Graphing
Logarithmic Functions: Examples (page
3 of 3)

Graph y
= log_{2}(x + 3).

This graph will be similar
to the graph of log_{2}(x),
but it will be shifted
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 point
(1, 0)
("basic" because it's neat and 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.

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
= log_{2}(x)
crawled 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 shouldn'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 know what I'm doing.

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

2^{0}
= 1, so log_{2}(1)
= 0; x
+ 3 = 1 for x
= 2: (2, 0)
2^{1}
= 2, so log_{2}(2)
= 1; x
+ 3 = 2 for x
= 1: (1, 1)
2^{2}
= 4, so log_{2}(4)
= 2; x
+ 3 = 4 for x
= 1: (1, 2)
2^{3}
= 8, so log_{2}(8)
= 3; x
+ 3 = 8 for x
= 5: (5, 3)

Then, working in the
other direction: Copyright
Š Elizabeth Stapel 2002-2011 All Rights Reserved

2^{1}
= 0.5, so log_{2}(0.5)
= 1;
x
+ 3 = 0.5 for x
= 2.5: (2.5, 1)
2^{2}
= 0.25, so log_{2}(0.25)
= 2;
x
+ 3 = 0.25 for
x
= 2.75: (2.75, 2)
2^{3}
= 0.125, so log_{2}(0.125)
= 3;
x
+ 3 = 0.125 for
x
= 2.875: (2.875, 3)

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:

If you check this in your
calculator, first, remember to put the "x
+ 3" inside parentheses,
or your calculator will think you mean "log_{2}(x)
+ 3", and you'll
get the wrong answer; second, remember that your calculator can only follow
its programming it can't think so the graph it displays will likely
be incorrect, even if you enter the function correctly.

If
you plug y
= log_{2}(x + 3)
into a graphing calculator (in the change-of-base formulation of
"ln(x
+ 3) / ln(2)"),
you will likely get a graph that looks something like this:

Now, you know full well
that the log doesn't just "end" there at the left, hanging uselessly
in space. Why is the calculator doing it wrong? Because it's just a machine,
and it's doing the best it can. The calculator graphs in a manner similar
to how you do: it picks x-values,
computes y-values,
plots the points, and connects the dots. But, whereas you know that the
log graph continues downward forever, getting infinitesimally close to
the y-axis
(or whatever the vertical asymptote happens to be), the calculator only
knows that it tried one x-value
on its list, got "ERROR" for an answer, tried the next x-value
on its list, and got a valid y-value.
Since it has no other dots before that first one, and because it can't
think, it starts the graph with that first dot. This is another instance
of "student smart; calculator stupid". Don't assume, just because
the calculator displays a graph a certain way, that this is what the graph
actually looks like. Use your head!

To review: below are some
different variations on the same basic logarithmic function, with the
associated graph below each equation. Note that, even if the graph is
moved left or right, or up or down, or is flipped upside-down, it still
displays the same curve: