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

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Answer 1 · 2017-11-11T15:48:55

See below.

$ln (x)$ $ln(x)$ is only defined for $x > - 2$ $x > -2$ 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]}

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