Question

5-8 Graphing absolute value functions.

36. Y = |x| + 2 37. Y= |x| -7

38. Y = |x + 3| 39. Y = |x – 5|

Write an equation for each translation of y = |x|

40. 5.5 units down. 41. 11 units left

42. 13 units up 43. 6.5 units right

44. Write an equation for the absolute value function at the right.
PLEASE HELP.

Answer 1

Rules of \(\large\color{black}{ \rm shifts }\) from \(\large\color{black}{ \rm f(x) }\) to \(\large\color{black}{ \rm g(x) }\).


\(\large\color{black}{ \rm f(x)=a\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=a\left| x \color{blue}{ -~\rm{c} }\right| }\)
\(\large\color{blue}{ ~\rm {c} }\) units to the \(\normalsize\color{blue}{ \rm right }\).


\(\large\color{black}{ \rm f(x)=a\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=a\left| x \color{blue}{ +~\rm{c} }\right| }\)
\(\large\color{blue}{ ~\rm {c} }\) units to the \(\normalsize\color{blue}{ \rm left }\).


\(\large\color{black}{ \rm f(x)=a\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=a\left| x \right| \color{blue}{ +~\rm{c} }}\)
\(\large\color{blue}{ ~\rm {c} }\) units \(\normalsize\color{blue}{ \rm up }\).


\(\large\color{black}{ \rm f(x)=a\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=a\left| x \right| \color{blue}{ -~\rm{c} }}\)
\(\large\color{blue}{ ~\rm{c} }\) units \(\normalsize\color{blue}{ \rm down }\).



Also,
the ` reflection across the X -axis. `

\(\large\color{red}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\color{blue}{ - }\left| x \right| }\)
(the \(\large\color{red}{ ~\rm{f(x)} }\) and \(\normalsize\color{red}{ \rm g(x) }\) are mirrors of each other over the \(\large\color{red}{ \rm{x-axis} }\). ) \(\LARGE\color{white}{ \rm │ }\)



And lastly,
\(\normalsize\color{black}{ \rm{ s~t~r~e~t~c~h~i~n~g} }\)

\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\color{blue}{ c }\left| x \right| }\)

For any real number \(\normalsize\color{blue}{ \rm{c} }\), (provided that \(\normalsize\color{blue}{ \rm{c\neq1~~or~~0} }\) )

\(\normalsize\color{black}{ \rm{1)} }\) When \(\normalsize\color{blue}{ \rm{\left| c \right| >1} }\) the (new function) \(\normalsize\color{black}{ \rm{g(x)} }\) is streched \(\normalsize\color{blue}{ \rm{ vertically} }\).
(if comparing to the initial function \(\normalsize\color{black}{ \rm{f(x)} }\). )

\(\normalsize\color{black}{ \rm{2)} }\) When \(\normalsize\color{blue}{ \rm{\left| c \right| <1} }\) the (new function) \(\normalsize\color{black}{ \rm{g(x)} }\) is streched \(\normalsize\color{blue}{ \rm{ horizontally} }\).
(if comparing to the initial function \(\normalsize\color{black}{ \rm{f(x)} }\). )

Answer 2

do you see weird codes or normal text?

Answer 3

Normal text.

Answer 4

the codes aren't working for me for some reason, so it loks like a fine set of rules, correct?

Answer 5

if you have some questions still, then ask away.

Answer 6

Yes it does.

Answer 7

the codes aren't working for me for some reason, so it loks like a fine set of rules, correct?

Answer 8

if you have some questions still, then ask away.

Answer 9

Still kind of confused about this, just so new to this subject.

Answer 10

you can see that you are adding +2 (in the first problem) right?
If you know what a graph of \(\large\color{black}{ y=\left| x \right| }\) is like, then shift it accoring to the rules I posted.

Answer 11

so number 1 will be -2?

Answer 12

no not minus 2, although you are getting the 2 correctly.

going from, \(\large\color{black}{ y= \left| x \right| }\), the function, \(\large\color{black}{ y=\left| x \right| \color{blue}{+2} }\) is shifted up, correct?

Answer 13

I guess.

Answer 14

looking at the third rule, your "a" is just 1, so we can re-write it as:

\(\large\color{black}{ \rm f(x)=\left| x \right| ~~~~~~~~~\rm{\Longrightarrow}~~~~~~~~\rm g(x)=\left| x \right| \color{blue}{ +~\rm{c} }}\)
\(\large\color{blue}{ ~\rm {c} }\) units \(\normalsize\color{blue}{ \rm up }\).
you see that your "+2", is the "+c" here?

Answer 15

Yes I see that.

Answer 16

would my answer be g (x) = x + 2

Answer 17

you mean, \(\large\color{black}{ y= \left| x \right| +2 }\) ?
and don't you have to graph it though?

are you allowed to use a graphing calculator or not?

Answer 18

Yes I am, and yeah i mean that just didn't punch the letters in. and i'm not really sure, its a chapter review on an online school..

Answer 19

okay, lets start from graphing the parent function, \(\large\color{black}{ y= \left| x \right| }\), do you know how to do this, or want me to help you doing this?

Answer 20

Need help. I mostly learn from a teacher, kinda hard with online school not having that extra help. please

Answer 21

okay, lets plug in some values.

you will need to tell me the following:

\(\large\color{black}{ \bullet }\) \(\large\color{black}{ \left| ~-1 ~\right| =? }\)
\(\large\color{black}{ \bullet }\) \(\large\color{black}{ \left| ~1 ~\right| =? }\)

\(\large\color{black}{ \bullet }\) \(\large\color{black}{ \left| ~-2 ~\right| =? }\)
\(\large\color{black}{ \bullet }\) \(\large\color{black}{ \left| ~2 ~\right| =? }\)

tell me all of the four, what they are equal to, receptively please.

Answer 22

you know what \(\large\color{black}{ \left| ~-1 ~\right| }\) is equal to, right?

Answer 23

Is it -1?

Answer 24

no, the absolute value is always positive.
because absolute value is basically the distance.

the distance from 0 to -1, is the same as the distance from 0 to 1. Correct?

So \(\large\color{black}{ \left| ~-1 ~\right| =1 }\)

(just like for any positive number \(\large\color{black}{ c }\),
\(\large\color{black}{ \left| ~-c~\right| =c }\) )

Answer 25

So you can say that: \(\large\color{black}{ \left| ~\pm1 ~\right| =1 }\)
(saying that \(\large\color{black}{ \left| ~-1 ~\right| }\) and \(\large\color{black}{ \left| 1 ~\right| }\) are both equal to \(\large\color{black}{ 1 }\) )

Answer 26

Oh okay! so 1=1,- 2=2?

Answer 27

Yeah

Answer 28

so can you tell me what \(\large\color{black}{ \left| ~-3 ~\right| }\) will be?

Answer 29

3.

Answer 30

Yes that is correct

Answer 31

So we can say the following statements: (I'll use a blue color)

\(\large\color{blue}{ \left| ~\pm1 ~\right| =1 }\)

\(\large\color{blue}{ \left| ~\pm2 ~\right| =2 }\)

\(\large\color{blue}{ \left| ~\pm3~\right| =3 }\)

\(\large\color{blue}{ \left| ~\pm4 ~\right| =4 }\)


yes?

Answer 32

and on...

Answer 33

Yep.

Answer 34

so when we have: \(\large\color{blue}{ y=\left| x~\right| }\)
let's plug in values:

when \(\large\color{blue}{ x=-1 }\), then \(\large\color{blue}{ y=1 }\).
when \(\large\color{blue}{ x=1 }\), then \(\large\color{blue}{ y=1 }\).

when \(\large\color{blue}{ x=-2 }\), then \(\large\color{blue}{ y=2 }\).
when \(\large\color{blue}{ x=2 }\), then \(\large\color{blue}{ y=2 }\).

Answer 35

so for number one would it be x=5.5 y = 5.5?

Answer 36

I mean number 40- oops

Answer 37

no, for your problem 36 the graph is shifted from the parent function, so let first graph the parent, and then see what happens when we shift it.

Answer 38

Okay.

Answer 39

https://www.desmos.com/calculator/xy3lqozckt
this si the graph of the parent function, \(\large\color{blue}{ y=\left| x \right| }\)

Answer 40

is this right?