At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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.
the quadratic on the numerator can be factorised
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
for restricted domain we have to find x such that f(x) is one-one
the domain is (-infinity,-5)U (-5,infinity)
I mean the restricted domain
ok, so how do i graph it?
could you show me please?
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
its just a straight line y = x -1 but place a small circle at x = -5, y = -6 where it is undefined
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?
yeah but omit the point x=-5 as alekos said
sorry i got lost, how'd you get 6? can you show me what i would put on the graph? please. sorry
When x = -5 then y = -6 because we have the line y = x -1 so that's the point that's undefined (-5,-6)
draw the line y = x -1 and draw a small circle at the point (-5, -6)
get it now?
is there a way to show me on an actual graph on here? please
my circle on (-5,-6) would be open right?
yes. it should be an open circle
that should give you some idea
would i put a closed circle on the (1,0)
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
@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.
@mathmate, so my only domain restriction would be the -5, right?
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.
dom f: \((-\infty,-5)\cup (-5,+\infty)\)
there are no closed circles on the graph just the open circle as i've indicated