Sketching Graphs (1 Viewer)

[let.me.die]

hi. im not too sure bout how to sketch things with asymptotes. so i was wondering if someone could show me how to do this question. if u wouldnt mind could u list out the method. any help is appreciated.

question: SKetch, showing any stationary points, points of inflexion or asymptotes:
1. y = x^2 / (2x + 3)

 

[sasquatch]

1. y = x^2
-------
2x+3

From your post it seems that you already know how to find stationary points and points of inflexions, but if you dont...reply.

To find the vertical assymptote all you simply do is look to see what values of x, would make the equation undefined. This simply means, what x value, makes the denominator equal 0. This can be done as such:

2x + 3 * 0
2x * -3
x * -3/2

Therefore the vertical asymptote is x = -3/2.


To find horizontal assymptotes, it is a little bit more complicated. The method i choose to use is using the limit as x approaches infinity. As i learnt last week from i think who_loves_maths, you need to simplify the equation so that if possible, you may split into seperate parts, so that it may be represented in the addition or subtraction of two functions.

An example is y = (x+3) / (x+5)

this can be simplified to:

y = (x + 5 - 2) / (x+5)
= (x+5) / (x+5) - 2 / (x+5)
= 1 - 2/ (x+5)

With every equation, this is not possible.

From y = (x^2) / (2x + 3), it is not possible to simplify using the method specified above. So limits can be applied straight away.

(divinging both the numerator and denominator by the highest power of x on the denominator)
y = lim x->inf (x^2/x) / (2x/x + 3/x)
(simplifying)
y = lim x-> inf (x) / (2 + 3/x)
(substituting lim x-> inf 1/x as 0)
y = x / (2 + 0)
y = x/ 2.

Thus the vertical or in this case oblique asymptote is y = x /2.

Hope you get all of this, and if you have any more problems, feel free to ask.

 

[let.me.die]

thanx. that was a really good explanation. i think i know what to do now.

 

[acmilan]

You can use division to split x²/(2x + 3) if you wanted.

 

[sasquatch]

hey um..i just checked with a graphing program i made, the vertical asymptote, yeah it looks correct, but the oblique asymptote..

Hmm im not sure if it is right then. But i use that method, and it works every time so thats weird.

Acmilan can you explain what you mean by dividing to split up, because again i think that is the problem.


Something that really annoys me is.. NO text book ive ever seen has a proper explination on finding asymptotes, most of them just write it not showing any steps, or use something that applies ONLY to that particular equation.

I know another method:

y = 1/x
y - 0 = 1/x
(y - 0)(x - 0) = 1

Therefore y cannot = 0, x cannot = 0, hence the asymptotes are y = 0, x = 0.

y = 10/(x+5) - 1
y + 1 = 10/(x+5)
(y + 1)(x + 5) = 10

Therefore y cannot = -1, x cannot equal -5, hence the asymptotes are y = -1, x = -5

This works for a range of equations but not all, so i dont ever use it.

If your given an equation as i shown before that can be split, y = (x+3) / (x+5) = 1 - 2/ (x+5)

y = 1 - 2/(x+5)
y - 1 = -2/(x+5)
(y - 1)(x + 5) = -2

As you can see it works for a range of them, but look!!

y = x^2 / (2x + 3)

(y - 0)(2x + 3) = x^2

Now what??

(y - 0)(2x + 3) = 1
-----------------
x^2

(y/x^2)(2/x + 3/x^2) = 1

that doesnt really help as y and x are not independant from each other.

 

[icycloud]

I shall try to explain:

y = (x^2)/(2x+3)

Using long division,

y = x/2 - 3/4 + (9/4)(1/(2x+3))

x --> infinity, y --> x/2 - 3/4

Thus, y = x/2 - 3/4 is the oblique asymptote.

Now, y is undefined at x = -3/2, thus that is a vertical asymptote.

You can then draw a guide graph y = 2x+3, which will show you that:

when x > -3/2, y > 0 and
when x < -3/2, y < 0

Then you can use calculus to find the turning points, etc...

Therefore the graph should be something like this:

Asymptotes in Blue
Guide graph in Green
Graph in Red

(Sorry about the crude nature of the graph, I did it in Paint )

 

[sasquatch]

hey at least i got the vertical one right and the x/2 bit of the oblique asymptote right..haha

icy dude howd you do that long division thing? can you show in more steps?

 

[icycloud]

You basically do the same thing as you do with any long division, except maybe this is a bit unfamiliar to you because it involves fractions. I'll try to explain: (once again forgive the dodgy diagram )

 

[sasquatch]

man..

whatever you did is harder to understand than calculus and even complex numbers!!!

is there any book or something that...or hmm actually ill ask my teacher tomorrow!!!!!!! but like my old teacher..maths didnt really explain it well...

so yeah i dunno if this one would be able to....

edit: is that polynomal division? if so.. i guess id figure it out soon since were up to 28.2 in polynomals next exercise 28.3 goes on to polynomial division so yeah!

 

[icycloud]

Oops sasquatch, I assumed that you knew polynomial long division already? Anyway, maybe this page will help (looks pretty detailed to me):

http://campus.northpark.edu/math/PreCalculus/Algebraic/Polynomial/Factoring/

 

[sasquatch]

Nup i have no idea what is talking bout..

man i feel stupid.. i dont understand that thingo and im doing 4 unit ( im gonna fail...

Ah well thanks alot anyway. I think itd be easier to be shown step by step how to do it. Hmm this is funny though, i didnt know how to do long division this way, dropping things down, and had to learn to do binary division. I learnt long division last week, and now all this other stuff relating is comming up.

 

[insert-username]

sasquatch, don't be disheartened. You'll pick up polynomial divison just fine I'm sure with a person explaining it to you face to face. It's not that hard a concept when there's someone there to help.

To find stationary points, derive the function. When the derivative = 0 for a value of x, there is a stationary point. Find the second derivative to determine what type of stationary points they are. Substitute in your x-value, and if the second derivative is positive, the stationary point is a local minimum. Second derivative negative is a local maximum point, and second derivative = 0 is usually a point of inflexion.


I_F

 

[icycloud]

Hmm well just think of it as normal long division, except with pronumerals in it.

 

[Riviet]

You can then draw a guide graph

Lol! That's a 4u technique! I know becuase i was reading the curve sketching chapter in the text book.