anonymous
  • anonymous
Piecewise function question
Mathematics
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anonymous
  • anonymous
Piecewise function question
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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misty1212
  • misty1212
\[f(x) = |x+4| = \left\{\begin{array}{rcc} x + 4 & \text{if} & x \geq -4 \\ - x - 4& \text{if} & x < -4 \end{array} \right. \]
anonymous
  • anonymous
yes i have a graph that i really need help on
anonymous
  • anonymous
could you please hhelp me out

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misty1212
  • misty1212
you need to graph something?
anonymous
  • anonymous
no ive got a graph and have some questions about it
misty1212
  • misty1212
ok i can try to help, post the picture
anonymous
  • anonymous
i can post a picture of the worksheet so it'll be easier to see the graph
anonymous
  • anonymous
okay pleasegive me a sec
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
this is the graph and the first question is f(1)=
anonymous
  • anonymous
and i got 3 as the answer
misty1212
  • misty1212
oh i thought you had to find the function
anonymous
  • anonymous
i do later on
misty1212
  • misty1212
yeah \(f(1)=3\)
anonymous
  • anonymous
and the next one is f(-11)
anonymous
  • anonymous
but on the graph its only till -10
misty1212
  • misty1212
then we better find the function for \(x\leq -5\)
misty1212
  • misty1212
you got the slope right, it is \(\frac{1}{2}\)
anonymous
  • anonymous
yes that is the slope
misty1212
  • misty1212
goes through the point \((-5,3)\) so it is \[y+5=\frac{1}{2}(x-3)\]
misty1212
  • misty1212
you can plug 11 in to that if you like
anonymous
  • anonymous
oh alright that makes sense
misty1212
  • misty1212
or else write it as \[y=\frac{1}{2}(x-3)+5\]
anonymous
  • anonymous
is it -2
misty1212
  • misty1212
whoah did i screw that up!!
anonymous
  • anonymous
im not sure haha
misty1212
  • misty1212
slope is \(\frac{1}{2}\) point is \((-5,3)\) equation is \[y-3=\frac{1}{2}(x+5)\] doe
anonymous
  • anonymous
so its y= 1/2(x+5)+3?
misty1212
  • misty1212
or \[y=\frac{x}{2}+\frac{11}{2}\]
misty1212
  • misty1212
whatever you like
anonymous
  • anonymous
alright let me solve one sec
misty1212
  • misty1212
cool henna btw
anonymous
  • anonymous
thank you :)
misty1212
  • misty1212
did you get zero yet?
anonymous
  • anonymous
yepp
anonymous
  • anonymous
okay so the next one says find x when f(x)=0
anonymous
  • anonymous
is it 5/2
misty1212
  • misty1212
lol you just did it
anonymous
  • anonymous
im not supposed to use decimals
misty1212
  • misty1212
you found \(f(-11)=0\) right?
anonymous
  • anonymous
yes i did
misty1212
  • misty1212
so that makes "find x when f(x)=0" rather obvious !!
anonymous
  • anonymous
lol idk why i put 5/2
misty1212
  • misty1212
plug in \(-11\) get out zero !!
misty1212
  • misty1212
me neither lol
anonymous
  • anonymous
i got it!
misty1212
  • misty1212
ok next...
anonymous
  • anonymous
alright then lol on to the next one
anonymous
  • anonymous
i have to find the x intercepts
misty1212
  • misty1212
ok we got one right? one is \(-11\)
anonymous
  • anonymous
and i got (5/2,0) and (-11,0)
misty1212
  • misty1212
there is another as well, because it is coming back up at the end and that little arrow indicates it continues up
misty1212
  • misty1212
lets just find the damn function and be done with it k?
anonymous
  • anonymous
lol alright
anonymous
  • anonymous
it actually says to write in point slope form and restricted domain in inequality form
anonymous
  • anonymous
so basically its asking me to write a fuction for each piece
misty1212
  • misty1212
right so far we are at \[f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}x+5 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.\]
misty1212
  • misty1212
what is the slope of the next piece? it is a little hard for me to see it
misty1212
  • misty1212
maybe \(-2\)?
anonymous
  • anonymous
you said in the first function theres a 3 ?
anonymous
  • anonymous
under where you write the 5 in the first equation
misty1212
  • misty1212
it is a constant (horizontal line) between -5 and 1 right?
anonymous
  • anonymous
yes thats correct
misty1212
  • misty1212
ooh damn i made a typo there forgot the parentheses
misty1212
  • misty1212
\[f(x) = \left\{\begin{array}{rcc} y-3=\frac{1}{2}(x+5) & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1 \end{array} \right.\]
anonymous
  • anonymous
im a little confused
misty1212
  • misty1212
ok lets go slow and make sure you get it
anonymous
  • anonymous
under the first equation you put a 3
anonymous
  • anonymous
but im not sure why
misty1212
  • misty1212
yeah we are not done you see how you have 4 long lines in your answer space?
misty1212
  • misty1212
because this piecewise function has four distinct parts
anonymous
  • anonymous
yes i do
misty1212
  • misty1212
4 different lines
anonymous
  • anonymous
okay yes i understand now thank you :)
misty1212
  • misty1212
on the interval from \(-5\) to \(1\) the function is a constant, it is \(f(x)=3\) a horizontal line
anonymous
  • anonymous
yes
misty1212
  • misty1212
so the second line down on the left you should have \(3,-5\leq x\leq 1\) because on that interval it is 3
anonymous
  • anonymous
i understand
anonymous
  • anonymous
oh okay
misty1212
  • misty1212
the next line has slope \(-2\) and goes through \((1,3)\) so its equation is \[y=y-3=-2(x-1)\] in point slope form
anonymous
  • anonymous
would the slope not be a 0?
misty1212
  • misty1212
yes, the slope is zero, the equation is \(y=3\)
anonymous
  • anonymous
okay i got it
anonymous
  • anonymous
okay i got it
misty1212
  • misty1212
\[f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1
misty1212
  • misty1212
one more to go
anonymous
  • anonymous
okay i got it
misty1212
  • misty1212
last one also has slope \(\frac{1}{2}\) right?
anonymous
  • anonymous
yes it does
misty1212
  • misty1212
and goes through the point \((6,-7\)?
anonymous
  • anonymous
yes i think so
misty1212
  • misty1212
\[f(x) = \left\{\begin{array}{rcc} \frac{1}{2}(x+5)+3 & \text{if} & x <-5 \\ 3& \text{if} & -5\leq x\leq 1\\ -2(x-1)+3 &\text{if}&1
misty1212
  • misty1212
that should do it
anonymous
  • anonymous
yes i got it i just have a couple of small questions that i wanted to confirm
misty1212
  • misty1212
ok
anonymous
  • anonymous
are the x intercepts (5/2,0) and (-11,0)?
misty1212
  • misty1212
yes, but there is one more
misty1212
  • misty1212
set \[\frac{1}{2}(x-7)-7=0\] for to find the last one
misty1212
  • misty1212
aggh!!
misty1212
  • misty1212
\[\frac{1}{2}(x-6)-7=0\]
misty1212
  • misty1212
that is the part on the right think you get \(20\) if i did it right
anonymous
  • anonymous
i was solving it one sec ill confirm
anonymous
  • anonymous
yes i got 20
anonymous
  • anonymous
so itll be (20,0)
misty1212
  • misty1212
yeah
anonymous
  • anonymous
okay and the range is (-infinity, infintiy)?
misty1212
  • misty1212
yeah that is what the arrows indicate
anonymous
  • anonymous
and the positives are (-11, 5/2)?
anonymous
  • anonymous
thast the last one haha
misty1212
  • misty1212
i don't know what that means
misty1212
  • misty1212
oh you mean the interval over which it is positive ?
anonymous
  • anonymous
the positives as in the parts in the upper part of the graph
anonymous
  • anonymous
yes
misty1212
  • misty1212
\[(-11,\frac{5}{2})\] is one interval but there is another
misty1212
  • misty1212
since it is zero at 20, and it keeps going up, the other one is \((20,\infty)\)
anonymous
  • anonymous
alrighty thank you so much
anonymous
  • anonymous
im just putting the functions in slope intercept form now and the doamins in interval notation
anonymous
  • anonymous
ill be done in a sec if you have time to check those
misty1212
  • misty1212
ok good sure
anonymous
  • anonymous
y= 1/2x+11/2
anonymous
  • anonymous
y=3 y=-2x+5 y=1/2x-10
misty1212
  • misty1212
slope intercept for the first one yes
anonymous
  • anonymous
yes i am doing slope intercept for all of them
misty1212
  • misty1212
ok let me scroll way up and see i know the first two are right
misty1212
  • misty1212
third is right too
anonymous
  • anonymous
okay annddd the last one
misty1212
  • misty1212
yeah that also looks good
anonymous
  • anonymous
okay and the interval notation is (infinity, -5)? for the first one?
anonymous
  • anonymous
[-5,-1] (1,6) (infinity, 6]
misty1212
  • misty1212
ooh mistake there
misty1212
  • misty1212
second interval should be \([-5,1]\)
misty1212
  • misty1212
and also do not put infinity on the left on the last interval!!
anonymous
  • anonymous
i think thats what i said for the second one
misty1212
  • misty1212
should be on the right, go from least to greatest i.e. \((6,\infty)\)
anonymous
  • anonymous
is the first interbal correct?
misty1212
  • misty1212
\((-\infty,-5)\) yes
anonymous
  • anonymous
and for the last one do i use a bracket before 6 since its greater than or equal to
misty1212
  • misty1212
since it meets up there it is your choice
misty1212
  • misty1212
btw usually these piecewise functions are written with inequalities not intervals but if you are supposed to use intervals that is fine too
anonymous
  • anonymous
yeah it asked us to write it nboth wwyas
anonymous
  • anonymous
do you think i should use a bracket?
misty1212
  • misty1212
kk makes no difference
anonymous
  • anonymous
alrigh thank you :)
misty1212
  • misty1212
yw ms kumar lol \[\huge \color\magenta\heartsuit\]

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