Question

PRECALCULUS HELP !

domain & range of rational function

2
f(x) = ---------
x^2-2x-3

i got domain = all real x except -1 and 3. i dont know how to find range please help

Answer 1

it has a horizontal asymptote at y=0 so y > 0 is part of range

Answer 2

but i dont know if thats correct way

Answer 3

To find the horizontal asymptotes, you look at:
\[\frac{\text{Degree of Numerator}}{\text{Degree of Denominator}}=\frac{0}{2}=0\]
So there is a horizontal asymptote at \(y=0\). Thus this is not included in the range, and the range is:
\[y\in(-\infty,0)\cup (0,\infty) \]
Does that make sense?

Answer 4

yes i thought the same initially... but it has vertical asymptotes also, when i graphed it in geogebra i get confuse because in the middle it is stopping at -1/2

Answer 5

i have been struggling to figure out how to get -1/2 as upper limit in range

y > 0 or, y <= -1/2

http://www.wolframalpha.com/input/?i=2%2F%28x%5E2-2x-3+%29

Answer 6

Sorry, it's early I seem to have made a mistake. This document may help you out:
http://cims.nyu.edu/~kiryl/Precalculus/Section_3.7-Rational%20Functions/Rational%20Functions.pdf
I'll have a look and see if I can explain it at all.

Answer 7

its okay... im going through the pdf thnks for the help xD

Answer 8

@sara12345
Can you show that -1/2 is the *only* relative maximum of the function? We know the function cannot have a value of 0. That will leave a gap in the Real range between [-1/2 , 0). Just thinking.

Answer 9

i got y-intercept at -2/3, wil it be any use ?

Answer 10

I don't think so unless you can show -2/3 is a relative maximum.
Can you get -1/2 as the relative maximum?

Answer 11

sorry m not sure what is relative maximum

Answer 12

to find the range put x=y in the equation you and then solve

Answer 13

2
-------
y^2-2y-3

Answer 14

@sara12345
See attached file.

Answer 15

oh yes that maximum only geogebra shows at -1/2, how to find it ?