# How do you graph y = -1 + log_2x?

Jun 21, 2016

Inverse relation $x = 2 \left({2}^{y}\right)$ is convenient for making a Table for the graph, for $y = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$
$\left(x , y\right) : \ldots . \left(\frac{1}{4} , - 3\right) \left(\frac{1}{2} , - 2\right) \left(1 , - 1\right) \left(2 , 0\right) \left(4 , 1\right) \left(8 , 2\right) \left(16 , 3\right) \ldots$
.

#### Explanation:

Rearrange to the form

${\log}_{2} x = y + 1$.

The inverse relation is

$x = {2}^{y + 1} = 2 \left({2}^{y}\right) > 0.$

Now, $\left(x , y\right) : \ldots . \left(\frac{1}{4} , - 3\right) \left(\frac{1}{2} , - 2\right) \left(1 , - 1\right) \left(2 , 0\right) \left(4 , 1\right) \left(8 , 2\right) \left(16 , 3\right) \ldots$

The graph cuts x-axis at x = 2.

x-axis, downwards, is the vertical asymptote.

As$x \to \infty , y \to \infty$.