SOLUTION: Seeking assistance. Describe transformations of the graph of f(x)=e^x. Note horizontal asymptote and y-intercept after the transformation. a)g(x) = e^x + 3 Graph: vertical

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Seeking assistance. Describe transformations of the graph of f(x)=e^x. Note horizontal asymptote and y-intercept after the transformation. a)g(x) = e^x + 3 Graph: vertical       Log On


   

Question 159204: Seeking assistance.
Describe transformations of the graph of f(x)=e^x. Note horizontal asymptote and y-intercept after the transformation.
a)g(x) = e^x + 3
Graph: vertical shift of 3 units up.
Equation for horizontal asymptote: x = 2.72 + 2, 4.72
y-intercept: (0,3)
b)h(x) = -e^x
Graph: reflection around y axis
Equation: -1e^x = 0, e^x = 1
y-intercept: (1,0)
The above is what has been attempted.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you started

a)

a)g(x) = e^x + 3


Graph: vertical shift of 3 units up. Correct.


Equation for horizontal asymptote: When "x" is a very large negative number, then e%5Ex becomes really small (and eventually approaches zero).

So what's left over is the term "3". So the horizontal asymptote is y=3


y-intercept:

g%28x%29=e%5Ex%2B3 Start with the given function


g%280%29=e%5E0%2B3 Plug in x=0


g%280%29=1%2B3 Raise "e" to the zeroth power to get 1


g%280%29=4 Add


So when x=0 then y=4 (g(x) and y are interchangeable)


So the y-intercept is (0,4)


Here's a graph to verify the answer.

Graph of g%28x%29=e%5Ex%2B3


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b)

Use the techniques used above to solve this problem.