Find verticle assymptotes (if any) f(x)=ln(x+3)?

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2007-10-26 09:01:03 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

ln(a), for a > 0

set the expression in side ln greater than 0

x + 3 > 0
x > -3

so the vertical asymptote is x = -3

2007-10-26 09:05:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋

Because the ln is undefined for a negative number, then anything above x= -3 makes f(x) defined so your vertical asymptote is x= -3.

2007-10-26 16:05:48 · answer #2 · answered by Anonymous · 0⤊ 0⤋

ln(x) has a vertical asymptotes at x = 0
so ln(x+3) would have an asym at x = -3

2007-10-26 16:06:42 · answer #3 · answered by norman 7 · 0⤊ 0⤋