Question

PLEASE, Can someone please help me? @pooja195 @michele_laino

Answer 1

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.

Answer 2

the quadratic on the numerator can be factorised

Answer 3

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

Answer 4

for restricted domain we have to find x such that f(x) is one-one

Answer 5

the domain is (-infinity,-5)U (-5,infinity)

Answer 6

I mean the restricted domain

Answer 7

ok, so how do i graph it?

Answer 8

could you show me please?

Answer 9

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

Answer 10

its just a straight line y = x -1 but place a small circle at x = -5, y = -6 where it is undefined

Answer 11

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?

Answer 12

yeah but omit the point x=-5 as alekos said

Answer 13

sorry i got lost, how'd you get 6?
can you show me what i would put on the graph? please. sorry

Answer 14

When x = -5 then y = -6 because we have the line y = x -1 so that's the point that's undefined (-5,-6)

Answer 15

draw the line y = x -1 and draw a small circle at the point (-5, -6)

Answer 16

get it now?

Answer 17

is there a way to show me on an actual graph on here?
please

Answer 18

my circle on (-5,-6) would be open right?

Answer 19

yes. it should be an open circle

Answer 20

that should give you some idea

Answer 21

would i put a closed circle on the (1,0)

Answer 22

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

Answer 23

@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.

Answer 24

@mathmate, so my only domain restriction would be the -5, right?

Answer 25

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.

Answer 26

dom f: \((-\infty,-5)\cup (-5,+\infty)\)

Answer 27

there are no closed circles on the graph
just the open circle as i've indicated