higher.

If your problem is actually y= -(ln|x+2|+ 3)= -ln|x+2|- 3, then the graph passes through (-1, -3), three places lower.