Question

values for y= |6-x| + |1+x|

Answer 1

hmmm, looked right in my head

Answer 2

Um, the graph should look something
I just need the values.. which I don't have...

Answer 3

its x=-1 to 6 and y = 7; but how to get to them

Answer 4

Yes.. Exactly... How to get them is the question...

Answer 5

\[\sqrt{(...)^2}\]
dunno if this would help

Answer 6

Um... It doesn't help much, sorry...

Answer 7

y = f(x) + g(x)

f(x) = |6-x|; g(x) = |x+1|

Answer 8

for x>=6 y= x-6 + 1+x= 2x-5
or y= 2x-5 (x>=6)

Answer 9

it seems to have a solution from vertex to vertex ... i loathe absvalue problems lol

Answer 10

Same Same... This is how it's meant to look http://www.wolframalpha.com/input/?i=plot+y%3D |6-x|+%2B+|1%2Bx|

Answer 11

yeah, the flat part is the solution for some reason

Answer 12

Oh well... The book I have doesn't explain this part properly... I mean the adding of the absolute values and graphing them, but only explains it for let's say, y=|x+2|... Textbooks hardly ever mention the further aspects, sadly.

Answer 13

for x<=-1 y= 6-x -x-1 or y= -2x+5

Answer 14

Ok... but I thought it was x>-1?

Answer 15

for -1 <=x <=6
y= 6-x + 1+ x
or y=7

Answer 16

How did you get that? I got 1< x< 6?

Answer 17

I started with the case x>=6 (my first post), which means |6-x| just becomes 6-x. same for |1+x|, it is just 1+x

Answer 18

And you do 6-x>=0
And you do -(6-x)<=0?

Answer 19

I mean -(6-x)<0?

Answer 20

Then, look at the case for x<= -1
|6-x| is 6-x (example |6 - (-2)| =|6+2| = 6+2
|1+x| is -1-x

Answer 21

I am trying to get rid of the absolute value operation. so for certain values of x you replace |6-x| with 6-x. you end up with y= f(x)

Answer 22

But I am confusing the issue, for x>=6, replace |6-x| with x-6

Answer 23

Yes, I understand that. But I don't understand why you got x<=-1

Answer 24

Oh I understand!

Answer 25

y = |6-x| + |1+x|
y has a value for all x, including x<= -1
when x<= -1, we can get rid of the abs ops.

Answer 26

You switch the value of 6-x to x-6! That makes more sense

Answer 27

Oh, after that, how do you get the values for the graph?

Answer 28

Yes, sorry about typing it wrong!

Answer 29

Do the problem for x<= -1

Answer 30

You have the 3 values.. or equation things.. yes, and end up with something like 2x-5 or so... but what does the 2x-5 stand for?

Answer 31

with three different things.. One will be a number, to will be something like 2x-5

Answer 32

y= 2x-5 for x>=6
it is a line, but only valid for x>=6. so from x>=6 you know y is 2x-5
for x<6, we need a different equation
(because |6-x| changes behaviour at x=6)

Answer 33

So, would the value be 6? And is the 2x-5 just the line?

Answer 34

you piece them together to get the whole solution

Answer 35

Ok, I understand.. but for the line 2x-5, what are the values?

Answer 36

if x=6, y= 12-5= 7

Answer 37

Do you just plug in random values?

Answer 38

no, the "value" is the equation. (unless they are asking for y at a specific x)

Answer 39

Hmm, so how would you know how to graph it accurately?

Answer 40

you can graph the line y= 2x-5,
then erase everything for x<6, because you use a different equation for that region

Answer 41

Ok, but how do you graph the line y= 2x-5 in the first place? how would you know at what degree it slopes upwards from that point?

Answer 42

pick an x value and find the y. plot the point. Repeat with another x. Connect the dots.
Erase the part of the line for x<6

Answer 43

Ok, so it's randome values?

Answer 44

random sorry...

Answer 45

Here's the entire solution in one place:
y= 2x-5 (x>=6)
y=7 (-1 <=x <=6)
y= -2x+5 (x<=-1 )

Answer 46

OK.. so 7 is the straight value... for how many x values?

Answer 47

Til 6 and -1 right?

Answer 48

Do you have to plot the solution? If so, I would use x=6 (y=7), and x=7 (y=9)

Answer 49

and then the lines slope upwards with random x- values that occur after or before -1 and 6?
I have to sketch the solution.. but later on I have to plot, so it's best I start knowing how

Answer 50

Yes, I understand how to plot!

Answer 51

I couldn't have solved this without you!!! Thanks so much!

Answer 52

It sounds like you have the gist of it. For the flat part (y=7), draw a horizontal line at y=7. But it is only valid between x=-1 and x=6

Answer 53

good luck!

Answer 54

Thanks again!