f(-3/2) = log_{2 }(-3/2 + 2) = log_{2
}(1/2) = log_{2 }(2^{-1}) = -1.

f(2) = log_{2 }(2 + 2) = log_{2
}(2^{2}) = 2.

We now have more information
on how to graph f. The graph increases as x increases. Close to the vertical
asymptote x = -2, the graph of f decreases without bound as x approaches -2 from
the right. The graph never cuts the vertical asymptote. We now join the
different points by a smooth curve.

**Matched Problem to Example 1:** f is a function given by

f (x) = log_{2 }(x + 3)

- Find the domain of f and range of f.
- Find the vertical asymptote of the graph of f.
- Find the x and y intercepts of the graph of f if there are any.
- Sketch the graph of f.

**Example 2:** f is a function given by

f (x) = -3ln_{}(x - 4)

- Find the domain of f and range of f.
- Find the vertical asymptote of the graph of f.
- Find the x and y intercepts of the graph of f if there are any.
- Sketch the graph of f.

**Answer to Example 2**

a - The domain of f is
the set of all x values such that

x - 4 > 0

or x > 4

The range of f is the
interval (-inf , +inf).

b - The vertical
asymptote is obtained by solving

x - 4 = 0

or x = 4

As x approaches 4 from the
right (x > 4) , f(x) increases without bound. How do we know this?

Let us take some values:

f(5) = ln(5-4) = -3ln(1) = 0

f(4.001) = -3ln(0.001) which is approximately
equal to 20.72.

f(4.000001) = -3ln(0.000001) which is
approximately equal to 41.45.

c - To find the x
intercept we need to solve the equation f(x) = 0

-3ln(x - 4) = 0

Divide both sides by -3 to
obtain

ln(x - 4) = 0

Use properties of logarithmic
and exponential functions to write the above equation as

e^{ln(x - 4) }= e^{0}

Then simplify

x - 4 = 1

x = 5

The x intercept is at
(5 , 0).

The y intercept is given by
(0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f.
There is no y intercept.

d - So far we have the
domain, range, x intercept and the vertical asymptote. We need extra points to
be able to graph f.

f(4.5) = -3ln(4.5 - 4) approximately equal to
2.08

f(8) = -3ln(8 - 4) approximately equal to -
4.16

f(14) = -3ln(14 - 4) approximately equal to -
6.91

Let us now sketch all the
points and the vertical asymptote. Join the points by a smooth curve and f
increases as x approaches 4 from the right.

**Matched Problem to Example 2: ** f is a function given by

f (x) = 2ln_{}(x + 5)

- Find the domain of f and range of f.
- Find the vertical asymptote of the graph of f.
- Find the x and y intercepts of the graph of f if there are any.
- Sketch the graph of f.

**Example 3:** f is a function given by

f (x) = 2ln_{}(| x |)

- Find the domain of f and range of f.
- Find the vertical asymptote of the graph of f.
- Find the x and y intercepts of the graph of f if there are any.
- Sketch the graph of f.

**Answer to Example 3**

a - The domain of f is
the set of all x values such that

| x | > 0

The domain is the set of all
real numbers except 0.

The range of f is the
interval (-inf , +inf).

b - The vertical
asymptote is obtained by solving

| x | = 0

which gives

x = 0

As x approaches 0 from the
right (x > 0) , f(x) decreases without bound. How do we know this?

Let us take some values:

f(1) =2 ln(| 1 |) = 0

f(0.1) = 2ln(0.1) which is
approximately equal to -4.61.

f(0.0001) = 2ln(0.0001) which is
approximately equal to -18.42.

f(0.0000001) = 2ln(0.0000001) which is
approximately equal to -32.24.

As x approaches 0 from the
left (x < 0) , f(x) decreases without bound. How do we know this?

Let us take some values:

f(-1) =2 ln(| -1 |) = 0

f(-0.1) = 2ln(| -0.1 |) which is
approximately equal to -4.61.

f(-0.0001) = 2ln(| -0.0001 |) which is
approximately equal to -18.42.

f(-0.0000001) = 2ln(| -0.0000001 |) which is
approximately equal to -32.24.

c - To find the x
intercept we need to solve the equation f(x) = 0

2ln(| x |) = 0

Divide both sides by 2 to
obtain

ln(| x |) = 0

Use properties of logarithmic
and exponential functions to write the above equation as

e^{ln(| x |) }= e^{0}

Then simplify

| x | = 1

Two x intercepts at
(1 , 0) and
(-1 , 0).

The y intercept is given by
(0 , f(0)). f(0) is undefined since x = 0 is not a value in the domain of f.
There is no y intercept.

d - So far we have the
domain, range, x intercept and the vertical asymptote.
By examining function f it is easy to show
that this is an even function and its graph is symmetric with respect to the y axis.

f( -x) = 2 ln(| -x |)

but

| -x | = | x |

and therefore

f( -x) = 2 ln(| x |) = f( x ),
this shows that f is an even function.

Let us find extra points.

f(4) = 2ln(| 4 |) approximately equal to 2.77.

f(0.5) = 2ln(| 0.5 |) approximately equal to
- 1.39.

Since f is even f(-4) = f(4)
and f(-0.5) = f(0.5).

Let us now sketch all the
points, the vertical asymptote and Join the points by a smooth curve.

**Matched Problem to Example 3: ** f is a function given by

f (x) = -2ln_{}(x^{2})

- Find the domain of f and range of f.
- Find the vertical asymptote of the graph of f.
- Find the x and y intercepts of the graph of f if there are any.
- Sketch the graph of f.

More references on logarithmic functions and graphing.

Graphing Functions

Logarithmic Functions.

Graphs of Basic Functions.