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

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)

i dont understand

which part doesn't make sense

where did you get a and all that?

y = |x - 6|

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

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

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

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

how?

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)

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

so why is -9 the x?

you mean k?

yea,

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

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

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

so k is x?

k is just the vertical shift

confused..

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

y \[y \ge -9\]

right?

yep, that's the range

oh wow, thanks.

hold on

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

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

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

or a different one

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 \[y \ge0\]

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 **

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

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

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

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

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

so k = what?

1

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

0

yep k = 0

so that means?

the range is based on k

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

so \[y \ge 0\]

yep

whoa,

lol yep

thanks,.

np

hold on one sec,

sure

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

the vertex is (h,k)

so k = 0, not 3

or -3

should it be \[y \ge0\]

yes it should be y >= 0

but on the example its

• the vertex is (-2, 0)

• the range is described by y ≥ 0.

range stays the same since k is still 0

shouldnt it just be the 2nd number?

what do you mean

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

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

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

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

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

now back to this y=|x-6|

how do we find the range?

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

you said its 0,infinity

yep [0, infinity)

so its \[y \ge infinity\]

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

oh!!

wow,

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

yea,

the first is in interval notation

the second is a simple inequality

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

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

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

is it always gonna be infinity?

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)

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

y= *********

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

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

i thought k was the first number?

so 6,0 ?

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 \[\Large y \ge 0\]

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 \[y \ge-9\]

yep, you nailed it

Join our real-time social learning platform and learn together with your friends!