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

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Answer 1 · 2018-02-20T21:36:01

The $ln x$ $lnx$ -function only accepts positive values of $x$ $x$

There is no upper limit to $x$ $x$ so the domain is:
$0 < x < \infty$ $0 < x < oo$

As $x$ $x$ gets smaller the value of $ln x$ $lnx$ gets more and more negative, or in 'the language':

$lim_(x->0) lnx=-oo$

Meaning $x=0$ is the (only) vertical asymptote

The $x-$ intercept is when $lnx+2=0->lnx=-2$
This means that $x=e^(-2)~~0.135...$

The range of the function is from $-oo$ to $+oo$

graph{ln x+2 [-7.78, 8.026, -3.534, 4.37]}

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