[MUSIC] Let's look at logarithmic graphs. For example, let's sketch the graph of

f(x)=1+ log base 2 of x+3, and then we'll find any x or y intercepts of its graph.

Let's sketch this by using graph transformations.

First of all, what does y = log base 2 of x look like? So let's say that this is

the y axis, and this is the x axis. Log base 2 of x looks like this,

where this x intercept is 1. And then when x = 2, y = log base 2 of 2

which is 1. Now what does this + 3 do here? By adding

3 to X, what that does is it rigidly shifts this graph 3 units to the left.

So the graph of Y is equal to log base 2 Of x plus 3 will look like the following.

Let's say this is the y axis and this is the x axis.

Now if were rigidly moving this entire graph 3 units to the left.

What's going to happen to this point here? This (1, 0).

And what's going to happen to this point here? The (2, 1).

Well (1, 0) is going to move to (1 -3, 0), or (-2,0).

And (2,1) will move to (2, -3, 1) or (-1,1).

So here is -1, -2, 0. And if this is 1, then here is the point

(-1,1). Now remember we originally had a vertical

asymter here as x is equal to zero.Suppose shifting everything to the

left 3 units then the protocol asymter will be shifted to x is equal to

negative. 3.

So therefore log base 2 of x + 3 will look like this.

Alright and finally what does this + 1 do here? What that does is it shifts this

graph rigidly up. One unit.

So lets say this is the y axis, and this is the x axis.

Now what is going to happen to this point here (-2,0).

It moves to - 2, 0 plus 1, or -2 1. And what happens to this point -1, 1? It

moves to negative 1. And then 1 + 1 or 2.

Let's say this is -1 and -2. Then our vertical asymptote is still at x

= -3. Let's say this is y = 1 and this is y =

2. The negative 2, 1, is right here.

And negative 1, 2, is right here. And our graph then looks like this.

So this is alpha is equal. To 1+log base 2 of x+3.

Now, we're also asked to find any x or y intercepts, and looking at our graph, we

have an x intercept here and a y intercept here.

So let's find out what these values are. Now to find the y intercept, we set x =

0. That is we have y is equal to one plus

log base two of zero plus three, or y is equal to one plus log base two of three.

This is the y intercept. Therefore the y coordinate of this point

here is 1 plus log base 2 of 3. And what about the x intercept?

To find the x intercept we set y equal to 0.

That is we have 0 is equal to 1 plus log base 2, of x + 3, or subtracting 1 from

both sides gives us, log base 2, of x + 3 is equal to negative 1.

And now rewriting this in exponential form gives us 2 raised to the negative

1st power is equal to X + 3, or 1/2 is equal to x + 3.

And then subtracting 3 from both sides gives us x is equal 1/2 minus 3 or x

equal to negative 5/2. So this is our x intercept.

That is the x co-ordinate of the point here is negative 5/2.

And, this is how we work with logarithmic graphs.

Thank you, and we'll see you next time. [SOUND]