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Mathematics 87 Online
OpenStudy (anonymous):

Without graphing, find the domain and range of each absolute-value function. y = |x - 6| Without graphing, find the domain and range of each absolute-value function. y = |x - 6| @Mathematics

jimthompson5910 (jim_thompson5910):

The domain of y = a|x-h|+k is the set of all real numbers (since you can plug in any real number for x) The range of y = a|x-h|+k is [k, infinity) when a > 0 Since a = 1 > 0 and k = 0, this means that the range is [0, infinity)

OpenStudy (anonymous):

i dont understand

jimthompson5910 (jim_thompson5910):

which part doesn't make sense

OpenStudy (anonymous):

where did you get a and all that?

OpenStudy (anonymous):

y = |x - 6|

jimthompson5910 (jim_thompson5910):

the general absolute value equation is y = a|x-h| + k In the case of y = |x - 6|, we can rewrite it as y = 1|x - 6|+0 So this shows us that a = 1, h = 6, and k = 0

jimthompson5910 (jim_thompson5910):

@fewscrewsmissing 0 is in the range, so it's really [0, infinity)

OpenStudy (anonymous):

would that be the case on \[y = |x| - 9\]

jimthompson5910 (jim_thompson5910):

domain for y = |x| - 9 is the same (-infinity, infinity)

OpenStudy (anonymous):

how?

jimthompson5910 (jim_thompson5910):

the range will be different y = |x| - 9 can be written as y = 1|x-0| + (-9) So a = 1, h = 0, and k = -9 So the range is [-9, infinity)

jimthompson5910 (jim_thompson5910):

The domain is the set of all real numbers since you can plug in any real number for x in y = |x| - 9

OpenStudy (anonymous):

so why is -9 the x?

jimthompson5910 (jim_thompson5910):

you mean k?

OpenStudy (anonymous):

yea,

OpenStudy (anonymous):

why is it the first in (-9, infinity) ?

jimthompson5910 (jim_thompson5910):

because y = 1|x-0| + (-9) matches up with y = a|x-h|+k

jimthompson5910 (jim_thompson5910):

notice the last terms are -9 and k, so k = -9

OpenStudy (anonymous):

so k is x?

jimthompson5910 (jim_thompson5910):

k is just the vertical shift

OpenStudy (anonymous):

confused..

jimthompson5910 (jim_thompson5910):

k is just a constant, and it's not a variable like x

OpenStudy (anonymous):

y \[y \ge -9\]

OpenStudy (anonymous):

right?

jimthompson5910 (jim_thompson5910):

yep, that's the range

OpenStudy (anonymous):

oh wow, thanks.

OpenStudy (anonymous):

hold on

jimthompson5910 (jim_thompson5910):

[-9, infinity) is another way of saying \[\Large y \ge -9\]

OpenStudy (anonymous):

so if a vertex is (-3,0) the range is \[y \ge0\]

jimthompson5910 (jim_thompson5910):

which one are you working on, y = |x| - 9 ???

jimthompson5910 (jim_thompson5910):

or a different one

OpenStudy (anonymous):

no, this is a different one, i find the vertex, the domain i sall real number and the vertext is -3,0 so is it \[y \ge0\]

jimthompson5910 (jim_thompson5910):

ok, if the vertex is (-3, 0), then the range is \[\Large y \ge 0\] so you are correct **Note: this is assuming that a > 0 **

OpenStudy (anonymous):

so the 2nd number in the vertex is always the range?

jimthompson5910 (jim_thompson5910):

if the vertex is (h, k) and a > 0, then the range is \[\Large y \ge k\]

jimthompson5910 (jim_thompson5910):

so in a way yes, but you have to be careful about the value of 'a'

OpenStudy (anonymous):

y = |x - 6| so that range is y\[\ge infinite\]

jimthompson5910 (jim_thompson5910):

y = |x - 6| really looks like y = |x - 6| + 0

jimthompson5910 (jim_thompson5910):

so k = what?

OpenStudy (anonymous):

1

jimthompson5910 (jim_thompson5910):

remember the form: y = a|x-h| + k

OpenStudy (anonymous):

0

jimthompson5910 (jim_thompson5910):

yep k = 0

OpenStudy (anonymous):

so that means?

jimthompson5910 (jim_thompson5910):

the range is based on k

jimthompson5910 (jim_thompson5910):

if a > 0, then the range is \[\Large y \ge k\]

OpenStudy (anonymous):

so \[y \ge 0\]

jimthompson5910 (jim_thompson5910):

yep

OpenStudy (anonymous):

whoa,

jimthompson5910 (jim_thompson5910):

lol yep

OpenStudy (anonymous):

thanks,.

jimthompson5910 (jim_thompson5910):

np

OpenStudy (anonymous):

hold on one sec,

jimthompson5910 (jim_thompson5910):

sure

OpenStudy (anonymous):

why is it when the vertex is -3,0 the range is \[y \ge3\]

jimthompson5910 (jim_thompson5910):

the vertex is (h,k)

jimthompson5910 (jim_thompson5910):

so k = 0, not 3

jimthompson5910 (jim_thompson5910):

or -3

OpenStudy (anonymous):

should it be \[y \ge0\]

jimthompson5910 (jim_thompson5910):

yes it should be y >= 0

OpenStudy (anonymous):

but on the example its

OpenStudy (anonymous):

• the vertex is (-2, 0)

OpenStudy (anonymous):

• the range is described by y ≥ 0.

jimthompson5910 (jim_thompson5910):

range stays the same since k is still 0

OpenStudy (anonymous):

shouldnt it just be the 2nd number?

jimthompson5910 (jim_thompson5910):

what do you mean

OpenStudy (anonymous):

on the vertex its -2,0 so the range is described by y ≥ 0. 0 is the 2nd number

jimthompson5910 (jim_thompson5910):

yes the first number of the vertex does not affect the range at all

jimthompson5910 (jim_thompson5910):

think of it graphically: the first number just shifts the graph side to side and it does not move it up and down

OpenStudy (anonymous):

so if iy was the vertex of (3,0) it would be y\[y \ge0\]

jimthompson5910 (jim_thompson5910):

yep, if the vertex was (223, 0), the range would still be the same

OpenStudy (anonymous):

now back to this y=|x-6|

OpenStudy (anonymous):

how do we find the range?

jimthompson5910 (jim_thompson5910):

y=|x-6| is the same as y=|x-6|+0

OpenStudy (anonymous):

you said its 0,infinity

jimthompson5910 (jim_thompson5910):

yep [0, infinity)

OpenStudy (anonymous):

so its \[y \ge infinity\]

jimthompson5910 (jim_thompson5910):

which is the same as \[\Large y \ge 0\]

OpenStudy (anonymous):

oh!!

OpenStudy (anonymous):

wow,

jimthompson5910 (jim_thompson5910):

no it's not greater than infinity, that doesn't make sense (and isn't possible)

OpenStudy (anonymous):

yea,

jimthompson5910 (jim_thompson5910):

the first is in interval notation

jimthompson5910 (jim_thompson5910):

the second is a simple inequality

OpenStudy (anonymous):

hold on so its y=|x-6| + 0

jimthompson5910 (jim_thompson5910):

yep, that's another way of saying y = |x-6|

OpenStudy (anonymous):

so where do we get the infinity from? in 0, infinity

OpenStudy (anonymous):

is it always gonna be infinity?

jimthompson5910 (jim_thompson5910):

when we say \[\Large y \ge 0\], we're saying that y can be any number larger than 0 (or it could equal 0) [0, infinity) is another way of expressing this interval Basically [0, infinity) is describing all the values of y by using the endpoints (ie 0 is the smallest y can be and infinity is the largest it can be)

OpenStudy (anonymous):

no cause y+ |x-6| would be a=1 h=6 and k=0

OpenStudy (anonymous):

y= *********

OpenStudy (anonymous):

so wouldnt it be 0,6 then the range is \[y \ge 6\]

jimthompson5910 (jim_thompson5910):

no, remember we're using k for the range and not h

OpenStudy (anonymous):

i thought k was the first number?

OpenStudy (anonymous):

so 6,0 ?

jimthompson5910 (jim_thompson5910):

the form is y = a|x-h|+k y = |x-6| really looks like y = 1|x-6| + 0 So h = 6, k = 0 So the vertex is (6, 0) and the range is \[\Large y \ge 0\]

OpenStudy (anonymous):

oh so in y=|x|-9 it would be y=1|x-h|-9 where a =1 h=0 and k=-9 so it's 0,-9 so tyhe range is \[y \ge-9\]

jimthompson5910 (jim_thompson5910):

yep, you nailed it

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