drawing graphs (1 Viewer)

[martin310015]

i need help with 3u curve sketching don't know the step :
can someone explain to me how to do it in steps
e.g

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

thanx.

 

[AGB]

i think the best way is to avoid using calculus...this is the way i do it for fairly simple ones

1. find asymptotes
2. find intercepts
3. test points if necessary

so for this graph, there are vertical asymptotes at x = 0 and x = 3, and there is a horizontal asymptote at y = 1....there are no x or y intercepts....from that it is pretty easy to figure out

 

[martin310015]

ok the problem is that can't find what the graph is when it is between the asymptote

 

[AGB]

Originally posted by martin310015
ok the problem is that can't find what the graph is when it is between the asymptote

it sort of looks like an upside down parabola, but it approaches the asymptotes negatively (if that is a word...)

 

[mojako]

>> but it approaches the asymptotes negatively (if that is a word...) <<
it's called: approaches from below (the asymptote)

I really don't think this question goes with 3U curve sketching. 3U curves should be reducible to (can be rearranged into.. I think reducible is the right term..) standard form of y = 1/(x+a) + b, where a and b are constants.
eg y=(3x-7)/(x-1) can be rearranged.

In fact the curve in your question is quite complex even for 4U students (like me..) as it has 2 turning points.

You should have in mind that curves of this form, (x^2+c)/(x^2+d) etc, will have some kind of branches (like hyperbola y=1/x), there may be 2 branches or more (well, I haven't seen one with 4 branches.. not sure if that's possible and am too lazy to find out).

Take the limits as x approaches positive and negative infinity. One way is to use calculator...
Another is to:
-> expand the denominator, which gives x^2-3x
-> Divide numerator and denominator by x^2, giving (1+3/x^2)/(1+3/x).
-> From that new notation, it's easy to see what y is when x is very large.
As x approaches positive infinity, the curve approaches y=1 from above. as x approaches negative infinity, it approaches y=1 from below.

Also, we know there are vertical asymptotes at x=0 and x=3

Use calculator to find where the curve is near those vertical asymptotes**(see notes below after you've read the rest). Substitute x=3.1 and x=-0.1 for example. Then, from intuition you can draw the two branches since you also know where the curve is for positive and negative infinities.

Then we can deduce that there must be "something" between x=0 and x=3. again we can substitute values at x=0.1 and x=2.9 and probably x=1.5 to see what happens (remember, that "something" will have the form like that kind of bowl or parabolic thing.. not exactly a parabola though).

Alternatively (and especially if you need to show where the turning point is), find dy/dx using quotient rule. determine nature of turning points. You'll find that there;s a maximum at x=1. Hence between x=0 and x=3 there's a kind of concave-down parabola.


**You'll also find that there's a minimum at x=-3. This is what makes it a bit tricky (for some people at least). If we don't find dy/dx and don't do that calculation at x=-0.1, we might think intuitively that since at x=negative infinity, the curve approaches y=1 from below, near x=0 y will have large negative values. But this is not the case since there's a turning point in between negative infinity and x=0.

 

[martin310015]

i got this question out of the new senior mathematics book for 3u course.....its quite hard

 

[mojako]

New Senior Maths.. wait.. is that the one by Fitzpatrick?
We (my maths class) use part of it in our extension2. It seems to be easier than Cambridge's Extension2 book...

 

[CM_Tutor]

Yes, New Senior Maths is Fitzpatrick's text book, and yes, this is a valid Extn 1 question. To expand on AGB's earlier post, a fuller cuve sketching strategy is:

1. Domain
2. Intercepts - both x and y
3. Symmetry - odd / even / neither
4. Asymptotes - both horizontal and vertical
5. Behaviour near asymptotes / sign (ie. if a vertical asymptote is at x = a, examine y as x --> a⁺ and as x --> a^-)
6. Calculus - this step is frequently unnecessary

NOTE: All problems should be approached (at least mentally) in this way. Most people do calculus first - bad idea. If you have thought about all of this first, you'll frequently be able to predict, at least in general terms, the outcome of the calculus.

 

[Grey Council]

oo k. So that is ALL we have to do? Nothing else to remember? Cause i'll be honest with you, my teacher last year (although he is dead and I shouldn't be talking about him ) wasn't all too good. He just drew y = 1/x, and then told us to do the whole 2u and 3u fitzpatrick questions by ourselves. I did the 2u ones (from what I remember. I think) but school ended and i didn't get around to doing 3u curve sketching.

in a nutshell. I'm crap at curve sketching. And from this point backward, i've always tried calculus first. >.<

Ta CM_Tutor.

 

[CM_Tutor]

Grey Council, this question, y = (x² + 3) / x(x - 3) is a good illustration:

1. Domain - clearly {x: x <> 0, x <> 3}

2. Intercepts - none, as y = 0 requires x² + 3 = 0, which has no solution, and x = 0 is outside the domain.

3. Symmetry - Neither odd nor even - this doesn't tell you much, but it tells you alot if it is odd or even

4. Asymptotes - Vertical, clearly x = 0 and x = 3
Horizontal - y = (x² + 3) / (x² - 3x) = (1 + 3 / x²) / (1 - 3 / x) ---> 1 as x ---> + / - inf

5. Behaviour near asymptotes / sign

As x ---> 0⁺, y has form (+) / (+)(-), so y ---> - inf
As x ---> 0^-, y has form (+) / (-)(-), so y ---> + inf

As x ---> 3⁺, y has form (+) / (+)(+), so y ---> + inf
As x ---> 3^-, y has form (+) / (+)(-), so y ---> - inf

As x ---> + inf, 3 > -3x, so numerator > denominator, so y ---> 1⁺
As x ---> - inf, 3 < -3x, so numerator < denominator, so y ---> 1^-

Now, before looking at 6, sketch what you know. Draw coordinate axes, vertical dashed lines at x = 0 and x = 3, and a horizontal dashed line at y = 1. Put a little arrow pointing up at the top of the x = 3 line on the RHS, and at the bottom of the x = 3 line on the LHS pointing down. Put a little arrow pointing up at the top of the x = 0 line on the LHS, and at the bottom of the x = 0 line on the RHS pointing down. Put a little arrow pointing to the right above the y = 1 line at its RH end, and below the y = 1 line pointing left at its LH end.

You know the curve has no intercepts (so it never meets / crosses an axis), and you need a smooth curve joing all the 'bits', as the curve is differentiable throughout its domain. There really is only one way to join the dots, and it should be obvious that calculus would show a MAX tp somewhere on 0 < x < 3, and a MIN TP, with 0 < y < 1 for some value x < 0. If this min is at x = a, then there must be an inflexion for some x < a, with
f(a) = (a² + 3) / a(a - 3) < y < 1.

IF you must do the calculus, this would be what it would show.

Now, consider if you had done calculus first. You would have a max on 0 < x < 3 with a y co-ordinate below that of the y co-ordinate of the min with x < 0. This is not a promissing start. You may also have the inflexion, but you've still got major problems.

I conclude - calculus is the last process to be considered in curve sketching (especially for an Extn 2 student), not the first.

As a further point, try and sketch y = x² / (x² - 4). You should be able to sketch it, showing all necessary features including the stationary point(s) without calculus.

 

[Grey Council]

ahh k, mad.

Would you recommend me to do the Fitzpatrick exercise using this "technique"?

and btw, sketching means roughly graphing, right? So no need for exact values of turning points etc.

 

[CM_Tutor]

Originally posted by Grey Council
ahh k, mad.

Would you recommend me to do the Fitzpatrick exercise using this "technique"?

and btw, sketching means roughly graphing, right? So no need for exact values of turning points etc.

Which Fitzpatrick exercise? Name or page ref please.

Yes, sketching does mean roughly, but for the above example I suggested, y = x² / (x² - 4), you should be able to give the exact coordinates of all stationary points without calculus...

 

[mojako]

>> and yes, this is a valid Extn 1 question. <<
Just curious, what syllabus point does this relate to?

>> x --> a+ and as x --> a- <<
Just in case somebody doesn't know, x --> a+ means as x approaches a from the right (so x is slightly larger than a).

>> this question, y = (x2 + 3) / x(x - 3) is a good illustration: 1. Domain - clearly {x: x <> 0, x <> 3} <<
CM_Tutor, is <> a proper symbol (on the net or computer) or is it yours only? I'm not offending. I want to know too...

 

[CM_Tutor]

Originally posted by mojako
Just curious, what syllabus point does this relate to?

Off hand, I have no idea, but I've seen variants of it on enough papers to know that it's examinable - after all, is the syllabus what is written in the syllabus document, or is it what is asked in exams? I personally favour (for practical matters) the second definition.

>> this question, y = (x² + 3) / x(x - 3) is a good illustration: 1. Domain - clearly {x: x <> 0, x <> 3} <<
CM_Tutor, is <> a proper symbol (on the net or computer) or is it yours only? I'm not offending. I want to know too...

<> is a standard symbol for not equal to in computer programming. Another standard symbol is !=, but that doesn't seem that common on this board, and I prefer <> in any case. (It's like => for equal to or greater than - <> means greater than or less than, and thus, not equal to). I would use an equal sign with a slash through it, ie. the correct symbol, if I could see an easy way to generate it.

 

[Grey Council]

exercise 24(a), page 70, 3u book. It's only 16 questions, maybe I should do em all. hmm

Although if you think that the cambridge book is better, by all means, i'd appreciate either of the books. I think its about time I learnt how to graph. hehhe

 

[mojako]

QUOTE: y = x2 / (x2 - 4), you should be able to give the exact coordinates of all stationary points without calculus...


Transform that into some other form. The goal for any type of fraction functions, roughly, is NOT to have x^n terms, where n>0 on both top and bottom of the fraction.
So, in this particular function, transform it to y = (x^2-4)/(x^2-4) + (4)/(x^2-4). Which then simplifies into y = 1 + 4/(x^2-4).
Then, you know there's a turning point somewhere between the two vertical asymptotes. But by examining that transformed form, you can sort of see that it's around x=0. Why? The function is even (due to x^2 being the only x-term). As x approaches zero, y approaches zero from below. So, maximum point at (0,0).

 

[mojako]

I knew != but not <>.

CM_Tutor... can I know what school you teach? (if any)

>> is the syllabus what is written in the syllabus document, or is it what is asked in exams <<
Proper HSc exam must not go beyond the syllabus. Part B of the syllabus document (downloadable from the "proper" Board of Studies website, of course) details what can and cannot be asked (give limits to the complexity of the questions). I sound like a teacher, huh? But I'm not...

 

[CM_Tutor]

Originally posted by Grey Council
exercise 24(a), page 70, 3u book. It's only 16 questions, maybe I should do em all. hmm

Although if you think that the cambridge book is better, by all means, i'd appreciate either of the books. I think its about time I learnt how to graph. hehhe

I think that exercise is worthwhile, although perhaps a little repetitive. I don't have Cambridge at hand, so I won't comment on its specifically, beyond saying I like Dr Pender's style.

And yes, you should learn how to sketch - it'll simplify 4u curve sketching enormously.

 

[CM_Tutor]

Originally posted by mojako
I knew != but not <>.

CM_Tutor... can I know what school you teach? (if any)

>> is the syllabus what is written in the syllabus document, or is it what is asked in exams <<
Proper HSc exam must not go beyond the syllabus. Part B of the syllabus document (downloadable from the "proper" Board of Studies website, of course) details what can and cannot be asked (give limits to the complexity of the questions). I sound like a teacher, huh? But I'm not...

Mojako, it isn't necessary to transform this equation to see a max tp at the origin - Any even function which is continuous and differentiable at x = 0 must have a turning point at its y-intercept. Similarly, any odd function that is continuous and differentiable at x = 0 must have an inflexion at the origin.

Also, I'm not a Capital-T Teacher - I don't work at a school. I am a PhD student at USyd, studying Chemistry Education, and at the same time doing a Masters degree in Education in the field of Teaching and Curriculum Studies. I have an enormous amoount of experience with HSC Maths and Chemistry, and I know the Chemistry syllabus document pretty well, but I don't worry much about the Maths one - I've never seen the need.

I agree with you that the syllabus document should be the final word, but it isn't. In the 2001 Chemistry HSC (first year of the new syllabus), there were two questions asked that were outside the syllabus. The markers argued they should be excluded, and the syllabus people agreed. They were overridden by the examination committee (who set the paper), and the questions were marked - although they did say they wouldn't be asked again. Should this have happened? No - but it did (or so I've heard) - and hence my 'operational' definition.

 

[Grey Council]

*adds the exercise to the to-do list over the weekend*

and i dunno about CM_Tutor not being qualified to teach, he prolly explains stuff in maths better than any person i've ever met. Its a wonder he can manage to do it over the net as well. heheh, so mojako, just take his word like the bible. if he says something, its right. ^____^

btw, how do you know if an equation is even or odd? I mean, I know that if f(x) = f(-x) then its even, but how do you see something like x^2 / (x^2 - 4) and know that its an even function?

And what about odd functions?