Welcome, ladies and gentlemen.
So what I'd like to do is show you
how to graph logarithmic equation with transformations,
other than just reflections and dilations, stretching
and compressing.
So basically what we're going to be doing
is we added back our two variables, h and k.
Which again, remember, h is going to be shifting your graph
left or right. k is going to be shifting your graph up or down.
Please also remember, though, that we
can write-- where is my blue?
I lost-- there's my blue.
Please also remember that we can write x minus h
like this, x minus h and put that h in parentheses
because h is technically in the positive.
So whenever doing this, that h is always
kind of like in the opposite direction, which
confuses students all the time.
And even though I've gone over it in different videos,
I wanted to just kind of highlight it again.
Let's see, do I have at least a-- OK,
so, I'll kind of take a stab here
at this one and a kind of explaining it
but then I'll kind of move forward
with it a little bit quicker for the rest of videos
because I don't want to be redundant for those of you that
understand it.
But basically when we're grappling
multiple transformations, the main important thing
we want to do is graph our paragraph.
Now, basically we're always going
to have-- our parent graph is always
going to have an x-intercept as long as the coefficient
of our x is 1, which we have in case of all of these,
except for our final last problem, which I again
will explain at the end.
But since all of the rest of our variables,
our x has a coefficient of 1 or negative 1, that's
just a reflection, we know that the paragraph
is going to look somewhat like this, 1 comma 0
and goes like that with a nice little vertical asymptote.
So basically what we're going to do
is we're going to take that graph
and we're going to transform it based on the transformations
or reflections.
So we're going to-- oh, did I choose
all-- oh, there's a couple different bases.
And remember, like numbers in front or the bases.
Those are going to affect the graph.
The graph could either be stretched vertically.
The graph could be compressed vertically.
And that's all dependent on the bases or the number that's
multiplying by a logarithm.
But again, I'm not going to be concerned
about the perfect slope, you might say, of the graph.
I just want to kind of get the main version of the graph
correct as well as apply the operations.
You can create table or you can use a graphing calculator
to kind of see how perfect your graph is compared
to what the actual answer is.
However, they're all going to look pretty close to the same.
OK, so let's get into y equals log base 5 of x minus 1.
Remember, I stated that this can be written as x minus h.
OK, so hopefully you see that I can put parentheses
around here, this one.
So it's x minus h.
x minus 1.
x minus h. x minus 1.
So therefore, h is equal to a positive 1.
What that's going to tell me to do
is I'm going to be shifting the graph to the right 1 unit.
Since the only thing I have is a base here of 5,
which is just going to be affecting kind
of my transformation of the graph,
I'm just going to sketch my normal parent graph.
And then I know I'm going to be shifting my graph one
unit to the right, so that's going to move over there.
Now, what's very important about transformations,
especially horizontal transformations
with logarithms, remember, we have this asymptote here.
Well, if I'm shifting the graph one unit to the right,
that means my asymptote has to be
shifted one unit to the right.
So that's very important because now what that's going to do
is that's going change our domain and range.
The domain is a set of all x values.
How far left does the graph go?
How far to the right does the graph go?
Well, you can see in our typical standard form--
in our [INAUDIBLE] graph, the asymptote was at 0.
So the farther left the graph went was at 0.
Well, now the farthest left graph goes is positive 1.
And the farthest it goes to the right is infinity.
So the domain is 1 comma 0.
The range still goes infinitely down, infinitely up,
so that's going to be from negative infinity to infinity.
All right, so now we have ln of x.
And again, ln of x is just a logarithm with base e.
I'll write it in for this one but I'm not going
to write it for the other ones.
Again, that's just going to be affecting-- it's basically
like 2.71 as your base.
So it might not-- it might increase.
It's not going to go out as far as 5.
But it's still going to look the exact same,
it's just a difference of how that graph kind of slopes,
but it's still you have an x-intercept of 1 comma 0.
However, so I'll graph 1 comma 0.
And then what's happening to the graph is this plus 4.
Well, the plus 4 is not inside the function.
It would be in parentheses if it was inside the function.
So since it's not inside the function,
that means we're just shifting the graph up or down.
Since that's a positive 4, plus k, plus 4,
it's not the opposite like negative h,
I'm just going to move it up 4 units, 1, 2, 3, 4.
I still have a vertical asymptote at 0.
And by shifting the graph up or down,
that does not affect my vertical asymptote.
So my domain and range are going to be unchanged.
0 to infinity, range is going to be from negative infinity
to infinity.
All right, in the next example I have y
equals log times x plus 3 minus 6.
Remember this x minus h idea that I did?
Well, remember, plussing or adding 3
is the same thing as subtracting a negative.
So I could rewrite that as x minus negative 3.
So therefore, x minus h.
So what you could say is h is equal to negative 3.
K is equal to negative 6.
Well, when h is equal to negative 3, h is left to right.
That means I'm shifting my graph 3 units to the left.
And when k is negative 6, that means
I'm shifting my graph 6 units down.
My base here in this case is not written,
so it's going to be 10.
Again, that's just going to be affecting
the compression of the graph.
But my standard parent graph is going
to be the same, which is going to be at 1 comma 0.
Still going on my vertical asymptote at x equals 0.
Now, what I'm going to do is I'm going to take this x-intercept
and shift it to the left 3, 1, 2, 3 and down 6, 1, 2, 3, 4, 5,
6.
Now to graph this, I think it's important to know
where the asymptote is.
I have a vertical asymptote.
The whole graph is being shifted to the left 3.
So this vertical asymptote needs to move 3, 1, 2, 3.
Shifting it down is not going to affect a vertical asymptote.
So therefore, the graph is going to look something like this.
But it's very important you can see
that now our domain and range are going to be-- are changed.
My domain is now, instead of it going to 0, or as far as it
going to 1, now the domain goes all the way to negative 3.
And it goes continues to infinity.
Whereas the range is going to be unchanged
negative infinity to infinity.
All right, next example here.
Again, this 3 is just give me affecting
my vertical compression and stretching.
My minus 5, we know, is now from my first examples,
is being shifted in the graph 5 units to the right.
But now I have a negative, which is
going to be reflecting about the x-axis.
Because remember, when you multiply
by a negative outside the function,
you reflect about the x-axis.
When you multiply a negative inside the function,
that reflects about the y-axis.
So let's just do the standard graph here.
So I'm going to shift this graph 5 units to the right, 1, 2, 3,
4, 5.
And then I'm going to reflect it about the y-axis.
Oh, what am I doing.
Whenever you're shifting, you should definitely
change the asymptote.
The asymptote is at 0.
That's been shifted to the right 5, 1 2, 3-- oh,
I didn't move this one.
1, 2, 3, 4, 5.
Asymptote gets moved 5, 1, 2, 3, 4, 5.
Then I just connect my graph.
My domain is now the graph goes only as far
as left as 1, 2, 3, 4, 5.
So it's 5 to infinity, whereas my range
is unchanged, negative infinity to infinity.
In the next one we have a fraction but that's OK,
it's just like the 3.
The 3 is going to stretch, while the 1/4 is just
going to compress it.
But again, I'm not worried about that.
Our coefficient of our x is 1, so we
know that the x-intercept is going to be 1.
And again, if you don't believe me, replace y with 0
and solve for x.
Convert it to exponential form and then solve for x.
And what you'll see is that the x-intercept is going to be 1.
So let's just go ahead and do that and then go
through our transformations.
So we have 1.
And again, remember, these two numbers
are just affecting how fast the graph grows.
I do this graph.
This 8 and the 1/4, stretch it, compress it, and so forth.
So I'm not worried about the accuracy of my graph.
I'm worried about where is the x-intercept, as well
as the transformations that need to be applied.
Again, you can use a table or use graphing technology
to see which of those three graphs
best represent that function.
Just not the purpose of my video,
it would take way too long for me to show all the points
and do all that work and so forth.
And I want to rather do a lot of examples
so you get some variety.
All right, so now we look at our transformations.
We have the x minus 2, which is going to shifting my graph 2
units to the right.
So we're going to be shifting the graph 2 units to the right
and then up 1.
So I'm just going to go from this point to the right 2,
up 1.
Since I'm shifting 2 units to right,
my x-intercept that was at x equals 0
is now going to be at x equals 2.
There's no transformation.
So now my domain is going to be from 2 to infinity.
And my range is going to be from negative infinity to infinity.
This next one we're multiplying by negative
inside the function, so now that's
going to be reflecting about the y-axis.
So it's going to be flipping about the y-axis rather
than the x-axis like I did over here.
So again, the main important thing-- oh
and also, I need to say I have a plus 3.
So let's just go ahead and graph our parent function here.
So I'm going to shift the graph up 3.
Now, I need to reflect it about the y-axis.
So it's going up 3 now to the left 1.
So the graph's going to look something like this.
Now by reflecting it, my asymptote has not changed.
My asymptote is going to remain at y equals 0.
However, my domain, you can see that now the graph
goes infinitely to the left.
There's no asymptote that's stopping
how far it's going to the left.
So my domain is going to be from negative infinity to 0.
And my range is from negative infinity to infinity.
All right, next, this is a good one
because now we have two reflections.
We have a reflection about the x-axis and reflection
about the y-axis.
Let me just label them x-axis reflection y-axis.
So the best thing to do is just do one at a time.
Doesn't matter which one you do first.
The two is just going to be compressing the graph.
So we have 1, 0.
I'll just work from left to right.
I'll reflect about the x-axis.
So that's going to look like this.
And then I'll reflect about the y-axis,
which would look like this.
Oops, I could probably fill that in, right?
OK, so that's what my graph looks like.
So it looks totally different than all the rest of them.
However, my domain is going to be exactly the same as this one
because the reflecting over the x-axis, if you remember,
did not affect the domain-- or where was that?
Oh, it was over there but I had a transformation.
Crap.
I didn't do just a regular one.
But reflecting over the x-axis doesn't change the domain.
If you were to take this one and flip it over,
the domain is still remaining the same.
But reflecting over the y-axis changes our domain.
So our domain is from negative infinity to 0.
And our range, in this case, is from negative infinity
to infinity.
OK, last example.
Again, they kind of talk about why we have our asymptote.
Let me see.
So to get this one, you really kind of got
to understand a couple points here.
And I'm actually going to erase these to go.
Why do we have our asymptotes?
Why do our asymptotes occur?
And I have a video for where the asymptotes come from.
But if we just look at a regular graph, log base 3 of x.
Now, the asymptotes occur, this one
we know that the asymptote, this graph,
looks something like this, with an asymptote at 0.
Now, why do we have an asymptote at 0?
Well, that means x value at 0 is undefined.
So what would happen if we put log base 3 of 0?
Well, rewriting this in exponential form,
3 to what power gives you 0?
And that's undefined.
There is no value that we can take 3, raise it to a power
and give it 0 because 3 to the 0 power is 1.
So there is no number.
So that's why we have this asymptote.
OK, so when x is equal to 0, or when inside our parentheses
is equal to 0, we have an asymptote.
Now, for each one of these equations
you could see that x minus 2, when that's equal to 0, x
equals 2.
Where is our asymptote?
At 2.
Over here, x minus 5 is equal to 0.
Solve for x.
X equals 5.
Where is our asymptote?
5.
So when you look at this problem,
I automatically understand that, you know what, my asymptote is
not going to occur at 9.
Because what we need to do is define the asymptote, what
we're going to need to do is graph it, or I'm sorry,
solve for x.
Just like I did here, you set whatever's
inside your function, set it equal to 0.
By doing that, x equals 9/5, or let me give you
the decimal approximation, 1.8.
Now, that's very important because if I'm
going to graph this, is I need to know
where this asymptote is.
Asymptote is not over 9 like it was here.
Here, there's no coefficient of x so we just
shifted it over 5 units.
That's where the new asymptote is.
Here, we have a coefficient of 5.
So it's not-- the graph is not being
shifted 9 units to the right.
And that's one of the major problems the students have
with this, the graph is not being shifted over 9 units
to the right.
It's only been shifted over 1.8.
So I'm just going to approximate here.
I'm not going to get too crazy with this.
OK, but now we still need to find the x-intercept.
And if you remember to find the x-intercept, what
I did in the previous problems, for instance y equals log base
3 to the x, the x-intercept is when y is equal to 0.
So when x is equal to 0, that's how we found the asymptote.
When y is equal to 0 is going to be our asymptote.
So if I did that here, log base 3x.
Now I rewrite this in exponential form, and I get 1
equals x.
That's why whenever there's no coefficient there
we always knew the x-intercept was equal to 1.
And you could have done that with all of these problems.
And for every answer you would have got
x equals 1, every single one.
But whenever you have a coefficient in front of that x,
it's going to change the x-intercept.
So let's see where that's going to change to.
So I'll take this equation and I'll replace y with 0.
We put 10 in as my base.
And I'll do 10 to the 0 power equals 5x minus 9,
because again, remember what I'm doing
is I'm rewriting it in exponential form.
That's 1 equals 5x minus 9 plus 9 plus 9.
10 equals 5x divide by 5 divide by 5. x equals 2.
So now my x-intercept is at 2.
And now I can go ahead and graph.
I know it's going to look just like everything else.
There's no other reflections.
I don't know the stretching or the compressing, which,
you know, the base 10.
But I can get a good idea of what
the graph's going to look like.
My domain is now going to be from my asymptote, which
was 9/5.
So I'll say, 9/5 to infinity and my range
is going to be unchanged, negative infinity to infinity.
So there you go, ladies and gentlemen,
that is how you graph the logarithmic function equations
when given multiple transformations.
Thanks.