SOLUTION: Graph the function, not by plotting points, but by starting from the graph of y = e^x. State the domain, range, and asymptote. y = e^(x+3) + 4 Shift down 4 units? Thanks

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Question 47735This question is from textbook College Algebra
: Graph the function, not by plotting points, but by starting from the graph of y = e^x. State the domain, range, and asymptote.
y = e^(x+3) + 4
Shift down 4 units?
Thanks for helping me. I feel so stupid when it comes to math, but you all are showing me the right steps to learn it!
This question is from textbook College Algebra

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+e%5E%28x%2B3%29+%2B+4
The +4 has the effect of shifting all the y values up 4 units. Every
single y value gets 4 added to it.
rewite the equation, noting that x%5E%28a%2Bb%29+=+x%5Ea%2Ax%5Eb
y+=+e%5E3%2Ae%5Ex+%2B+4
e^3 is a constant, so k+=+e%5E3 and rewrite as
y+=+k%2Ae%5Ex+%2B4
Note that y can never be negative, so you can forget about everthing
below the x-axis. No part of the curve is there. In fact, y can never
be less than +4 because that means k%2Ae%5Ex would have to be some
negative value, and that is impossible, so forget about everything below
the line y+=+%2B4. No part of the curve can be there.
As x become more and more negative and approaches minus infinity,
e%5Ex approaches zero, so k%2Ae%5Ex approaches zero, and
y approaches +4. y+=+%2B4 is an asymptote.
find (0 , y)
------------------------
y+=+k%2Ae%5Ex+%2B4
y+=+k%2Ae%5E0+%2B4
y+=+k+%2B+4
y+=+e%5E3+%2B+4
find (-3 , y)
------------------------
y+=+k%2Ae%5Ex+%2B4
y+=+k%2Ae%5E%28-3%29+%2B4
y+=+e%5E3%2Ae%5E%28-3%29+%2B+4
y+=+e%5E0+%2B+4
y+=+1+%2B+4
y+=+5
Using my calculator, e%5E3 is about 20.09, so the points found
are (0,24.09) and (-3,5)
The range is + infinity > y > +4
The domain is + infinity > x > - infinity