Does the function represented by the graph have an inverse function? The three coordinates are (0,0), (4,8), and (8,0).

- j2lie

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- 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

I cannot, because the line is slanted on the graph.

- jim_thompson5910

did you plot the three points?

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## More answers

- j2lie

yes. It's been drawn for me.

- jim_thompson5910

ok is it possible to draw a flat horizontal line through more than one point?

- j2lie

Yes, between zero and eight.

- 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

Why? I think there should be an inverse. I just don't know the function for the graph.

- jim_thompson5910

did you learn about the vertical line test?

- j2lie

Not yet. We're talking about functions and graphs. Also domain and range.

- 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

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

Ok. But how do I find the function of the graph?

- 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

can you post a screenshot of the full thing?

- j2lie

I will check my notes, but they're confusing.

- 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

I haven't learned that yet, so would it count if put the vertical and horizontal tests as my answer?

- 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

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

can you post a screenshot of the full thing?

- j2lie

Ok. Sorry.

- j2lie

I cannot attach a picture.

- jim_thompson5910

the "attach file" button isn't working?

- jim_thompson5910

maybe host it on a site like imgur and then paste the link

- j2lie

I have instagram.

- jim_thompson5910

that might work, just as long as you can host the image somewhere and share the link

- j2lie

My username is princess5062. You can see the picture that I just uploaded.

- j2lie

https://instagram.com/p/7Ja6_CmGRgNTAiNKsvvsYhyWXzo4VLMRi9m1A0/?taken-by=princess5062

- jim_thompson5910

it's telling me "Sorry, this page isn't available"

- j2lie

you have to have an account. It worked when I was logged on.

- jim_thompson5910

oh sorry I don't have an account. There's no way to share to non-account holders?

- j2lie

maybe I should i create an imgur account

- jim_thompson5910

you don't need to create one, they allow you to upload without an account

- j2lie

##### 1 Attachment

- jim_thompson5910

thanks

- j2lie

Sorry, it took forever.

- jim_thompson5910

ok so it's a V shaped line, I see now

- j2lie

That's what I said like a parabola.

- 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

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

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

`Can I use any information I wrote down for the function?` I'm not sure what you mean

- j2lie

Like the average rate of change, can I put that in an equation that is the function of the graph?

- jim_thompson5910

oh, yes you can to describe the slope of each piece of the V shape

- jim_thompson5910

so for #3, you'll have 2 answers there

- j2lie

That's the average rate of change also known as slope.

- j2lie

I got 2 and -2 for the rate of change. There is an inverse, because the signs are the opposite.

- 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

Is it going to be an equation or inequality?

- jim_thompson5910

the graph you mean?

- j2lie

Yes, the function of the graph.

- jim_thompson5910

the function is an equation

- j2lie

I thought it was an equality, because 2 and -2 are opposite of each other.

- j2lie

so the equation is y=|x| + 2 = -2?

- 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

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

The function is not an equation with the letter y. We have to find the equation for the function.

- jim_thompson5910

you can replace y with f(x)

- j2lie

I might be wrong, because I looked at problem six, it's asking for an equation of the function.

- jim_thompson5910

yeah you start with y = |x| and do a bunch of transformations to end up with y = -2|x-4| + 8

- j2lie

Where did you get the eight?

- jim_thompson5910

that's the y coordinate of the vertex

- jim_thompson5910

|dw:1441239501121:dw|

- j2lie

Thank you. How did you get the equation? Sorry, I'm not understanding your equation.

- jim_thompson5910

I started with the template y = a|x-h|+k

- jim_thompson5910

have you learned about y = a|x-h|+k ?

- j2lie

Did you multiply 2 and -2?

- j2lie

I haven't learned that yet.

- j2lie

I did, I'm so sorry. I am silly sometimes.

- jim_thompson5910

so because the vertex is (4,8), this means h = 4 and k = 8. I plugged those in

- jim_thompson5910

along with a = -2

- j2lie

so you moved -2 to both sides and multiplied 2 and -2?

- 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

Ok. Thank you for that clarification. So the function is -2|x-4|+8?

- jim_thompson5910

y = -2|x-4|+8
or
f(x) = -2|x-4|+8

- j2lie

f(x)=-2|x-4|+8

- jim_thompson5910

here's the graph to check
https://www.desmos.com/calculator/m8dav9jses

- jim_thompson5910

ignore the portions that aren't in quadrant 1

- j2lie

I don't know what to do on there

- jim_thompson5910

it's basically a graphing calculator
do you see the red V shape graph?

- j2lie

No. It's blank.

- jim_thompson5910

hmm strange

- jim_thompson5910

you should see this

##### 1 Attachment

- j2lie

I'm sorry. I never used a graphing calculator before.

- j2lie

That's correct, but on my paper the number eight is not labeled.

- jim_thompson5910

probably because it's cut off? not sure

- 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

To find the inverse graphically, you take each point and swap the coordinates
eg:
(4,8) ---> (8,4)
(8,0) ---> (0,8)

- j2lie

I need it in an equation.

- jim_thompson5910

later on when you learn about the vertical line test, you'll find that the inverse is NOT a function

- j2lie

I'm sorry, but I'm doing it the way I learned.

- jim_thompson5910

have you learned that absolute value functions are made up of two linear equations?

- j2lie

I don't know.

- j2lie

Do you mean like x=2 and x=-2?

- jim_thompson5910

so the term "piecewise function" doesn't sound familiar?

- j2lie

No. sorry. I don't know what level math you're in right now.

- jim_thompson5910

hmm I'm trying to figure out what the teacher wants and the method s/he wants you to use

- j2lie

He wants me to use composition. Have you ever heard about that?

- jim_thompson5910

yes

- j2lie

Ok, I need help with making the function conforming to the composition rules.

- j2lie

I don't need the composition. I think, but it's to check the inverse function.

- 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

I know what you mean.

- j2lie

##### 1 Attachment

- 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

How do I get that?

- jim_thompson5910

focus on the left half of the V shape
what two points are on this line?

- j2lie

(0,0) and (4,8)

- jim_thompson5910

what is the equation of the line through those two points

- j2lie

I don't know. I know the slope.

- jim_thompson5910

what's the slope

- j2lie

2

- jim_thompson5910

what's the y intercept?

- j2lie

I think I'm going to be wrong, but is it (2,4)

- jim_thompson5910

where does that line cross the y axis?

- j2lie

Ok. I'm feeling silly, but if the line continued wouldn't it be (0,0)

- jim_thompson5910

the left line goes through (0,0), yes

- jim_thompson5910

slope = 2 ----> m = 2
y intercept = 0 ---> b = 0

- jim_thompson5910

the left piece of the line is y = 2x+0 or just y = 2x

- j2lie

The line would just 2x

- jim_thompson5910

|dw:1441241810907:dw|

- j2lie

So the line is 2x without 0?

- 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

Ok. I'm going to try and find the inverse.

- j2lie

The y intercept is a zero. It cannot have an inverse.

- jim_thompson5910

why not?

- j2lie

you need to have an equation like y=x-4/3

- j2lie

It could work.

- jim_thompson5910

you can find the inverse of y = 2x though

- jim_thompson5910

swap x and y, solve for y

- j2lie

I found it and I made the equation y=2x+0.

- jim_thompson5910

you should find the inverse to y = 2x is y = x/2

- j2lie

Ok, because the inverse cannot have zero?

- jim_thompson5910

I don't know what you mean by not having 0

- j2lie

Never mind.

- j2lie

Should I use composition?

- jim_thompson5910

I think composition is only used to verify you have the correct inverse (as your attached image shows)

- j2lie

I'll use it. I need to make sure they are inverses.

- jim_thompson5910

ok so f(x) = 2x and g(x) = x/2
confirm that f( g(x) ) = x and g( f(x) ) = x

- j2lie

I got -2(x/2). Is that correct?

- jim_thompson5910

you should have 2(x/2) now simplify that

- j2lie

I got -x and -x for both compositions?

- jim_thompson5910

you should get +x

- jim_thompson5910

2(x/2) = 2x/2 = x

- j2lie

but f(x)=-2x.

- j2lie

I'm sorry. I confused myself.

- jim_thompson5910

no, f(x) = 2x

- j2lie

I wrote that when I tried it my way.

- j2lie

I fixed it and wrote x and x.

- jim_thompson5910

so that confirms x/2 is the inverse of 2x

- j2lie

Yes. I understand that now

- jim_thompson5910

now find the equation of the right half of the V

- j2lie

Ok.

- j2lie

There is no y intercept for the right side. There is a x intercept.

- jim_thompson5910

extend the line until it crosses the y axis

- jim_thompson5910

or you can use y = mx+b and solve for b

- jim_thompson5910

slope = -2 ---> m = -2
point on the line is (8,0) ---> x = 8 and y = 0

- 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

why do you have to replace x with 8?

- jim_thompson5910

because we know (8,0) is on that right half of the V

- jim_thompson5910

we could use (4,8) as well

- j2lie

Ok. Now I will multiply -2 and 8.

- jim_thompson5910

then isolate b

- j2lie

to isolate b I will cross cancel b by adding -b.

- jim_thompson5910

why not move the -16 over

- j2lie

f(x)=b-16

- jim_thompson5910

0 = -2*8+b
0 = -16+b
16 = b ... Add 16 to both sides.
so b = 16
agreed?

- j2lie

I'm starting to forget basic stuff, I'm sorry. I agree.

- jim_thompson5910

slope = -2
m = -2
y-intercept = 16
b = 16

- jim_thompson5910

y = mx+b turns into y = -2x+16

- jim_thompson5910

|dw:1441243656946:dw|

- j2lie

Ok. I got it. Now I need to find the inverse.

- j2lie

I got the answer. The answer is x and x.

- jim_thompson5910

what inverse did you get

- j2lie

The inverse is x and x.

- jim_thompson5910

those are the results after you do the composition

- j2lie

Yes.

- jim_thompson5910

what is the actual inverse

- j2lie

I don't know. I'm sorry.

- jim_thompson5910

did you solve x = -2y+16 for y?

- j2lie

Yes I did.

- jim_thompson5910

and you got y = ??

- j2lie

I got x.

- jim_thompson5910

forget about the composition stuff for now

- jim_thompson5910

isolate y in `x = -2y+16`

- j2lie

I'm sorry I did not solve it, but if you mean y=x-16/-2

- j2lie

then I did solve it.

- jim_thompson5910

good

- jim_thompson5910

you can also write that as y = (16-x)/2

- j2lie

Yes. I'm right. I got the composition correctly right?

- jim_thompson5910

yes if you ended up with x for each composition, you confirmed you have the correct inverse

- j2lie

Yes. Can you help with the rest of page? Or are you tired?

- j2lie

Me^

- jim_thompson5910

how much is left?

- j2lie

three problems and I have the answers I just need to make sure they're right.

- jim_thompson5910

ok I can check what you have

- j2lie

Ok. Please wait.

- jim_thompson5910

ok, I'll be right back.

- j2lie

Ok.

- j2lie

##### 1 Attachment

- jim_thompson5910

for #5 they want you to describe how the parent function is changing

- jim_thompson5910

how is y = |x| transformed into that V shape you see?

- j2lie

by points?

- jim_thompson5910

you agree that the graph of y = |x| looks like this right?
|dw:1441247206827:dw|

- j2lie

I don't agree. It's upside down.

- jim_thompson5910

well the graph you have is upside down, but that isn't the graph of y = |x|

- jim_thompson5910

the two are similar though
which is why we apply a bunch of transformations

- j2lie

Mine is not upside down. Mine is right, you're looking at it wrong.

- jim_thompson5910

this is what |x| looks like
http://demo.activemath.org/ActiveMath2/LeAM_calculusPics/AbsValue.png?lang=en

- j2lie

No it's not, it's going down.

- jim_thompson5910

the teacher is trying to get you to transform what I posted into what you have

- jim_thompson5910

at some point, the graph will flip

- j2lie

no. It won't we haven't learned that yet.

- jim_thompson5910

hmm let me think

- 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

You're taking harder Algebra.

- jim_thompson5910

yeah I'm thinking of a different algebra class maybe

- j2lie

I'm taking Algebra 2. How about you?

- jim_thompson5910

your best bet is to ask the teacher since he'll know what he wants

- jim_thompson5910

I'm not in algebra, but I'm probably thinking of another algebra class

- j2lie

So your other answers are wrong? There are only two algebra classes.

- 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

Then I am going to cry.

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