ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Timtasticle on November 08, 2007, 04:27:51 pm
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Can someone explain how these things work as dumbed down as possible? :)
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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.
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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:
(http://img181.imageshack.us/img181/1341/absgraphrf7.png)
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 :P
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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...
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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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yep. and make sure if there is a domain defined you draw the endpoints on the abs graph too.
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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...
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)
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what about absolute functions that are like this:
y= |x+1| - |x+2|
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addition of ordinates?
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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.
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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.
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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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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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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..