Man I hate homework, who else is with me?

Anyways how do you graph absolute value functions?

f(x)=a|x-h|

ex. y= 6|x-7|
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just work out where (x-h) is larger or smaller than zero

eg: solve x-h > 0

then on one side of that number draw it normally
and on the other side draw the same function but without regular parentheses and -x instead of x

you usually end up with a V shaped graph
y=6 *(x+7)
because the absolute value of -7 is 7
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i think i would shoot myself in the head if i had that problem.
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you kids should be paying attention in class, not passing notes and listening to your stereos. damn youths.

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Quote by sporkface
y=6 *(x+7)
because the absolute value of -7 is 7

fail

the absolute value of -7 is 7
but x is a variable
Quote by CrazyDavey
you kids should be paying attention in class, not passing notes and listening to your stereos. damn youths.

Actually I was reading Animal Farm

so far I've graphed (0, 7), is that right haha?
And we will weave in and out of sanity unnoticed
Swirling in blissfully restless visions of all our bleary progress
Quote by seljer
fail

the absolute value of -7 is 7
but x is a variable

i know
y=6*(x +7)

x is still a variable
In diesem Herz hab ich die Macht.

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Quote by TDKshorty
Actually I was reading Animal Farm

so far I've graphed (0, 7), is that right haha?

6|x-7|

x-7 > 0
x > 7

so that means the graph changes at x=7

for all values of x that are right of 7 you draw y = 6(x-7) = 6x-42

for all values of x that are left of 7 your draw y = 6*(-x-7) = -6x-42

Quote by sporkface
i know
y=6*(x +7)

x is still a variable

but put in x=-8
then whats in the parentheses is -8+7=-1 = negative = that can't be an absolute value
You fake it and pray to god the teacher dosent look too closely.
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he he, Animal Farm, i finished reading that for my english class like a month ago. The ending is awesome, as for your maths, i am clueless. That reminds me i need to type out a essay about it right now.
Hold on I think I've got it, when X is 0 Y = 7 or is it 42? and then like if X is 1 then Y = 48?
And we will weave in and out of sanity unnoticed
Swirling in blissfully restless visions of all our bleary progress
Quote by seljer
6|x-7|

x-7 > 0
x > 7

so that means the graph changes at x=7

for all values of x that are right of 7 you draw y = 6(x-7) = 6x-42

for all values of x that are left of 7 your draw y = 6*(-x-7) = -6x-42

but put in x=-8
then whats in the parentheses is -8+7=-1 = negative = that can't be an absolute value

woops it seems i may have lost something over the summer lol
In diesem Herz hab ich die Macht.

Gear:
Fender MIM HSS Strat (Wine Red)
VOX Valvetronix VT20+
Kay K390 Acoustic
vertex is at (7,0) so go from there and graph points for x=6 and x=8. then extend the lines. the graph of absolute value is a V
Quote by seljer
6|x-7|

x-7 > 0
x > 7

so that means the graph changes at x=7

for all values of x that are right of 7 you draw y = 6(x-7) = 6x-42

for all values of x that are left of 7 your draw y = 6*(-x-7) = -6x-42

but put in x=-8
then whats in the parentheses is -8+7=-1 = negative = that can't be an absolute value

Actually that isn't quite correct.

If x>or=7, y=6(x-7) y=6x-42
if x<7, y=6(-(x-7)) y=-6x+42

The logic is that when x is greater than or equal to 7, anything within the abs value would be greater than or equal to zero, thus eliminating the need for the bars.

If x is less than 7, however, anything within them would be negative, so to drop the bars set whatever is within the abs value to it's negative.

Don't believe me? Graph the abs value and then the two separate equations.
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Quote by emagdnimasisiht
haha
This is the funniest thing i've ever read on UG.
lespaulrocks39, you sir are awesome.
AHHHH you guys are confusing.
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Quote by lespaulrocks39
Actually that isn't quite correct.

If x>or=7, y=6(x-7) y=6x-42
if x<7, y=6(-(x-7)) y=-6x+42

The logic is that when x is greater than or equal to 7, anything within the abs value would be greater than or equal to zero, thus eliminating the need for the bars.

If x is less than 7, however, anything within them would be negative, so to drop the bars set whatever is within the abs value to it's negative.

Don't believe me? Graph the abs value and then the two separate equations.

bah, thats what using Mathematica for everything does
Quote by seljer
bah, thats what using Mathematica for everything does

Mathematica?

I just know this because I'm tutoring for Calculus and they generally always ask questions about the limits of quotients involving absolute value functions.

Not trying to be a douche, but there were a whole bunch of incorrect answers floating around this thread, lol.
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Quote by emagdnimasisiht
haha
This is the funniest thing i've ever read on UG.
lespaulrocks39, you sir are awesome.
Quote by lespaulrocks39
Mathematica?

I just know this because I'm tutoring for Calculus and they generally always ask questions about the limits of quotients involving absolute value functions.

Not trying to be a douche, but there were a whole bunch of incorrect answers floating around this thread, lol.

fancy math program

you just type it in and it draws pretty graphs

it excellent if you're lazy