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September 18, 2019, 07:23:33 am

### AuthorTopic: Absolute Value Functions  (Read 1675 times) Tweet Share

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#### Timtasticle

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##### Absolute Value Functions
« on: November 08, 2007, 04:27:51 pm »
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Can someone explain how these things work as dumbed down as possible?

#### kingmar

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##### Absolute Value Functions
« Reply #1 on: November 08, 2007, 04:36:31 pm »
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Think of it as taking it both ways. :shock:

No seriously. Imagine a function. Then, anything below the x-axis is reflected above it. So, take |x|.

If y = x, at -1, y = -1.

If y = |x|, at -1, ordinarily y should equal -1. But as all negative y values are reflected in the x-axis, thus y = 1 instead. Instead of a straight line graph, you get a sorta V shaped graph.

The graph is defined separately - so for y = |x|, for domain (-infinity, 0] the equation is -x. For domain (0,infinity), the equation is x.

Incidentally, there is no derivative at cusps, or sharp points. Usually, it is at the x-intercepts the derivative is not defined, but check the graph just in case.  If it looks sharp, it ain't defined. Open circles.
ENTER: Incomprehensibly high

#### BenBenMan

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##### Absolute Value Functions
« Reply #2 on: November 08, 2007, 04:42:07 pm »
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Absolute just means, if it's negative, you make it positive. That's really all there is to it. If it's positive, then nothing happens to it, if it's negative, you make it positive.

So if your graph was y=|x^2|, then it would look exactly the same. However if it was y=|x^2 - 1|, then for where it's below the x axis, you would draw on top of the x axis:

Here the red graph is x^2-1, and the blue graph is |x^2-1|. Note that the blue graph has gone over the top of the red graph for the part of the graph on the left of the left x intercept and the right of the right x intercept.

EDIT: Beaten to it! Oh well

#### Timtasticle

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##### Absolute Value Functions
« Reply #3 on: November 08, 2007, 04:42:24 pm »
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Oww I see. Thanks!

This is probably a stupid question...But how about when you have like y= 5+ |x^2-1|

Oh how I wish I had have actually done some work in methods...

#### kingmar

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##### Absolute Value Functions
« Reply #4 on: November 08, 2007, 04:54:56 pm »
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Oh, but yours looks so much better!

A further note: When drawing, say |f(x)|, you need to draw the part that isn't reflected as well (i.e. trace over the positive part if they give you the graph).
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#### melanie.dee

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##### Absolute Value Functions
« Reply #5 on: November 08, 2007, 04:59:21 pm »
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yep. and make sure if there is a domain defined you draw the endpoints on the abs graph too.

#### Eriny

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##### Absolute Value Functions
« Reply #6 on: November 08, 2007, 05:27:33 pm »
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Quote from: "Timtasticle"
Oww I see. Thanks!

This is probably a stupid question...But how about when you have like y= 5+ |x^2-1|

Oh how I wish I had have actually done some work in methods...

You do the absolute value part (as already explained) and then move it up five places. So, your range, instad of being [0, infinity) becomes [5, infinity)

#### asa.hoshi

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##### Absolute Value Functions
« Reply #7 on: November 08, 2007, 05:45:11 pm »
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what about absolute functions that are like this:
y= |x+1| - |x+2|
I KNOW WHAT YOU DID LAST SUMMER!!

#### bilgia

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##### Absolute Value Functions
« Reply #8 on: November 08, 2007, 05:46:56 pm »
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My Subjects:
2006 I.T Systems --> 42
2007 English --> 40
Methods --> 41
Spec --> 38
Chem --> 36
Physics --> 37
Unimaths --> 5.5

ENTER: 97.35

#### asa.hoshi

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##### Absolute Value Functions
« Reply #9 on: November 08, 2007, 05:48:19 pm »
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and one more thing. i was reading the assessment report last year about the absolute function question they had.

you know when u reflect everything on the x-axis right, remember to place a "open" circle on the cusp if they have any. apparently students lost a mark because of that. the question was worth 2 marks.
I KNOW WHAT YOU DID LAST SUMMER!!

#### Collin Li

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##### Absolute Value Functions
« Reply #10 on: November 08, 2007, 06:04:43 pm »
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Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

If you differentiate a modulus function, and the limits do not agree, then open circles will reside on the x-coordinate of the cusp.

#### joechan521

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##### Absolute Value Functions
« Reply #11 on: November 08, 2007, 06:17:26 pm »
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Quote from: "asa.hoshi"
what about absolute functions that are like this:
y= |x+1| - |x+2|

ok, this graph is kinda hard to draw by hand, but it is possible.

for x>-1 both x+1 and x+2 is positive
therefore
y=(x+1)-(x+2)
y=-1 for values  x>-1

for    -2<x<-1   we can see that x+1 is negative, and x+2 is positive
y=-(x+1)-(x+2)
y=-2x-3
x+2>0 x+1<0    -1>x>-2

for    x<-2
both x+1 and x+2 is negative
y=-(x+1)+(x+2)
y=1.

therefore the whole graph would have 3 sections
y=1 for  x<-2
y=-2x-3 for -2<=x<-1
y=-1 for -1<=x

and they will intesect at points -1and-2, so no need to worry open or close circle
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##### Absolute Value Functions
« Reply #12 on: November 08, 2007, 06:32:09 pm »
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Quote from: "coblin"
Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

Coblin is right.
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#### Freitag

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##### Absolute Value Functions
« Reply #13 on: November 08, 2007, 06:35:42 pm »
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Quote from: "coblin"
Not sure if I missed something here, but:

Cusps do not have to be open circles. They can be closed circles. For example: y = |x| has a cusp at (0,0) and it certainly exists (a closed/full circle).

Coblin is right.

Is he ever wrong? Damn your 80/80 for methods paper 2 Coblin..
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