VCE Methods Question Thread!

[lzxnl]

Firstly, the guide was written for spesh, so it might be a little above the Methods level (the part you have attached to your post regarding 'm' and 'n' is a restriction on the graphs they offer in questions in spesh, not relevant for you).

Well say we had the equation:

We can see that there is obviously a vertical asymptote at as can't exist at that point.

To see if there are other asymptotes, we say
Hence,
Hence, which is your other asymptote!

As x approaches infinity, y becomes unbounded so the limit doesn't exist xP
I think saying y-x^2-1 approaches 0 is better

[pi]

Yeah my mathematical notation is very rusty (/I never knew it LOL), I'd take my notation with a grain of salt and just focus on the ideas haha, thanks lzxnl!

[Splash-Tackle-Flail]

As x approaches infinity, y becomes unbounded so the limit doesn't exist xP
I think saying y-x^2-1 approaches 0 is better

Sorry this is probably branching out to spesh but would we lose marks for saying y approaches x^2 +1? (My spesh teacher is more of a physics teacher and doesn't know

[knightrider]

What would be the rule of the following graphs in the form assuming a=1 or a=-1

[Conic]

What would be the rule of the following graphs in the form assuming a=1 or a=-1

You can use the fact that you know the turning point. For the first, the turning point is (2,-3), so you can write the rule in the form for some real k. This gives . We are given that the coefficient of must either be 1 or -1, so k must be either 1 or -1. The parabola has a minimum, so the coefficient of is positive, and hence k is 1. So the rule is . You can repeat the same logic for the second question, but in this case the coefficient of the term is negative, as the parabola has a maximum.

[knightrider]

You can use the fact that you know the turning point. For the first, the turning point is (2,-3), so you can write the rule in the form for some real k. This gives . We are given that the coefficient of must either be 1 or -1, so k must be either 1 or -1. The parabola has a minimum, so the coefficient of is positive, and hence k is 1. So the rule is . You can repeat the same logic for the second question, but in this case the coefficient of the term is negative, as the parabola has a maximum.

Thanks conic 

but you know when writing the rules aren't you meant to include the domain restriction as the domain isn't R

[Conic]

From memory, in exams when you are asked for the rule of a function you only need to write the equation of the function. You give the domain when you are asked for a function (I may be mistaken, I finished Methods over a year ago). In this case the graph is shown with a resitricted domain, so I would write it in function form (something like ).

[keltingmeith]

Sorry this is probably branching out to spesh but would we lose marks for saying y approaches x^2 +1? (My spesh teacher is more of a physics teacher and doesn't know

As x approaches infinity, y approaches x^2+1... But x is approaching infinity, so y is approaching infinity as well, so it's not really approaching x^2+1.

As lzxnl mentioned earlier, it's just an odd thing to say - saying y-x^2-1 is approaching zero is more correct.

[Conic]

Yes that is an error

[cosine]

Strawberries: Yeah I was referring to the [, ] brackets instead of (, )

Conic: thank god.... 

[strawberries]

Strawberries: Yeah I was referring to the [, ] brackets instead of (, )

Conic: thank god.... 

Omg sorry I didn't get your question haha.

but yeah, infinity or negative infinity cannot be 'included' in anything, it must always be a round bracket

[SE_JM]

As x approaches infinity, y approaches x^2+1... But x is approaching infinity, so y is approaching infinity as well, so it's not really approaching x^2+1.

As lzxnl mentioned earlier, it's just an odd thing to say - saying y-x^2-1 is approaching zero is more correct.

Hi Eulerfan
Could you explain this in more detail? Sorry, I don't understand
Would you still say that that the oblique asymptote is y=x²+1, though?
Also, how you notate that as x approaches infinity, y-x^2-1 approaches zero?
lim
x ---> infinity    y-x^2-1 --->0

is this right?


Thanks

[Splash-Tackle-Flail]

Not really a methods question but is there a reason why the scaling was so high in 2O13? Is it just because it was a harder exam but some people still smashed it raising the standard deviation? How did they calculate it so a 46 in 2013 scaled higher than a 50 last year?

[SE_JM]

Hello,
I've come across another thing I didn't understand...
Let's say that I'm graphing this function:
f(x) = x^1/2+ 1/x
I can clearly see that f(0) cannot exist so that my vertical asymptote would be at x=0
Then I can see as limit of x as it approaches infinity is x^1/2

So I would say that the oblique asymptote is at y=x^1/2

But when I looked at the answer, there are asymptotes at y=x^1/2 and y=1/x

It's the second question in the attachment file

Could you tell me where I am wrong?

[Splash-Tackle-Flail]

Hello,
I've come across another thing I didn't understand...
Let's say that I'm graphing this function:
f(x) = x^1/2+ 1/x
I can clearly see that f(0) cannot exist so that my vertical asymptote would be at x=0
Then I can see as limit of x as it approaches infinity is x^1/2

So I would say that the oblique asymptote is at y=x^1/2

But when I looked at the answer, there are asymptotes at y=x^1/2 and y=1/x

Could you tell me where I am wrong?

I think your answer is correct- but probably better to get a second opinion