Question

Absolutely no idea. |Graphing Rational Functions|
http://gyazo.com/672aec44b53596cda4a51d80bb40cfe5

Answer 1

@Hero

Answer 2

@agent0smith

Answer 3

@Loser66

Answer 4

@e.mccormick

Answer 5

Well, do you know what a linear binomeal is?

Answer 6

Not quite.

Answer 7

The equation of a line. So some mx+b thing.

Answer 8

Now, do you know what a rational function is?

Answer 9

Isn't it just a regular function?

Answer 10

No. The rational comes from ratio, which also sometimes means fraction. That is where the numerator and denominator part come in.

Answer 11

So am I going to be using proportions later on?

Answer 12

No. Just going to make a fraction with a line equation on both the top and bottom, then graph it.

Answer 13

y = (6x-5) / (3x-2) would that work?

Answer 14

Yes.

Answer 15

Is it y = or F(x) ?

Answer 16

I assume you mean f(x) because f(x) and F(x) have technically different meanings...

Anyhow, either works and they said use a graphing utility, so:
https://www.desmos.com/calculator/m6kzzwmhvu

Answer 17

That's my graph?

Answer 18

Yep. As you can see, it uses the function you selected.

Answer 19

Now just do the whole asymptote and intercept calculations.

Answer 20

@e.mccormick It says Include the horizontal and vertical asympotes and the x- and y- intercepts on your graph. How would I do that?

Answer 21

I have no idea what they mean by that.

Answer 22

Well, do you know what the intercepts are?

Answer 23

X intercept, where the line goes through the X axis, Y is where it goes through the Y intercept.

Answer 24

Yes. Another way of saying that is (x?,0) and (0,y?), with the key points being that in each case you know what either x or y is: 0.

Answer 25

If you put 0 in for x and solve for y, you have the y intercept.

If you put 0 in for y and solve for x, you have the x intercept.

Answer 26

And how do I find the asympotes?

Answer 27

Well, have they gone over any of the asymptote calculations with you?

Answer 28

Not at all.

Answer 29

They are actually not too hard. One has to do with the coefficents and the other has to do with where the denominatior (bottom) would be 0 (undefined).

Answer 30

See, asumptotes are lines that are not crossed. Now, horozontal ones might get crossed at some points, but the \(\infty\) and \(-\infty\) tails of the equation do not cross them. The vertical ones flat out do not get crossed.

So to find the vertical one, when would the bottom of the fractio be invalid (0)?

Answer 31

No idea. Would 0 be a choice? Since 0 is something involving extraenous equations?

Answer 32

Well, how you find it is set the bottom = to 0. So:

3x-2=0

What is that x value?

Answer 33

0.6 repeating

Answer 34

Might as well leave it as a fraction, so 2/3.

OK. On the graph, click on line 2 in the equaition list on the left. Then type in:

x=2/3

When you do that, it will add the first asymptote.

Answer 35

Now, the other asymptote, since these have the same power on the first term, is y = the ratio of the coefficents.

Let me write that out with letter.

Answer 36

OK, so in this form:

\(\Large y=\dfrac{m_1x_1+b_1}{m_2x_2+b_2}\)

You have the two linar equations there. The ratio of the coefficents of the first term is:

\(\Large y=\dfrac{m_1}{m_2}\)

That is the horozontal asymptote. So if you pull those out of your one, you plug them in there and presto you have it.

Answer 37

Thanks. :>

Sorry for the long responses, my internet keeps dying out.

Answer 38

So how would my equation look with the asympotes put in?

Answer 39

I'm still super confused.

Answer 40

=(

Anyhow, if you follow those instuctions and add the horozontal and vertical to that graph, you will have it all on one.

OH! And you can put in (x,y) point in the parentisis and it will put them as a point on the graph. Also, if you click the gear icon (Edit List) it brings up the settings so you could change the asymptotes to dashed ines. They are typically dashed on a drawing.

Answer 41

Well, you found the vertical one, x=2/3

What did you get for the horozontal one?

Answer 42

.83 repeating, which is 5/6

Answer 43

Ummm... that is the top, which would be the x intercept, not ratio of the coefficents. It is the m part of the top over the m part of the bottom when both are lines. y=mx+b. That m.

Answer 44

I have to input 5/6 into an equation, don't I?

Answer 45

It's not just standalone.

Answer 46

OH. It's 2!

Answer 47

That is the x valuse of the x intercept. (0,5/6)

y = (6x-5) / (3x-2)

0 = (6x-5) / (3x-2)

0 (3x-2) = (6x-5)

0= (6x-5)

0= 6x-5

5= 6x

5/6= x

Therefore, when y=0, x = 5/6, which is the point:

(5/6, 0)

That is an intercept.

What I was talking about was the ratio of the coefficents for the asymptote.

Answer 48

@e.mccormick One question about your lesson before i leave you aloen

Answer 49

http://gyazo.com/977980a8e353515ae2f31a20ab8bd2cb

Answer 50

In the first equation with m1x1 + b1/ m2x2 + b2

Answer 51

I can use y 1 and y2 as well?

Answer 52

And also, what's b2?

Answer 53

There is no \(y_2\). It is a single function. I just did the right side with sub one and sub 2 because it was two linear equations in a ratio.

Answer 54

\(y=\dfrac{\text{Linear equation one}}{\text{Linear equation two}}\)

Answer 55

Are you sure I need to do all of this for my question? This is all foreign to me.

Answer 56

I've never learned this.

Answer 57

That part was just an example. To show you what was needed.

Odd that they never went over all this. Your question asks for it all.

Answer 58

To recap: When it asks me to include my vertical and horizontal asympotes on the graph, I would label them? Would that work?

And the X and Y intercepts, I would do the same.

Answer 59

"Create a rational funtion with a linear binomeal in the numerator and denominator" That is the y = (6x-5) / (3x-2) part.

"Graph using technology." That was the original graph.

"Include the horozontal and vertal asymtotes" Well, to include them, you must find them. The horozontal is the \(m_1/m_2\) part and the vertical is solving the bottom for 0.

"and the x- and y-intercepts) Well, that is the set x = 0 and solevm then set y = 0 and solve.

Part 2: Show all your work... well... that should not be too hard if you do the work to find them. =)

Answer 60

:)

Answer 61

I can't thank you enough.

Answer 62

You might neeed to take a screen shot to label them.

Now, did you ever find the ratio of \(m_1/m_2\)?

Answer 63

Oh, and if you want a reference to all of this, here is one:
http://www.purplemath.com/modules/grphrtnl.htm

If you go over that, you will see it is the same basic stuff I was doing. =)

Answer 64

I never did. I'm going to bed, since it's a bit late in the UK. I'll be sure to contact you via PM to let you know how I do. :>

Answer 65

I'm also going to leave this post open so I can use it for refernce

Answer 66

You can also come back to closed ones. They are in the "Asked Questions" on your profile page.

https://www.desmos.com/calculator/olj8mf6erf