How do you find the domain, x intercept and vertical asymptotes of #f(x)=ln(x+2)#?

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Answer 1 · 2017-11-11T15:48:55

See below.

#ln(x)# is only defined for #x > -2# for real numbers.

So the domain is:

#color(blue)({x in RR|-2 < x < oo})#

#x# axis intercepts occur when #y = 0#:

#y=ln(x+2)#

#ln(x+2)=0#

This has no solution, because #e^x# can never equal 0.

No x axis intercepts.

Y axis intercept occurs when #x =0#

#ln(0+2)=ln(2)~~color(blue)(0.69314718)#

Vertical asymptotes occur where #ln(x+2)# is undefined:

#ln(x+2)# is undefined when #x=-2#

So the line #color(blue)(x=-2)# is a vertical asymptote. graph{ln(x+2) [-22.81, 22.8, -11.4, 11.41]}