Question

reflect this function over y axis

|2x - 1| + 3

what would be the f(-x) equailvalent?

im thinking it would've been |-2x - 1 | + 3 but.. thats not it.. whats the reflection of y axis?

Answer 1

should work
replace x by -x

Answer 2

that would reflect about the x axis
what is up goes down, what is down goes up

Answer 3

@jackthegreatest , @satellite73
|-2x - 1 | + 3 isn't an answer choice for some reason for me. Am i over looking this? Is there a simiplfied verison of my solution

Answer 4

if i distribute that it becomes |2x+1| - 3 right?, if so then yes

Answer 5

whatever they are, if there is not \(|-2x-1|+3\) then they are wrong

Answer 6

A. |2x+1| + 6
B. |2x+1| - 3
C. |2x - 1| - 3
D. |2x+1| + 3

Answer 7

check out the nice pictures here
http://www.wolframalpha.com/input/?i=%7C2x-1%7C%2B3,+%7C-2x-1%7C%2B3

Answer 8

aah i see

Answer 9

which one is the same as \[|-2x-1|+3\]?

Answer 10

im guessing it's the one |2x+1| + 3? care to explain it though?

Answer 11

i think the answer is |2x+1| + 3, but it doesnt make sense to me

how is that equalivent to |-2x + 1| + 3?

if i plug in for example 1 for x, that becomes

|-2(1) + 1| + 3 = 1 = 4

and |2(1) + 1| + 3 = 6... so im thinking it doesnt make sense
@satellite73

Answer 12

anyone?

Answer 13

this was your original \[|2x-1|+3\] right?

Answer 14

replace \(x\) by \(-x\) and get \[|-2x-1|+3\] however...

Answer 15

\[|-2x-1|=|-(2x+1)|=|2x+1|\]

Answer 16

so your answer is \[|2x+1|+3\]

Answer 17

How come when you distribute that negative, everything becomes positive @satellite73 , whenever u dsitribute the negative, don't you get |-2x - 1|?

Answer 18

lets go back and make sure we have the origninal one correct
it was \(|2x-1|+3\) yes?

Answer 19

the orginial function is |2x-1|+3, yes. we are reflecting this over the y-axis

Answer 20

ok one step at a time
first \[f(x)=|2x-1|+3\] so \[f(-x)=|2(-x)-1|+3\] or just \[f(-x)=|-2x-1|+3\] so far so good?

Answer 21

Yes, so far so good.

Answer 22

Plug in -x for x.

Answer 23

right so are we cool with \[f(-x)=|-2x-1|+3\]?

Answer 24

Yes, but that wasn't one of the answer choices.

Answer 25

i know, we are getting there

Answer 26

so lets look at \(|-2x-1|\) and factor a minus one out of it to make it \(|-2x-1|=|-(2x+1)|\) you ok with that?

Answer 27

gotcha, u factored a minus sign. I never knew you could do that, will keep that in note next time!. Now what?

Answer 28

then \[|-(2x+1)|=|2x+1|\] for any number of reasons
one is to write \[|-(2x+1)|=|-1||2x+1|=1|2x+1|=|2x+1|\]

Answer 29

a more simple way to think about it is that \[|whatever|=|-whatever|\]

Answer 30

like \(|-5|=|5|\) etc

Answer 31

hm, that's a bit tricky. I always understood absoulute value was the postiive number of a negative, bascially. Why does -1 gets its own absoulute value but the others dont?

Answer 32

A little analysis of the problem might save time and effort. Reflecting the graph about the y-axis does not affect the vertical offset, +3. I's suggest focusing on y=|2x-1 first.
Ideally the problem can be solved either graphically or analytically. I think it'd be instructive to graph y=|2x-1| and then to reflect the graph about the y-axis. You should be able to revert to an analytical equation for the new graph.|

Answer 33

Our teacher is asking us to solve it alegbraticially which is why I haven't tried to graph it.

Answer 34

Better review "absolute value." What does this mean?
What does the "absolute value function" do? If the argument is negative, the abs. val. fn. returns a positive result; if the argument is already positive, the abs. val. fun returns exactly the same result as if there were no abs. val. operator present.

Answer 35

Generally, abs. val. fns. return TWO separate results, one for each of the 2 cases I've mentioned above.

Answer 36

so whenever i try to solve |-2x+1| i dont solve whats inside the absoulute value? I just convert it to positive numbers? I think maybe thats where my confusion is coming from.

Answer 37

no no not at all

Answer 38

i.e |-3x + 5 | = |3x+5|?

Answer 39

no no (no)

Answer 40

holy crap im so confused lol

Answer 41

it wasn't \(-2x+1\) it was \(-2x-1\)

Answer 42

it is clear that \(|x|=|-x|\)?

Answer 43

i understand the part where u factored -1, but im lost there

Answer 44

would it better to say |-x| = |x|? thats how i always learned it

Answer 45

lol sure , but equal sign doesn't care

Answer 46

yes i understand that part @satellite

Answer 47

so have it your way \(|-x|=|x|\) fine

Answer 48

now lets try \[|-\spadesuit|=|\spadesuit|\] true?

Answer 49

yes

Answer 50

Perhaps this interpretatioon will make sense:
1. If the quantity within the abs. val. symbols is already positive, simply drop the abs. val symbols ||.
2. If the quant with in the abs. val. symbols is neg., change the sign of everything within those symbols. If 2x-1 is neg, then |2x-1| becomes -2x + 1.

Answer 51

In the present problem, your job is not to "solve" the expression, but rather to come up with the correct equation for the given function AFTER its been refl. in the y-axis.

Answer 52

how about \[|-(\spadesuit+\heartsuit)|=|\spadesuit+\heartsuit|\] is that ok?

Answer 53

I don't understand what you did with that negative

Answer 54

you have to think a bit abstractly here \\(x\) is not "ex" it is just a place holder
so if \(|-x|=|x|\) then i can replace \(x\) by \(\xi\) and write \(|-\xi|=|\xi|\)

Answer 55

and we can replace \(x\) by \(whatever\) and write \[|-whatever|=|whatever|\]

Answer 56

OH. I see.

Answer 57

we can replace \(x\) by \(x+y\) and write \[|-(x+y)|=|x+y|\]

Answer 58

or mover card like \[|-(\spadesuit+\heartsuit)|=|\spadesuit+\heartsuit|\]

Answer 59

or anything at all really

Answer 60

so now is it more or less clear that \[|-(2x+1)|=|2x+1|\]?

Answer 61

yes, i think i understand. As well with the tips @mathmale gave me, it is much easier to understand..

Just to make sure I understand, heres another problem:
-3 + |x-11| reflected over y axis is -3 + |x+11| ?

because:
step 1. Plug in for -x, -3 + |-x - 11|
step 2. factor out the negative, -3 + |-(x+11)|
step 3 = -3 + |-(x+11)| = -3 + |x+11| because |-whatever| = |whatever|
-3 + |-x-11| = -3 + |-(x+11)| which is same as -3 + |x+11|

OR
using @mathmale way
Step 1. plug in for -x, -3 + |-x-11|
Step 2. replace change sign with oppposite in absolute value because it has negative
-3 + |x+11|

am i correct? I think im understanding now

Answer 62

looks good to me

Answer 63

i love this website

Answer 64

Why not graph both equations and determine that way whether or not you have the correct reflection? Adding another skill to your repetoire can only be a good thing.

Answer 65

Delighted, absolutely delighted, to hear that you "love this website."

Answer 66

btw, you certainly don't have to "do" steps 1, 2,3 whatever, that was just my explanation
it is always the case that \[|ax+ay|=|a||x+y|\] no matter what \(a\) is

Answer 67

@mathmale , i don't graph often but i definitely should've.

Could you explain the graphing side?:

If I had graphed it and plugged in 1,
lets say |2(1)-1| + 3 which is 4. Coordinate (1,4)
and |-2(1)-1| + 3 = 6. Coordinate (1,6)

This is my coordinates right? I was taught if i reflected over the y axis, my y coordinate stays the same and my X coordinate changes. How come this is not the case?