Piecewise function question

- anonymous

Piecewise function question

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- 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

yes i have a graph that i really need help on

- anonymous

could you please hhelp me out

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## More answers

- misty1212

you need to graph something?

- anonymous

no ive got a graph and have some questions about it

- misty1212

ok i can try to help, post the picture

- anonymous

i can post a picture of the worksheet so it'll be easier to see the graph

- anonymous

okay pleasegive me a sec

- anonymous

##### 1 Attachment

- anonymous

this is the graph and the first question is f(1)=

- anonymous

and i got 3 as the answer

- misty1212

oh i thought you had to find the function

- anonymous

i do later on

- misty1212

yeah \(f(1)=3\)

- anonymous

and the next one is f(-11)

- anonymous

but on the graph its only till -10

- misty1212

then we better find the function for \(x\leq -5\)

- misty1212

you got the slope right, it is \(\frac{1}{2}\)

- anonymous

yes that is the slope

- misty1212

goes through the point \((-5,3)\) so it is
\[y+5=\frac{1}{2}(x-3)\]

- misty1212

you can plug 11 in to that if you like

- anonymous

oh alright that makes sense

- misty1212

or else write it as
\[y=\frac{1}{2}(x-3)+5\]

- anonymous

is it -2

- misty1212

whoah did i screw that up!!

- anonymous

im not sure haha

- misty1212

slope is \(\frac{1}{2}\) point is \((-5,3)\) equation is
\[y-3=\frac{1}{2}(x+5)\] doe

- anonymous

so its y= 1/2(x+5)+3?

- misty1212

or
\[y=\frac{x}{2}+\frac{11}{2}\]

- misty1212

whatever you like

- anonymous

alright let me solve one sec

- misty1212

cool henna btw

- anonymous

thank you :)

- misty1212

did you get zero yet?

- anonymous

yepp

- anonymous

okay so the next one says find x when f(x)=0

- anonymous

is it 5/2

- misty1212

lol you just did it

- anonymous

im not supposed to use decimals

- misty1212

you found \(f(-11)=0\) right?

- anonymous

yes i did

- misty1212

so that makes "find x when f(x)=0" rather obvious !!

- anonymous

lol idk why i put 5/2

- misty1212

plug in \(-11\) get out zero !!

- misty1212

me neither lol

- anonymous

i got it!

- misty1212

ok next...

- anonymous

alright then lol on to the next one

- anonymous

i have to find the x intercepts

- misty1212

ok we got one right? one is \(-11\)

- anonymous

and i got (5/2,0) and (-11,0)

- misty1212

there is another as well, because it is coming back up at the end and that little arrow indicates it continues up

- misty1212

lets just find the damn function and be done with it k?

- anonymous

lol alright

- anonymous

it actually says to write in point slope form and restricted domain in inequality form

- anonymous

so basically its asking me to write a fuction for each piece

- 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

what is the slope of the next piece? it is a little hard for me to see it

- misty1212

maybe \(-2\)?

- anonymous

you said in the first function theres a 3 ?

- anonymous

under where you write the 5 in the first equation

- misty1212

it is a constant (horizontal line) between -5 and 1 right?

- anonymous

yes thats correct

- misty1212

ooh damn i made a typo there forgot the parentheses

- 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

im a little confused

- misty1212

ok lets go slow and make sure you get it

- anonymous

under the first equation you put a 3

- anonymous

but im not sure why

- misty1212

yeah we are not done
you see how you have 4 long lines in your answer space?

- misty1212

because this piecewise function has four distinct parts

- anonymous

yes i do

- misty1212

4 different lines

- anonymous

okay yes i understand now thank you :)

- misty1212

on the interval from \(-5\) to \(1\) the function is a constant, it is \(f(x)=3\) a horizontal line

- anonymous

yes

- 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

i understand

- anonymous

oh okay

- 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

would the slope not be a 0?

- misty1212

yes, the slope is zero, the equation is \(y=3\)

- anonymous

okay i got it

- anonymous

okay i got it

- 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

one more to go

- anonymous

okay i got it

- misty1212

last one also has slope \(\frac{1}{2}\) right?

- anonymous

yes it does

- misty1212

and goes through the point \((6,-7\)?

- anonymous

yes i think so

- misty1212

- misty1212

that should do it

- anonymous

yes i got it i just have a couple of small questions that i wanted to confirm

- misty1212

ok

- anonymous

are the x intercepts (5/2,0) and (-11,0)?

- misty1212

yes, but there is one more

- misty1212

set
\[\frac{1}{2}(x-7)-7=0\] for to find the last one

- misty1212

aggh!!

- misty1212

\[\frac{1}{2}(x-6)-7=0\]

- misty1212

that is the part on the right
think you get \(20\) if i did it right

- anonymous

i was solving it one sec ill confirm

- anonymous

yes i got 20

- anonymous

so itll be (20,0)

- misty1212

yeah

- anonymous

okay and the range is (-infinity, infintiy)?

- misty1212

yeah that is what the arrows indicate

- anonymous

and the positives are (-11, 5/2)?

- anonymous

thast the last one haha

- misty1212

i don't know what that means

- misty1212

oh you mean the interval over which it is positive ?

- anonymous

the positives as in the parts in the upper part of the graph

- anonymous

yes

- misty1212

\[(-11,\frac{5}{2})\] is one interval but there is another

- misty1212

since it is zero at 20, and it keeps going up, the other one is \((20,\infty)\)

- anonymous

alrighty thank you so much

- anonymous

im just putting the functions in slope intercept form now and the doamins in interval notation

- anonymous

ill be done in a sec if you have time to check those

- misty1212

ok good sure

- anonymous

y= 1/2x+11/2

- anonymous

y=3
y=-2x+5
y=1/2x-10

- misty1212

slope intercept for the first one yes

- anonymous

yes i am doing slope intercept for all of them

- misty1212

ok let me scroll way up and see
i know the first two are right

- misty1212

third is right too

- anonymous

okay annddd the last one

- misty1212

yeah that also looks good

- anonymous

okay and the interval notation is
(infinity, -5)? for the first one?

- anonymous

[-5,-1]
(1,6)
(infinity, 6]

- misty1212

ooh mistake there

- misty1212

second interval should be \([-5,1]\)

- misty1212

and also do not put infinity on the left on the last interval!!

- anonymous

i think thats what i said for the second one

- misty1212

should be on the right, go from least to greatest i.e. \((6,\infty)\)

- anonymous

is the first interbal correct?

- misty1212

\((-\infty,-5)\) yes

- anonymous

and for the last one do i use a bracket before 6 since its greater than or equal to

- misty1212

since it meets up there it is your choice

- 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

yeah it asked us to write it nboth wwyas

- anonymous

do you think i should use a bracket?

- misty1212

kk
makes no difference

- anonymous

alrigh thank you :)

- misty1212

yw ms kumar lol \[\huge \color\magenta\heartsuit\]

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