Graph y = log_3(x + 1) - 1

Step by step

The question

Question: graph y = log₃(x + 1) − 1.

A log is a power.

log₃ of c is the power of 3 that makes c.

1 Find the domain

3 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 + 1.

So x + 1 > 0.

To remove the + 1, we take away 1 on both sides.

So the domain is x > −1.

Books also write it as (−1, ∞).

The round bracket means −1 itself is not in.

∞ means it goes on for ever.

2 Vertical asymptote

At the edge, the inside is 0: x = −1.

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 = −1.

3 Find some points

The log is easy when the inside is a power of 3: 1/3, 1, 3 and 9.

We choose x to make that happen.

Make the inside 1/3: x is −2/3.

log₃ of 1/3 is −1, because 3 to the power −1 is 1/3.

So y is −2.

Inside 1: x is 0.

log₃ of 1 is 0, because 3 to the power 0 is 1.

So y is −1.

Inside 3: x is 2.

log₃ of 3 is 1, because 3 to the power 1 is 3.

So y is 0.

Inside 9: x is 8.

log₃ of 9 is 2, because 3 to the power 2 is 9.

So y is 1.

4 Draw the graph

Draw the grid.

The asymptote x = −1, dashed.

Plot the points.

Join them with a smooth curve.

Near x = −1 it goes down along the dashed line, but never touches it.

So: domain x > −1, asymptote x = −1.

5 Check

Let's check: at x = −1 the inside is 0, and no power of 3 is 0.

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

The purple curve is y = log₃ x, the plain log.

Ours is that curve moved left 1 and down 1.

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.