Another Example of Graphing Logs

This graph will be similar to the graph of log₂(x) but, because the 3 that is added inside the function, its graph will be shifted three units sideways.

Since the +3 is inside the log's argument, the graph's shift cannot be up or down. This means that the shift has to be to the left or to the right. But which way?

You can keep track of the direction of the shift by looking at the basic graph's point (1, 0) (which I'm calling "basic" because it's both standard and also easy to remember). The log will be 0 when the argument, x + 3, is equal to 1. When is x + 3 equal to 1? When x = −2. Then the basic log-graph point of (1, 0) will be shifted over to(−2, 0) on this graph; that is, the graph is shifted three units to the left.

(If you are not comfortable with this concept or these manipulations, please review how to work with translations of functions directly from the rules.)

Since a log cannot have an argument of zero or less, then I must have x + 3 > 0. This tells me that, for this graph, x must always be greater than −3.

The graph of the basic log function y = log₂(x) crawls up the positive side of the y-axis to reach the x-axis, with the line never going to the left of the limitation that x must be greater than zero. To remind myself of the similar limitation of this log (where x must always be greater than −3), I will insert a dashed line at x = −3:

A line like this, which marks off territory where the graph can't go, is called a vertical asymptote, or simply an asymptote. I don't have to add this to the graph, but it can be very helpful, and might convince the grader that I do indeed know what I'm doing.

After I dash in the asymptote, I plot some points:

• 2⁰ = 1, so log₂(1) = 0; x + 3 = 1 for x = −2, so (−2, 0) is a point on the graph
• 2¹ = 2, so log₂(2) = 1; x + 3 = 2 for x = −1, so (−1, 1) is a point on the graph
• 2² = 4, so log₂(4) = 2; x + 3 = 4 for x = 1, so (1, 2) is a point on the graph
• 2³ = 8, so log₂(8) = 3; x + 3 = 8 for x = 5, so (5, 3) is a point on the graph

Then, working in the other direction:

• 2⁻¹ = 0.5, so log₂(0.5) = −1; x + 3 = 0.5 for x = −2.5, so (−2.5, −1) is a point on the graph
• 2⁻² = 0.25, so log₂(0.25) = −2; x + 3 = 0.25 for x = −2.75, so (−2.75, −2) is a point on the graph
• 2⁻³ = 0.125, so log₂(0.125) = −3; x + 3 = 0.125 for x = −2.875, so (−2.875, −3) is a point on the graph

Note that, to find each of these points, I did not start with an x-value and then puzzle my way to a y-value; that would be too hard, and I'm too lazy. Instead, I started with a simple exponential statement, switched it around to the corresponding logarithmic statement, and then figured out, for that exponent (which is also my y-value), what the x-value needed to be. This method is, in my view, much the simpler way to work these problems.

Plotting the points I've calculated, I get:

...and connecting the dots gives me the following graph: