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Sketching the addition of absolute values (1 Viewer)
- Thread starter planino
- Start date
Take the first question as an example: y=|x|+|x+2|
1. Write down the definition of y=|x|. i.e.
|x|= x, x>=0
=-x, x<0
2. Do the same for y=|x+2|
|x+2|= x+2, x>=-2
= -(x+2), x<-2
Then using these regions, add the functions up so that u get a separate function for the following regions:
(x<-2) , (-2< x< 0) and (x > 0.)
From there, u can just read of your working out and graph it.
* I dont want to write the solution down because I want you to have a go at it first with these hints. If you need further help, maybe someone who is proficient with latex can type my steps nicely (carrot)?
Alternatively you can draw both function separately then add/subtract ordinates (this method is sort of extension 2 so do it if your teacher has told you about it)</x<0>
1. Write down the definition of y=|x|. i.e.
|x|= x, x>=0
=-x, x<0
2. Do the same for y=|x+2|
|x+2|= x+2, x>=-2
= -(x+2), x<-2
Then using these regions, add the functions up so that u get a separate function for the following regions:
(x<-2) , (-2< x< 0) and (x > 0.)
From there, u can just read of your working out and graph it.
* I dont want to write the solution down because I want you to have a go at it first with these hints. If you need further help, maybe someone who is proficient with latex can type my steps nicely (carrot)?
Alternatively you can draw both function separately then add/subtract ordinates (this method is sort of extension 2 so do it if your teacher has told you about it)</x<0>
Last edited:
Carrotsticks
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I actually have my own special way of drawing such curves for all cases:
y = |ax+b| plus/minus |cx+d|, for any real a,b,c,d.
Such that you can draw them within a matter of seconds straight after solving 1 (sometimes two) pairs of simultaneous equations.
You just need to recognise that depending on the question, all the curves inevitably come down to 4 cases.
The question now is identifying which of the cases it is.
My sugggestion, get a graphing program like Graphmatica, input random cases and observe what happens to the curves as you change the numbers and the signs. Be sure to draw the component graphs as well ie: y=|x+1| + |2x+3| has component functions y=|x+1| and y=|2x+3|.
You will soon see a pattern.
y = |ax+b| plus/minus |cx+d|, for any real a,b,c,d.
Such that you can draw them within a matter of seconds straight after solving 1 (sometimes two) pairs of simultaneous equations.
You just need to recognise that depending on the question, all the curves inevitably come down to 4 cases.
The question now is identifying which of the cases it is.
My sugggestion, get a graphing program like Graphmatica, input random cases and observe what happens to the curves as you change the numbers and the signs. Be sure to draw the component graphs as well ie: y=|x+1| + |2x+3| has component functions y=|x+1| and y=|2x+3|.
You will soon see a pattern.
Autonomatic
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LOL Didn't you say you learnt addition of ordinates in class?
What?
Autonomatic
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A-d-d-i-t-i-o-n of O-r-d-i-n-a-t-e-s.
So for y=|x|+|x+2|,
You would sketch |x| and |x+2| individually first in one colour. (Don't tell me you can't do these)
Now you start adding the y-values or ordinates of both graphs in order to find a new point.
e.g. At x=1 (imagine a line), y=|x| would cross this line at y=1, y=|x+2| would cross this line at y=3,
Add these two ordinates and you get y=4, Therefore (1,4) would be a point on y=|x|+|x+2|
You do this for all the points you need in order to produce the finished function, y= |x| + |x+2|
Thanks!! Finally I get this after like 4 months