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ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 unit (3 Viewers)

SpiralFlex

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Jesus, no wonder I didn't get 100 in my 3U haha.
 

SpiralFlex

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

the inspection method? haha yeah it's pretty cool :)
No I should be more clear. The inspection method allows you to factorise the cubic with ease. Rather than finding u-1 as a factor.
 
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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

No I should be more clear. The inspection method allows you to factorise the cubic with ease. Rather than finding u-1 as a factor.
but then that's not really a "trick" is it? Seems quite obvious...
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

I currently am not understanding anything about the consequences of the factor theorem/special results.

How did everybody get to understand this?
 

Sy123

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Dont worry I dont know how to do that polynomial via inspection either, either way just use u=x-1, find a factor by substituting different factors of the constant, then use polynomial division, factorise and get the polynomial as linear factors of u, find the values of u, sub in x-1, then you have x.
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Finding Domain is obvious for hyperbola, its pretty much what x can be, so what we are looking for is what x cant be.
For
x cannot be 3.

Therefore our domain is all real x,

For Range however, in hyperbolas, what you can do is subsitute negative and positive infinity values into your calculator (sub in very large numbers), and see where they approach, for example in that above function, when you sub in values you will notice that you will get very close to 1,
our range is hence y is all real except y cannot equal to 1.

A useful tip, that will help you in 3U further curve sketching and finding range of hyperbolas is:


Our horizontal asymptote (and the value that y cannot be) is:

This works to find horizontal asympote for any function:



The rule is:

If n<.m, the horizontal asymptote is y=0
If n=m, horizontal asypmtote is y=a/b

If n>m, use polynomial division (do not worry about this right now)



That is just the logic behind it that is all.</m,>
Sorry if I'm going backwards here, but can someone explain what to do when the degree of "x" in the numerator is greater than the degree of "x" in the denominator? I know it's something to do with expanding the x's in the numerator to whatever is in the denominator so that they can cancel out or something, but I don't know exactly what to do. E.g. x^2/(2x+3)
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Sorry if I'm going backwards here, but can someone explain what to do when the degree of "x" in the numerator is greater than the degree of "x" in the denominator? I know it's something to do with expanding the x's in the numerator to whatever is in the denominator so that they can cancel out or something, but I don't know exactly what to do. E.g. x^2/(2x+3)
You will probably never meet this in preliminary unless your teacher decides to be really nasty and teach it, and then put it in the test.
But here is how it works, referring to my above 'formula' if you will. Since the degree of x on the numerator is higher than the denominator, then we use polynomial division to find oblique asymptotes (or slant, curved, whatever). What we do is, we divide the denominator by the numerator, our result becomes the equation for the oblique asymptote (ignore remainder), for the example you gave:

Long Division

As you can see here, our oblique asymptote is
What this means is, as we approach infinity, our function approaches this line, so we draw this line on our Cartesian plane (in dotted form to illustrate asymptote), we can then take limits of either sides to infinity, find vertical asymptotes, find limits of either sides of vertical asymptotes (if you want to be absolutely sure find differentiate and find stationary points). Then sketch.

You never really see it in 3U either (I think 4U), but yeah.
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Oh right, I get it. Yeah our teacher decided to go over it actually, but did say that it wasnt part of the 3U syllabus but it was studied in 4U. Seems a bit easy for 4U though
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Eliminate the parameter and hence find the Cartesian equation of the curve:
(a) x=3-t, y=2t+1
(c) x=1+2tanx, y=3sedx-4

I do not understand how this works and why it is even in a form like that (since the standard paramtetrisation is x=2at and y=at^2)
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

What you are trying to do in any parametrics question is to, obviously, eliminate the parameter. So we have to get rid of t somehow. By expressing t in terms of x, we can then sub this into the y equation, eliminating the t's and leaving just an equation with x's and y's. For example:

t=3-x
y=2t+1
y=2(3-x)+1
y=7-2x

Now do a similar thing to c: work out how you can remove the thetas...
 

D94

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Eliminate the parameter and hence find the Cartesian equation of the curve:
(a) x=3-t, y=2t+1
(c) x=1+2tanx, y=3sedx-4

I do not understand how this works and why it is even in a form like that (since the standard paramtetrisation is x=2at and y=at^2)
With (a), make t the subject in one equation, and substitute that into the other question. In (c) I think that should be tan(t) and sec(t) because it would be self-implicitly defined but anyway, I would rearrange y=3sec(t)-4 so that sec(t) is the subject, then square both sides of that new equation. Now you have a sec2t which is just tan[sup2[/sup]t+1. Then rearrange the first equation so that tan(t) is the subject then substitute that into the equation with sec2t.

That parametrisation is for the parabola.
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Lol oops (probably should of read the definition of cartesian beforehand)
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Dont worry I dont know how to do that polynomial via inspection either, either way just use u=x-1, find a factor by substituting different factors of the constant, then use polynomial division, factorise and get the polynomial as linear factors of u, find the values of u, sub in x-1, then you have x.
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Okay. I am sticking with LaTeX, Math Type is weird...
 

SpiralFlex

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Oh right, I get it. Yeah our teacher decided to go over it actually, but did say that it wasnt part of the 3U syllabus but it was studied in 4U. Seems a bit easy for 4U though
You should still know it. Our head teacher may sneakily include it in your yearlies like he did last year with our group. It doesn't hurt to learn.
 

Sy123

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Thanks for the heads up spiral
 

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Probably a rookie concept, though I guess that's why I'm asking it just to confirm.

For inequalities, do you have to square both sides, or is there an alternative way (I think there is, though I don't fulply remember it - I think it is where you just multiply it though you have two cases.
 

deswa1

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Re: ITT we help examine to get 85 in his Maths extension 1 yearlies so he can do 4 un

Probably a rookie concept, though I guess that's why I'm asking it just to confirm.

For inequalities, do you have to square both sides, or is there an alternative way (I think there is, though I don't fulply remember it - I think it is where you just multiply it though you have two cases.
Give an example question?
 

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