#1

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|

Anyways how do you graph absolute value functions?

f(x)=a|x-h|

ex. y= 6|x-7|

#2

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

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

#3

y=6 *(x+7)

because the absolute value of -7 is 7

because the absolute value of -7 is 7

#4

i think i would shoot myself in the head if i had that problem.

#5

you kids should be paying attention in class, not passing notes and listening to your stereos. damn youths.

#6

you kids should be paying attention in class, not passing notes and listening to your stereos. damn youths.

I agree. It's because of that damned rock and roll music!

#7

^

#8

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

#9

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?

#10

fail

the absolute value of -7 is 7

but x is a variable

i know

y=6*(

**x**+7)

x is still a variable

#11

Homework Only Thread!!!

Damn kids....

Damn kids....

#12

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

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

#13

You fake it and pray to god the teacher dosent look too closely.

#14

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.

#15

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?

#16

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

#17

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

#18

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.

#19

AHHHH you guys are confusing.

#20

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

#21

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.

#22

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