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Question 623937: What is the domain and range of y = x+2/2x+1 ?
the answer must be in {x/x blah blah} form - don't know the name XD
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! equation is y = (x+2) / (2x+1)
your domain can be any value of x except x = -1/2.
when x = -1/2, the denominator of (2x+1) becomes (2*-1/2 + 1) which becomes (-1 + 1) which becomes 0.
this is a nono.
so, your domain is the set of all real numbers except x = -1/2
this would be the set of x such that x is an element of the real number system and x is not equal to -1/2.
in set notation, that would be shown as:
domain = {x|x element of R, x <> -1/2}
i think that's what you're referring to.
here's a reference on set notation.
http://www.purplemath.com/modules/setnotn.htm
since the equation is:
y = (x+2) / (2x+1)
when x is not equal to -1/2, the range is defined as the set of all real numbers since there are no restrictions on what the value of y could be.
in set notation, you get:
range = {y|y element R|
a graph of your equation is shown below:
looking at this graph, i see that it appears to have a vertical asymptote as x = -1/2.
this i expected.
the graph, however, also appear to have a horizontal asymptote at y = 1/2.
i drew a horizontal line at y = 1/2 to confirm that.
the graph now looks like this:
i also drew a vertical line at x = -1/2 to confirm the vertical asymptore.
the graph looks like this:
the vertical line is not perfectly vertical because the graphing software doesn't allow that, but it's pretty close to vertical and enough for you to see that there is a vertical asymptore at x = -1/2 as calculated earlier.
the domain is still:
domain = {x|x element of R, x <> -1/2}
the range becomes:
range = {y|y element of R, y <> 1/2}
you can never exactly equal the value of each asymptote. you can only approach it from either side, getting closer and closer, but never equal.
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