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absolute value graphs (1 Viewer)

Y

yeh

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i'm having trouble trying to understand some of these absolute value graphs. can some please answer the following questions? i know how to answer the ones without |y|'s but that's it.

i. why does |x| + |y| = 1 look like a diamond? (i think the y axis is part of the graph too. can't see the answer properly)
ii. why does |y| = |x + 1| look like a cross?
iii. how does making y an absolute value change the graph?
iv. why does |x + y| = 2 have two parrallel lines.
 

Ozz^E

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1. y= /f(x)/......All negative y values become positive.
2. y= f(/x/)......All positive x values are mirrored about y-axis
3. /y/= f(x)......All positive y values are mirrored about x-axis.

So for ur number one: Simlpy draw 1-/x/ and then treating that as f(x), use number 3 above....ull end up with a diamond with the four points on (1,0), (-1,0), (0,1), (0,-1). Similarly, for number 2, draw /x+1/ and then apply 3 from above, i.e reflect all the positive values about the x-axis.

:D
 
Y

yeh

Guest
ooooh, that simple!

thanx Ozz^E.

but i still don't understand iv.
 
Y

yeh

Guest
:confused: still can't figure it out Ozz^E. sorry, i'm dumb!

and one last question:

in logx how does replacing x with something like |x| or (x^2 - 1) change the curve?
 

Ozz^E

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Well, its pretty easy: /x+y/ = 2 simply means that x+y = 2 or -2, therefore it represents two parallel lines. U could have used the property i gave in the hint whereby the root of (x+y)^2 would equal 2. But u cant simply cancel the root and the squared because a square is the opposite of a + or - square root not simply a + square root. So wat u do is square the right hand side making the equation (x+y)^2 = 4 and then take the root of both sides giving x+y = 2 or -2.

For ur other question: each change has to be dealt with independently. Replacing x with /x/ would warrant a reflection of the entire logx curve in the y-axis, leaving the logx curve as well, giving a discontinous graph of two parts. Replacing x with something like (x^2-1) makes it a composite function of the form log u where u = x^2 -1. The best way to draw this is to sketch u and then log all its f(x) values.

Btw, have u finshed the course? and when's ur trial? Do u do the CSSA one.

:D :cool:
 
Y

yeh

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ohhhhhh ic ic ic ic.

nah we haven't finished the course. still got the inverse trig section of integration, polynomials, harder 3u and half of mechanics. my teacher's pretty crap actually. he doesn't even understand resisted motion!

my trials start week 4. what's cssa?
 
Y

yeh

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ohhh nah, i go to a public school. my teacher's probably going to buy the independant school's trial paper. he's been doing that since last year.
 

-=«MÄLÅÇhïtÊ»=-

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"ohhh nah, i go to a public school"

juz thought u'd like to know that cssa doesn't mean only catcholic skools do it. Public skools can buy it too.
 

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