Question 116576
<font size = 6 color = "red">
STANBON'S SOLUTION IS WRONG 
ABOUT THE ASYMPTOTES AND THE 
x-INTERCEPTS.
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Solution by Edwin:
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Describe the transformation on the following graph of f(x) = log(x). State the placement of the horizontal asymptote and x-intercept after the transformation. For example, "left 1" or "stretched vertically by a factor of 2" are descriptions.
a) g(x) = log(x+2)
b) g(x) = -log(x)
<pre>

These two don't have HORIZONTAL asymptotes, but they have do have VERTICAL
asymptotes.

Here are all the rules you'll ever need for such problems. There are 10 of 
them:

1. When +k is added to the right side of f(x), the graph of the new function
g(x) is the graph of f(x) shifted "UP k".

2. When -k is added to the right side of f(x), the graph of the new function
g(x) is the graph of f(x) shifted "DOWN k".

3. When x+k is substituted for x in the right side of f(x), the graph of the
new function g(x) is the graph of f(x) shifted "LEFT k".

4. When x-k is substituted for x in the right side of f(x), the graph of the
new function g(x) is the graph of f(x) shifted "RIGHT k".

5. When the right side of f(x) is multiplied by -1, the graph of the
new function g(x) is the graph of f(x) reflected into (or across) the x-axis

6. When -x is substituted for x in the right side of f(x), the graph of the
new function g(x) is the graph of f(x) reflected into (or across) the y-axis.

7. When the right side of f(x) is multiplied by k, where k > 1 the graph of 
the new function g(x) is the graph of f(x) stretched vertically by a factor 
of k.

8. When the right side of f(x) is multiplied by k, where k < 1 the graph of 
the new function g(x) is the graph of f(x) shrunk vertically by a factor 
of k.

9. When k is replaced by kx in the right side of f(x), where k > 1 the graph 
of the new function g(x) is the graph of f(x) shrunk horizontally by a factor
of k.

10. When k is replaced by kx in the right side of f(x), where k < 1 the graph
of the new function g(x) is the graph of f(x) stretched horizontally by a
factor of k.

(a)
For your first problem, we want to go from

f(x) = log(x)
 
to 

g(x) = log(x+2)

We see that the right side of g(x), which is log(x+2), is the result of
substituting x+2 for x in the right side of f(x), so by rule 3, the
graph of g(x) is the graph of f(x) shifted "LEFT 2"

Here is the graph of f(x) = log(x). Notice that the y-axis is the
vertical asymptote, and that the x-intercept is (1,0)

{{{graph(400,400,-4,4,-4,4,log(10,x))}}}

And here is the graph of g(x) = log(x+2).  Note that the vertical
asymptote has been also shifted left by 2 units from x = 0 (the
y-axis) to x = -2. Notice also that the x-intercept has been
shifted left by 2 units from (1,0) to (-1,0). 

{{{graph(400,400,-4,4,-4,4,0,log(10,x+2),999(x+2))}}}

Here they are both on the same graph:

{{{graph(400,400,-4,4,-4,4,log(10,x),log(10,x+2),999(x+2))}}}

b) g(x) = -log(x)

For your second problem, we want to go from

f(x) = log(x)
 
to 

g(x) = -log(x)

We see that the right side of g(x), which is -log(x), is the result of
multiplying the right side of f(x) by -1, so by rule 5, the
graph of g(x) is the graph of f(x) reflected across the x-axis.

Here is the graph of f(x) = log(x) again

{{{graph(400,400,-4,4,-4,4,log(10,x))}}}

And here is the graph of g(x) = -log(x)

{{{graph(400,400,-4,4,-4,4,0,-log(10,x))}}}

Here they are both on the same graph:

{{{graph(400,400,-4,4,-4,4,log(10,x),-log(10,x))}}}

Notice that the vertical asymptote is the same for both
because a reflection across the x-axis of a vertical line
is the same vertical line.  Also notice that their 
x-intercepts are the same point (1,0).  Any point on
the x-axis remains the same when a graph is reflected
across the x-axis.

Edwin</pre>