This video is provided as supplementary
material for
courses taught at Howard Community
College and in this video
I'm going to show how to do a
transformation of a logarithmic function.
So the basic function I'm going to work with
is y equals log-base-2 of x.
I've drawn a graph of that basic function.
It's an increasing function - we can tell
that from the logarithm itself.
Whenever the base is greater than one you
will have an
increasing function. Like all basic
logarithmic functions
it has a vertical asymptote along the
y-axis
and it has an x-intercept at (1, 0).
So let's look at some 
transformations of this.
As with other functions you've seen, if I
take the basic function and add a constant
to it,
like a 3, that's going to
raise or lower the entire graph
by that many units - a positive constant
will raise the graph and a negative constant
would lower the graph.
So in this case I'm going to
raise the graphic up three units. The
vertical asymptote will remain the same.
I'm just gonna add 3 to each of the 
y-coordinates for points on the graph.
So I can take the point (1, 0) and make
that (1, 3).
I take the point (2, 1) and make that
(2, 4), and (4, 2)  would
be
(4, 5) and then I could draw a curve
to sketch a graph.
Besides shifting the graph vertically,
I can stretch it vertically. If I put a
coefficient,
like a 2, in front of the logarithm
then, as with other functions you've seen,  that's going to 
stretched the graph vertically. It's
going to take all the y-coordinates
and multiply them by 2. So this point I
have here
at (2, 1) will become (2, 2).
The point at (4, 2) will be (4, 4).
The x-intercept can stay the same
since the
y-coordinate is 0, and 2 times 0 is still 0.
I could take the negative coordinates -- I
have a point at 
(1/2, -1) -- that's gonna become
(1/2, -2). The vertical asymptote
stays the same.
I'll sketch the graph in.
We can see that it's stretched out vertically
by a factor of 2. If I put in a negative sign
in fact that the coefficient, that
would
stretch the graph by a factor of 2,
that's what the 2 would do,
and also flip or reflect it across the
x-axis.
So we'd have a graph that looks something
like this,
stretched and reflected.
Now besides
transforming the graph vertically with
reflections and shifts and stretches,
we can also move the graft horizontally.
So if we have the 
log-base-2 of
x + 3, with that entire x + 3 in parentheses,
as you've see with other transformations,
when we add a constant to the x-value,
if that constant is positive, it shifts the
graph over
to the left that many units. So this
(x + 3) is going to shift
everything over to the left
2 units. We'll have a new vertical asymptote
at x = -3. The x-intercept, which
was at (1, 0)
would be at (-2, 0).
The point that was at (2, 1) would-be
at (-1, 1)
and this point at (4, 2)
would-be at (1, 2).
And I can
sketch that also. So that's
just shift it 3 units to the left.
If I wanted to reflect the graph
across the y-axis, all I'd have to do
is take the x and put a negative
in front of it.
The vertical asymptote
would be the same, but the
graph
would be reflected
across the y-axis.
 
Now let's see what we can do about
combining some of these
different transformations into one
function. Let's try 
y equals 3 times
log-base-2 of (x + 3)
-4. Let's get an idea of what that's
going to do.
This (x + 3) is gonna shift the graph
over three units to the left.
That's what we just saw. So I can draw
that vertical asymptote in now
at x = -3. The 3
in front of the logarithm is going to
stretch everything out
by a factor of 3, and the -4
after the logarithm is going to drop
everything down by four units.
Now let's make a table of values
and find some points on this graph. I'm going to use
x equals -2 to look for 
a point.
I'm dong that because when x is -2
then (x + 3) is 1, and the log of 1,
one no matter what the base is, is always
going to be 0.
So this will be easy. I'll have log-base-2 of 1,
which is zero. I'll multipy 0 times 3
which is still 0, and subtract 4.
So that's going to be -4.
So I'll have a point at
(-2, -4). Next I'll use
-1 for an x-value
because that'll take this (x + 3)
and turn it into
-1 + 3, which is 2,
and whenever the base and the number you're taking the logarithm of
are the same, the whole logarithm
equals 1.
So I'll have log-base-2 of 2,
which equals 1. I'll multiply that by 3
and get a 3. I'll subtract 4
and I'll have -1. Which means we'll have a point
 
And the last value I want to use is 1.
That's going to mean this 
x + 3
will equal 4, and it's always nice
when the number you're taking a logarithm
of is a power of the base.
So we'll have log-base-2 of 4
and that's 2. So we'll take the 2 and
multiply it by 3. That's gonna be 6,
and 6 - 4
is 2.  So the last point I'll plot
is going to be at (1, 2). That will
be up here.
Given these three points
and my vertical asymptote, I can 
sketch in the curve.
It's going to go up like this
and this is basically the curve of the original graph
shifted over 3 units to the left --
that's what the (x + 3) does --
and then stretched by a factor of 3 --
that's what this 3 coefficient does --
and dropped down 4 units --
that's what -4
at the end of function does.
I hope this helped. Take care. I'll see you next time.
