How do you sketch the graph of $y=log_2(x+2)$ ?

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Answer 1 · 2016-11-29T02:09:25

The graph of $y$ is the standard graph of $lnx$ transformed 2 units left and scaled by $1/ln2$

To change the base of the log to base e:

$log_2 x = ln(x)/ln2$

In this example $y=log_2(x+2)$

$:. y= ln(x+2)/ln2$

The graph of $y$ is the standard graph of $lnx$ transformed 2 units left and scaled by $1/ln2$

$y$ is defined for $x> -2$
$y$ has a single zero at $x=-1$
$y=1$ at $x=0$

The graph of $y$ is shown below

graph{ln(x+2)/ln2 [-10, 10, -5, 5]}

Answer 2 · 2016-11-29T06:50:58

x-intercept is -1; y-intercept is 1.
As $x to-2, y to -oo$ ; as $x to oo, y to oo$ . Graph for the inverse $x=2^y-1$ is inserted.

Cuts axes at (-1, 0) and (0, 1). x = -2 is the asymptote.

For graphing, I have used the inverse $x = 2^y-1$ .

graph{2^y-x-2=0 [-10, 10, -5, 5]}

How do you sketch the graph of y=log_2(x+2)y=log_2(x+2)?