Consider the logarithmic equation f(x) = ln (x + 2) – 1. round all values to the tenths place.
(a) determine any x- and y-intercepts.
(b) determine the domain and the equation of the vertical asymptote.
(c) make a table of values to find three other points on the graph.
(d) graph the function. label the three points you found in part (c).

Hello,

A) x=0==> f(0)=ln(2)-1=-0,30685281944005469058276787854182
y=0==>ln(x+2)-1=0
==>ln(x+2)=1
==>e^ln(x+2)=e^0
==>x+2=0
==>x=-2

B)
dom f(x)={x∈IR | x>2}
x=-2 is the vertical asymptote

c)
A=(-1,-1)
B=(0,ln(2)-1=-0,30685281944005469058276787854182 )
C=(1,ln(3)-1=0,09861228866810969139524523692253)
D=(2,ln(4)-1)=0,38629436111989061883446424291635)

d) see picture

Cthe residual plot for model 1 has a non random pattern and is a good fit for the data

Idon’t see the picture