j2lie
  • j2lie
Does the function represented by the graph have an inverse function? The three coordinates are (0,0), (4,8), and (8,0).
Mathematics
katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
Plot all of the points. If you can draw a horizontal straight line through any two points, then this function fails the horizontal line test. Any time you fail the horizontal line test, the function won't have an inverse.
j2lie
  • j2lie
I cannot, because the line is slanted on the graph.
jim_thompson5910
  • jim_thompson5910
did you plot the three points?

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j2lie
  • j2lie
yes. It's been drawn for me.
jim_thompson5910
  • jim_thompson5910
ok is it possible to draw a flat horizontal line through more than one point?
j2lie
  • j2lie
Yes, between zero and eight.
jim_thompson5910
  • jim_thompson5910
yeah through (0,0) and (8,0) so this means the horizontal line test fails and we won't have an inverse function
j2lie
  • j2lie
Why? I think there should be an inverse. I just don't know the function for the graph.
jim_thompson5910
  • jim_thompson5910
did you learn about the vertical line test?
j2lie
  • j2lie
Not yet. We're talking about functions and graphs. Also domain and range.
jim_thompson5910
  • jim_thompson5910
well the rule about functions is that a relation must pass the vertical line test in order to be a function
jim_thompson5910
  • jim_thompson5910
if you can pass a vertical line through more than one point, then the vertical line test fails and we do NOT have a function
j2lie
  • j2lie
Ok. But how do I find the function of the graph?
jim_thompson5910
  • jim_thompson5910
In this case, we just have 3 points. It seems like there isn't enough information to find the actual function
jim_thompson5910
  • jim_thompson5910
can you post a screenshot of the full thing?
j2lie
  • j2lie
I will check my notes, but they're confusing.
jim_thompson5910
  • jim_thompson5910
I recommend watching this video about the vertical line test https://www.youtube.com/watch?v=-xvD-n4FOJQ hopefully it will help
j2lie
  • j2lie
I haven't learned that yet, so would it count if put the vertical and horizontal tests as my answer?
jim_thompson5910
  • jim_thompson5910
if you haven't learned it, then the teacher probably won't accept it. But it's handy to learn nonetheless (chances are the teacher will teach it later down the road)
j2lie
  • j2lie
I need the exact answer, because I have a test tomorrow and I never got a question like this before. I can find the inverse of a function if it's given.
jim_thompson5910
  • jim_thompson5910
can you post a screenshot of the full thing?
j2lie
  • j2lie
Ok. Sorry.
j2lie
  • j2lie
I cannot attach a picture.
jim_thompson5910
  • jim_thompson5910
the "attach file" button isn't working?
jim_thompson5910
  • jim_thompson5910
maybe host it on a site like imgur and then paste the link
j2lie
  • j2lie
I have instagram.
jim_thompson5910
  • jim_thompson5910
that might work, just as long as you can host the image somewhere and share the link
j2lie
  • j2lie
My username is princess5062. You can see the picture that I just uploaded.
j2lie
  • j2lie
https://instagram.com/p/7Ja6_CmGRgNTAiNKsvvsYhyWXzo4VLMRi9m1A0/?taken-by=princess5062
jim_thompson5910
  • jim_thompson5910
it's telling me "Sorry, this page isn't available"
j2lie
  • j2lie
you have to have an account. It worked when I was logged on.
jim_thompson5910
  • jim_thompson5910
oh sorry I don't have an account. There's no way to share to non-account holders?
j2lie
  • j2lie
maybe I should i create an imgur account
jim_thompson5910
  • jim_thompson5910
you don't need to create one, they allow you to upload without an account
j2lie
  • j2lie
1 Attachment
jim_thompson5910
  • jim_thompson5910
thanks
j2lie
  • j2lie
Sorry, it took forever.
jim_thompson5910
  • jim_thompson5910
ok so it's a V shaped line, I see now
j2lie
  • j2lie
That's what I said like a parabola.
jim_thompson5910
  • jim_thompson5910
ok so the domain you have (0,8) is close but not quite there the actual domain is [0,8]. What's the difference? The difference is that the endpoints are included when you use square brackets.
j2lie
  • j2lie
Oh. It's because my friend said that the brackets are for the range only. Ok, thank you for that. Can I use any information I wrote down for the function?
jim_thompson5910
  • jim_thompson5910
no, as long as there is a solid point on the graph, it is included. The endpoints have solid points (closed circles). if you saw open circles, then you would exclude those points
jim_thompson5910
  • jim_thompson5910
`Can I use any information I wrote down for the function?` I'm not sure what you mean
j2lie
  • j2lie
Like the average rate of change, can I put that in an equation that is the function of the graph?
jim_thompson5910
  • jim_thompson5910
oh, yes you can to describe the slope of each piece of the V shape
jim_thompson5910
  • jim_thompson5910
so for #3, you'll have 2 answers there
j2lie
  • j2lie
That's the average rate of change also known as slope.
j2lie
  • j2lie
I got 2 and -2 for the rate of change. There is an inverse, because the signs are the opposite.
jim_thompson5910
  • jim_thompson5910
yes for the first half from 0 to 4, the slope is +2 the second half from 4 to 8 has the slope -2
j2lie
  • j2lie
Is it going to be an equation or inequality?
jim_thompson5910
  • jim_thompson5910
the graph you mean?
j2lie
  • j2lie
Yes, the function of the graph.
jim_thompson5910
  • jim_thompson5910
the function is an equation
j2lie
  • j2lie
I thought it was an equality, because 2 and -2 are opposite of each other.
j2lie
  • j2lie
so the equation is y=|x| + 2 = -2?
jim_thompson5910
  • jim_thompson5910
equality is the same as equation both have "equa" in them an inequality is something like y > 2x which would involve shading
jim_thompson5910
  • jim_thompson5910
the vertex is at (4,8) so h = 4, k = 8 the value of 'a' is a = -2 because the V shape is stretched by a factor of 2. The negative is to mean "flip the V upside down" y = a|x-h|+k y = -2|x-4| + 8 is your equation
j2lie
  • j2lie
The function is not an equation with the letter y. We have to find the equation for the function.
jim_thompson5910
  • jim_thompson5910
you can replace y with f(x)
j2lie
  • j2lie
I might be wrong, because I looked at problem six, it's asking for an equation of the function.
jim_thompson5910
  • jim_thompson5910
yeah you start with y = |x| and do a bunch of transformations to end up with y = -2|x-4| + 8
j2lie
  • j2lie
Where did you get the eight?
jim_thompson5910
  • jim_thompson5910
that's the y coordinate of the vertex
jim_thompson5910
  • jim_thompson5910
|dw:1441239501121:dw|
j2lie
  • j2lie
Thank you. How did you get the equation? Sorry, I'm not understanding your equation.
jim_thompson5910
  • jim_thompson5910
I started with the template y = a|x-h|+k
jim_thompson5910
  • jim_thompson5910
have you learned about y = a|x-h|+k ?
j2lie
  • j2lie
Did you multiply 2 and -2?
j2lie
  • j2lie
I haven't learned that yet.
j2lie
  • j2lie
I did, I'm so sorry. I am silly sometimes.
jim_thompson5910
  • jim_thompson5910
so because the vertex is (4,8), this means h = 4 and k = 8. I plugged those in
jim_thompson5910
  • jim_thompson5910
along with a = -2
j2lie
  • j2lie
so you moved -2 to both sides and multiplied 2 and -2?
jim_thompson5910
  • jim_thompson5910
I got a = -2 from the fact that the slope of each piece of y = |x| is +1 and -1 but the slopes here are +2 and -2, so I doubled things to go from y = |x| to y = 2|x| then I flipped the V over to get y = -2|x|
j2lie
  • j2lie
Ok. Thank you for that clarification. So the function is -2|x-4|+8?
jim_thompson5910
  • jim_thompson5910
y = -2|x-4|+8 or f(x) = -2|x-4|+8
j2lie
  • j2lie
f(x)=-2|x-4|+8
jim_thompson5910
  • jim_thompson5910
here's the graph to check https://www.desmos.com/calculator/m8dav9jses
jim_thompson5910
  • jim_thompson5910
ignore the portions that aren't in quadrant 1
j2lie
  • j2lie
I don't know what to do on there
jim_thompson5910
  • jim_thompson5910
it's basically a graphing calculator do you see the red V shape graph?
j2lie
  • j2lie
No. It's blank.
jim_thompson5910
  • jim_thompson5910
hmm strange
jim_thompson5910
  • jim_thompson5910
you should see this
1 Attachment
j2lie
  • j2lie
I'm sorry. I never used a graphing calculator before.
j2lie
  • j2lie
That's correct, but on my paper the number eight is not labeled.
jim_thompson5910
  • jim_thompson5910
probably because it's cut off? not sure
j2lie
  • j2lie
Something like that. I need help finding the inverse of the function. The function is not in the regular form to create the inverse of a function.
jim_thompson5910
  • jim_thompson5910
To find the inverse graphically, you take each point and swap the coordinates eg: (4,8) ---> (8,4) (8,0) ---> (0,8)
j2lie
  • j2lie
I need it in an equation.
jim_thompson5910
  • jim_thompson5910
later on when you learn about the vertical line test, you'll find that the inverse is NOT a function
j2lie
  • j2lie
I'm sorry, but I'm doing it the way I learned.
jim_thompson5910
  • jim_thompson5910
have you learned that absolute value functions are made up of two linear equations?
j2lie
  • j2lie
I don't know.
j2lie
  • j2lie
Do you mean like x=2 and x=-2?
jim_thompson5910
  • jim_thompson5910
so the term "piecewise function" doesn't sound familiar?
j2lie
  • j2lie
No. sorry. I don't know what level math you're in right now.
jim_thompson5910
  • jim_thompson5910
hmm I'm trying to figure out what the teacher wants and the method s/he wants you to use
j2lie
  • j2lie
He wants me to use composition. Have you ever heard about that?
jim_thompson5910
  • jim_thompson5910
yes
j2lie
  • j2lie
Ok, I need help with making the function conforming to the composition rules.
j2lie
  • j2lie
I don't need the composition. I think, but it's to check the inverse function.
jim_thompson5910
  • jim_thompson5910
I'm not sure how he wants you to find the inverse equation if you haven't learned about how an absolute value function is composed of 2 linear equations.
j2lie
  • j2lie
I know what you mean.
j2lie
  • j2lie
1 Attachment
jim_thompson5910
  • jim_thompson5910
all that shows how to find the inverse of a linear equation. So we just need to break up the V shape into 2 linear equations, then find the inverse of each piece
j2lie
  • j2lie
How do I get that?
jim_thompson5910
  • jim_thompson5910
focus on the left half of the V shape what two points are on this line?
j2lie
  • j2lie
(0,0) and (4,8)
jim_thompson5910
  • jim_thompson5910
what is the equation of the line through those two points
j2lie
  • j2lie
I don't know. I know the slope.
jim_thompson5910
  • jim_thompson5910
what's the slope
j2lie
  • j2lie
2
jim_thompson5910
  • jim_thompson5910
what's the y intercept?
j2lie
  • j2lie
I think I'm going to be wrong, but is it (2,4)
jim_thompson5910
  • jim_thompson5910
where does that line cross the y axis?
j2lie
  • j2lie
Ok. I'm feeling silly, but if the line continued wouldn't it be (0,0)
jim_thompson5910
  • jim_thompson5910
the left line goes through (0,0), yes
jim_thompson5910
  • jim_thompson5910
slope = 2 ----> m = 2 y intercept = 0 ---> b = 0
jim_thompson5910
  • jim_thompson5910
the left piece of the line is y = 2x+0 or just y = 2x
j2lie
  • j2lie
The line would just 2x
jim_thompson5910
  • jim_thompson5910
|dw:1441241810907:dw|
j2lie
  • j2lie
So the line is 2x without 0?
jim_thompson5910
  • jim_thompson5910
y=2x+0 is the same as y = 2x it's better to write the most simplified form to make things easier
j2lie
  • j2lie
Ok. I'm going to try and find the inverse.
j2lie
  • j2lie
The y intercept is a zero. It cannot have an inverse.
jim_thompson5910
  • jim_thompson5910
why not?
j2lie
  • j2lie
you need to have an equation like y=x-4/3
j2lie
  • j2lie
It could work.
jim_thompson5910
  • jim_thompson5910
you can find the inverse of y = 2x though
jim_thompson5910
  • jim_thompson5910
swap x and y, solve for y
j2lie
  • j2lie
I found it and I made the equation y=2x+0.
jim_thompson5910
  • jim_thompson5910
you should find the inverse to y = 2x is y = x/2
j2lie
  • j2lie
Ok, because the inverse cannot have zero?
jim_thompson5910
  • jim_thompson5910
I don't know what you mean by not having 0
j2lie
  • j2lie
Never mind.
j2lie
  • j2lie
Should I use composition?
jim_thompson5910
  • jim_thompson5910
I think composition is only used to verify you have the correct inverse (as your attached image shows)
j2lie
  • j2lie
I'll use it. I need to make sure they are inverses.
jim_thompson5910
  • jim_thompson5910
ok so f(x) = 2x and g(x) = x/2 confirm that f( g(x) ) = x and g( f(x) ) = x
j2lie
  • j2lie
I got -2(x/2). Is that correct?
jim_thompson5910
  • jim_thompson5910
you should have 2(x/2) now simplify that
j2lie
  • j2lie
I got -x and -x for both compositions?
jim_thompson5910
  • jim_thompson5910
you should get +x
jim_thompson5910
  • jim_thompson5910
2(x/2) = 2x/2 = x
j2lie
  • j2lie
but f(x)=-2x.
j2lie
  • j2lie
I'm sorry. I confused myself.
jim_thompson5910
  • jim_thompson5910
no, f(x) = 2x
j2lie
  • j2lie
I wrote that when I tried it my way.
j2lie
  • j2lie
I fixed it and wrote x and x.
jim_thompson5910
  • jim_thompson5910
so that confirms x/2 is the inverse of 2x
j2lie
  • j2lie
Yes. I understand that now
jim_thompson5910
  • jim_thompson5910
now find the equation of the right half of the V
j2lie
  • j2lie
Ok.
j2lie
  • j2lie
There is no y intercept for the right side. There is a x intercept.
jim_thompson5910
  • jim_thompson5910
extend the line until it crosses the y axis
jim_thompson5910
  • jim_thompson5910
or you can use y = mx+b and solve for b
jim_thompson5910
  • jim_thompson5910
slope = -2 ---> m = -2 point on the line is (8,0) ---> x = 8 and y = 0
jim_thompson5910
  • jim_thompson5910
y = mx+b y = -2x+b ... replace m with -2 0 = -2*8+b ... replace x with 8, replace y with 0 solve for b
j2lie
  • j2lie
why do you have to replace x with 8?
jim_thompson5910
  • jim_thompson5910
because we know (8,0) is on that right half of the V
jim_thompson5910
  • jim_thompson5910
we could use (4,8) as well
j2lie
  • j2lie
Ok. Now I will multiply -2 and 8.
jim_thompson5910
  • jim_thompson5910
then isolate b
j2lie
  • j2lie
to isolate b I will cross cancel b by adding -b.
jim_thompson5910
  • jim_thompson5910
why not move the -16 over
j2lie
  • j2lie
f(x)=b-16
jim_thompson5910
  • jim_thompson5910
0 = -2*8+b 0 = -16+b 16 = b ... Add 16 to both sides. so b = 16 agreed?
j2lie
  • j2lie
I'm starting to forget basic stuff, I'm sorry. I agree.
jim_thompson5910
  • jim_thompson5910
slope = -2 m = -2 y-intercept = 16 b = 16
jim_thompson5910
  • jim_thompson5910
y = mx+b turns into y = -2x+16
jim_thompson5910
  • jim_thompson5910
|dw:1441243656946:dw|
j2lie
  • j2lie
Ok. I got it. Now I need to find the inverse.
j2lie
  • j2lie
I got the answer. The answer is x and x.
jim_thompson5910
  • jim_thompson5910
what inverse did you get
j2lie
  • j2lie
The inverse is x and x.
jim_thompson5910
  • jim_thompson5910
those are the results after you do the composition
j2lie
  • j2lie
Yes.
jim_thompson5910
  • jim_thompson5910
what is the actual inverse
j2lie
  • j2lie
I don't know. I'm sorry.
jim_thompson5910
  • jim_thompson5910
did you solve x = -2y+16 for y?
j2lie
  • j2lie
Yes I did.
jim_thompson5910
  • jim_thompson5910
and you got y = ??
j2lie
  • j2lie
I got x.
jim_thompson5910
  • jim_thompson5910
forget about the composition stuff for now
jim_thompson5910
  • jim_thompson5910
isolate y in `x = -2y+16`
j2lie
  • j2lie
I'm sorry I did not solve it, but if you mean y=x-16/-2
j2lie
  • j2lie
then I did solve it.
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
you can also write that as y = (16-x)/2
j2lie
  • j2lie
Yes. I'm right. I got the composition correctly right?
jim_thompson5910
  • jim_thompson5910
yes if you ended up with x for each composition, you confirmed you have the correct inverse
j2lie
  • j2lie
Yes. Can you help with the rest of page? Or are you tired?
j2lie
  • j2lie
Me^
jim_thompson5910
  • jim_thompson5910
how much is left?
j2lie
  • j2lie
three problems and I have the answers I just need to make sure they're right.
jim_thompson5910
  • jim_thompson5910
ok I can check what you have
j2lie
  • j2lie
Ok. Please wait.
jim_thompson5910
  • jim_thompson5910
ok, I'll be right back.
j2lie
  • j2lie
Ok.
j2lie
  • j2lie
1 Attachment
jim_thompson5910
  • jim_thompson5910
for #5 they want you to describe how the parent function is changing
jim_thompson5910
  • jim_thompson5910
how is y = |x| transformed into that V shape you see?
j2lie
  • j2lie
by points?
jim_thompson5910
  • jim_thompson5910
you agree that the graph of y = |x| looks like this right? |dw:1441247206827:dw|
j2lie
  • j2lie
I don't agree. It's upside down.
jim_thompson5910
  • jim_thompson5910
well the graph you have is upside down, but that isn't the graph of y = |x|
jim_thompson5910
  • jim_thompson5910
the two are similar though which is why we apply a bunch of transformations
j2lie
  • j2lie
Mine is not upside down. Mine is right, you're looking at it wrong.
jim_thompson5910
  • jim_thompson5910
this is what |x| looks like http://demo.activemath.org/ActiveMath2/LeAM_calculusPics/AbsValue.png?lang=en
j2lie
  • j2lie
No it's not, it's going down.
jim_thompson5910
  • jim_thompson5910
the teacher is trying to get you to transform what I posted into what you have
jim_thompson5910
  • jim_thompson5910
at some point, the graph will flip
j2lie
  • j2lie
no. It won't we haven't learned that yet.
jim_thompson5910
  • jim_thompson5910
hmm let me think
jim_thompson5910
  • jim_thompson5910
it's so strange how they want you to find the inverse when you haven't learned about horizontal/vertical line tests and how they ask about transformations when you haven't covered that either
j2lie
  • j2lie
You're taking harder Algebra.
jim_thompson5910
  • jim_thompson5910
yeah I'm thinking of a different algebra class maybe
j2lie
  • j2lie
I'm taking Algebra 2. How about you?
jim_thompson5910
  • jim_thompson5910
your best bet is to ask the teacher since he'll know what he wants
jim_thompson5910
  • jim_thompson5910
I'm not in algebra, but I'm probably thinking of another algebra class
j2lie
  • j2lie
So your other answers are wrong? There are only two algebra classes.
jim_thompson5910
  • jim_thompson5910
No I'm quite sure the answers I got are correct. It's just sometimes the teacher wants you to do things a very specific way
j2lie
  • j2lie
Then I am going to cry.

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