Question

How do you graph `y=ln abs(x+3)`?

Answer 1

Please see below.

Explanation:

Note that `ln(x+3)` is not defined if `x+3<0` i.e. `x<-3`.

Further if `x=-2`, `g(-2)=ln(-2+3)=ln1=0` and therefore in the interval `(-2,oo)` `ln(x+3)` is above `x`-axis and in interval `(-3,-2)`, it is below `x`-axis.

Also `g'(x)=1/(x+3)` and hence in the interval `(-3,oo)`, `g'(x)` is always positive and hence `g(x)` is a continuously increasing function and its graph looks like
graph{ln(x+3) [-10, 10, -5, 5]}

Hence, for `y=|ln(x+3)|`,

While in the interval `(-2,oo)`, it is exactly same as for `ln(x+3)`, in interval `(-3,-2)` it will be positive too and a reflection of curve above. Obviously, we have a discontinuity at `x=-2`
graph{|ln(x+3)| [-10, 10, -5, 5]}