SOLUTION: I have to describe the transformation on the following graph of f(x)=e^x. I also have to state the placement of the horizontal asymptote and y-intercept after the transformation.

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Question 152622: I have to describe the transformation on the following graph of f(x)=e^x. I also have to state the placement of the horizontal asymptote and y-intercept after the transformation. I'm lost and do not understand. The first transformation is g(x)=e^x-5 and
the second is h(x)= -e^x
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
have to describe the transformation on the following graph of f(x)=e^x. I also have to state the placement of the horizontal asymptote and y-intercept after the transformation. I'm lost and do not understand. The first transformation is g(x)=e^x-5 and the second is h(x)= -e^x

I cannot tell whether that first one is 

g%28x%29+=+e%5E%28x-5%29 or g%28x%29+=+e%5Ex-5

So I'll do it both ways.

Here are the transformation rules:

Assume +k is a positive number and -k is a 
negative number.

1. Vertical shifts: 

A. Adding +k to the right side of the equation of
   a function shifts its graph UPWARD by k units.

B. Adding -k to the right side of the equation of
   a function shifts its graph DOWNWARD by k units.

2. Horizontal shifts:

A. Replacing x in the right side of the equation of
   a function by x+k shifts its graph to the LEFT by
   k units.

B. Replacing x in the right side of the equation of
   a function by x-k shifts its graph to the RIGHT by
   k units.

3. Reflecting across axes:

A. Multiplying the right side of the equation of a
function through by -1 reflects its graph in (or
across) the x-axis  

B. Replacing x in the right side of the equation of
a function by -x reflects its graph in (or across) 
the y-axis  

--------------------
f%28x%29+=+e%5Ex has y-intercept (0,1) and asymptote the
vertical line y=0 which is the x-axis.

--------------------

f%28x%29+=+e%5Ex to
g%28x%29+=+e%5E%28x-5%29

Here g(x) is formed by replacing x in the right side of f(x)
by x-5, so this is case 2B above, and thus the graph of g(x)
is the graph of f(x) shifted to the RIGHT by 5 units.  

The y-intercept is shifted from (0,1) 5 units RIGHT to the
point (5,1).  The horizontal asymptote is not affected when 
shifting right, so it is still y=0

--------------------

f%28x%29+=+e%5Ex
g%28x%29+=+e%5Ex-5

Here g(x) is formed by adding -5 to the right side of f(x),
so this is case 1B above, and thus the graph of g(x)
is the graph of f(x) shifted DOWNWARD by 5 units.

The y-intercept is shifted from (0,1) 5 units DOWNWARD to the
point (0,-4).  The horizontal asymptote is shifted DOWNWARD
from y=0 (the x-axis) to the horizontal line y=-5.
  

--------------------

f%28x%29+=+e%5Ex
h%28x%29+=+-e%5Ex

Here h(x) is formed by multiplying the right side of f(x)
through by -1, a case of 3A. This reflects its graph in 
(or across) the x-axis.  The y-intercept (0,1) is reflected 
to its "image point" (0,-1) [thinking of the x-axis as a 
mirror].  The asymptote y=0 (the x-axis) is not affected by
a reflection in the x-axis.

Edwin