anonymous
  • anonymous
PLEASE, Can someone please help me? @pooja195 @michele_laino
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
2. For the function, : c(x) = (x^2 +4x - 5)/(x + 5) (1) Identify the restricted domain value(s). (2) Simplify the function. Show your work. (3) Graph the function.
alekos
  • alekos
the quadratic on the numerator can be factorised
anonymous
  • anonymous
I've already factored the numerator and got (x-1)(x+5)/(x+5) afterwards, I crossed out the like terms and am left with just x-1

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jango_IN_DTOWN
  • jango_IN_DTOWN
for restricted domain we have to find x such that f(x) is one-one
jango_IN_DTOWN
  • jango_IN_DTOWN
the domain is (-infinity,-5)U (-5,infinity)
jango_IN_DTOWN
  • jango_IN_DTOWN
I mean the restricted domain
anonymous
  • anonymous
ok, so how do i graph it?
anonymous
  • anonymous
could you show me please?
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah sure... see the function cannot take x=-5 or else the denominator is 0. When x is not equal to -5, the function reduces to x-1, which is a one-one function
alekos
  • alekos
its just a straight line y = x -1 but place a small circle at x = -5, y = -6 where it is undefined
anonymous
  • anonymous
so would it just be a straight line on my graph running through the points 1,0 for the x axis and -1,0 for the y axis?
jango_IN_DTOWN
  • jango_IN_DTOWN
yeah but omit the point x=-5 as alekos said
anonymous
  • anonymous
sorry i got lost, how'd you get 6? can you show me what i would put on the graph? please. sorry
alekos
  • alekos
When x = -5 then y = -6 because we have the line y = x -1 so that's the point that's undefined (-5,-6)
alekos
  • alekos
draw the line y = x -1 and draw a small circle at the point (-5, -6)
alekos
  • alekos
get it now?
anonymous
  • anonymous
is there a way to show me on an actual graph on here? please
anonymous
  • anonymous
my circle on (-5,-6) would be open right?
alekos
  • alekos
|dw:1446926660236:dw|
alekos
  • alekos
yes. it should be an open circle
alekos
  • alekos
that should give you some idea
anonymous
  • anonymous
would i put a closed circle on the (1,0)
anonymous
  • anonymous
so for 1, I would put x cannot equal 1, -5 or no? I'm sorry if it seems like i'm not paying attention
mathmate
  • mathmate
@tristian23 In the future, I suggest whenever you \(cancel\) common factors (like (x+5) in the present case), you write down on the side the restriction x+5\(\ne\)0, or equivalently x\(\ne\)-5. This way, by the end of your calculations, you have a list of restrictions, and you will also know where to put the open circle, if applicable.
anonymous
  • anonymous
@mathmate, so my only domain restriction would be the -5, right?
mathmate
  • mathmate
exact, x\(\ne\)-5 is the only restriction, which explains the previous response of dom f : \((-\infty,5)\cup (5,+\infty)\) which means the same as \(R\)\5.
mathmate
  • mathmate
dom f: \((-\infty,-5)\cup (-5,+\infty)\)
alekos
  • alekos
there are no closed circles on the graph just the open circle as i've indicated

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