Graph f(x) = log_2(x + 3) - 2

Step by step

The question

Question: graph y = log₂(x + 3) − 2.

A log is a power.

log₂ of c is the power of 2 that makes c.

1 Find the domain

2 to any power is always more than 0.

So a log only takes positive numbers.

The inside must be more than 0.

The inside is x + 3.

So x + 3 > 0.

To remove the + 3, we take away 3 on both sides.

So the domain is x > −3.

Books also write it as (−3, ∞).

The round bracket means −3 itself is not in.

∞ means it goes on for ever.

2 Vertical asymptote

At the edge, the inside is 0: x = −3.

Close to it, the graph goes down and down.

This is the vertical asymptote.

Careful!

The asymptote is where the inside is 0, not where x is 0.

Here that is x = −3.

3 Find some points

The log is easy when the inside is a power of 2: 1/2, 1, 2 and 4.

We choose x to make that happen.

Make the inside 1/2: x is −5/2.

log₂ of 1/2 is −1, because 2 to the power −1 is 1/2.

So y is −3.

Inside 1: x is −2.

log₂ of 1 is 0, because 2 to the power 0 is 1.

So y is −2.

Inside 2: x is −1.

log₂ of 2 is 1, because 2 to the power 1 is 2.

So y is −1.

Inside 4: x is 1.

log₂ of 4 is 2, because 2 to the power 2 is 4.

So y is 0.

4 Draw the graph

Draw the grid.

The asymptote x = −3, dashed.

Plot the points.

Join them with a smooth curve.

Near x = −3 it goes down along the dashed line, but never touches it.

So: domain x > −3, asymptote x = −3.

5 Check

Let's check: at x = −3 the inside is 0, and no power of 2 is 0.

So there is no point there: the asymptote is right.

The purple curve is y = log₂ x, the plain log.

Ours is that curve moved left 3 and down 2.

The same shape.

It works!

Let's look back.

First: find the domain.

Then: vertical asymptote.

Then: find some points.

Then: draw the graph.

Last: check.