VCE Methods Question Thread!

[Mattjbr2]

I'm not 100% certain but if that point where the r intersects with x, I'm pretty sure that would have to be less than 3 so it would definitely be restricted between 0 and 3 but I don't have my cas here so I'm not exactly sure how you'd go about it but I'd assume it would be something regarding the angle at the top right corner being less than 90 degrees or something.
I could be wrong but logically that makes sense

No worries! Found that it has a horizontal asymptote at r=1.47728
For some reason, doing fMax(f1(x),x) gave me x=infinity, so I just assumed (without zooming out) that r's range is infinity. Silly mistake!

[snowisawesome]

So i guess i could have a go at this.

For the inverse u swap x and y values. So what used to be the domain becomes the range. And vice versa. 
You cant have a negative square root. Eg what is the square root of -4? There is not one ; no 2 same numbers multiply to make a negative number.
By multiplying the negative 1 by negative number what happens? Well what happens when -1 is divided by -1? They become positive and the reason we do this is because we CANT HAVE A NEGATIVE  SQUARE ROOT. 

For ur other question,  domain of function is range of inverse. So range is what uv given above. Domain of inverse is range of function. Find the range of the function. U should get the domain of the inverse which uv listed above. 

It has minus in front of square root sign because the domain is from -4 to 0. Remember this transformation is in the y axis?

So since the domain is [-4,0] it's only -√(16-x^2) and not +-√(16-x^2)
Would this work in any situation?

[Sine]

So since the domain is [-4,0] it's only -√(16-x^2) and not +-√(16-x^2)
Would this work in any situation?

range of inverse = domain of original = [-4,0]
So range of inverse = [-4,0]
From the two possible functions you have stated -√(16-x^2) and not +√(16-x^2) think about the shape of the graph the first is the lower half of a circle (semi circle below the x-axis) the 2nd is a upper half of a circle (semi circle above the x-axis) the range is [-4,0] so we choose the semi circle below the x-axis.

[snowisawesome]

range of inverse = domain of original = [-4,0]
So range of inverse = [-4,0]
From the two possible functions you have stated -√(16-x^2) and not +√(16-x^2) think about the shape of the graph the first is the lower half of a circle (semi circle below the x-axis) the 2nd is a upper half of a circle (semi circle above the x-axis) the range is [-4,0] so we choose the semi circle below the x-axis.

Thanks sine
Also, do you how 1/(x-1)^2+2 is a many-to-one function?

[Sine]

Thanks sine
Also, do you how 1/(x-1)^2+2 is a many-to-one function?

your attatchment isn't 1/(x-1)^2+2, looks like 1/x^2 so just a couple translations away.

Many to one in basic terms means many x terms can go to produce the same y term.

As you can see from your attatchment multiple x-values produce the same y-value.

For exampe from your attatchment f(1) = f(-1) = 1, two x-values (1 and -1) produce the same y-value which in this case is 1.

[snowisawesome]

your attatchment isn't 1/(x-1)^2+2, looks like 1/x^2 so just a couple translations away.

Many to one in basic terms means many x terms can go to produce the same y term.

As you can see from your attatchment multiple x-values produce the same y-value.

For exampe from your attatchment f(1) = f(-1) = 1, two x-values (1 and -1) produce the same y-value which in this case is 1.

Thanks again sine
Do you know how to change y = (4x-7)/(x-2) into the standard form to work out the asymptotes?

[Sine]

Thanks again sine
Do you know how to change y = (4x-7)/(x-2) into the standard form to work out the asymptotes?

y = (4x-7)/(x-2)
y = 4(x-7/4)/(x-2)
y=4(x-2 +1/4)/(x-2)
y = (4(x-2) +1)/(x-2)
y = 4(x-2)/(x-2) + 1/(x-2)
y = 4 + 1/(x-2)

[snowisawesome]

y = (4x-7)/(x-2)
y = 4(x-7/4)/(x-2)
y=4(x-2 +1/4)/(x-2)
y = (4(x-2) +1)/(x-2)
y = 4(x-2)/(x-2) + 1/(x-2)
y = 4 + 1/(x-2)

Thanks again
Does anyone know how hard it actually is to finish exam 2 in 2 hours? Since I've heard that over 50% of students can't finish exam 2 within 2 hours?

[Sine]

Thanks again
Does anyone know how hard it actually is to finish exam 2 in 2 hours? Since I've heard that over 50% of students can't finish exam 2 within 2 hours?

yes for the average methods student I would think it's quite hard to finish the exam (especially if they want to answer the "seperator questions") - if that's ever the case make sure you know how to identify the "easy" marks so that you can answer everything you know how to do. However, right now its too early to think about not finishing

[snowisawesome]

yes for the average methods student I would think it's quite hard to finish the exam (especially if they want to answer the "seperator questions") - if that's ever the case make sure you know how to identify the "easy" marks so that you can answer everything you know how to do. However, right now its too early to think about not finishing

But isn't 2 hours a lot of time for an exam?

[Sine]

But isn't 2 hours a lot of time for an exam?

yup but also a lot of work to do the time needed will differ between many people. When i did exams I could get it down to 55mins  but chose to slow it down to get stuff correct on the first go so ended up going 60-70mins most of the time.

[TheAspiringDoc]

yup but also a lot of work to do the time needed will differ between many people. When i did exams I could get it down to 55mins  but chose to slow it down to get stuff correct on the first go so ended up going 60-70mins most of the time.

Are you saying you could complete the 2 hour section 2 exam in 55 minutes?!

[RuiAce]

Are you saying you could complete the 2 hour section 2 exam in 55 minutes?!

It's not impossible; in the HSC some people were doing 3 hour math exams in 2 hours. Just takes heaps of practice.

(Some overly talented students that also wrote fast could go down to well below 1.5 hours)

[Sine]

Are you saying you could complete the 2 hour section 2 exam in 55 minutes?!

yeah but definitely wasn't my average or anything I have heard others who have done the same or maybe even faster, but 55 min would be for the optimal easy exam after a lot of practice exams. I'd say to cut down time my biggest asset would've been knowing how to use the CAS quickly (more instinctive than looking/checking for buttons etc) Also remember in the actual exam you'll probably end up going a lot slower.

[lzxnl]

yeah but definitely wasn't my average or anything I have heard others who have done the same or maybe even faster, but 55 min would be for the optimal easy exam after a lot of practice exams. I'd say to cut down time my biggest asset would've been knowing how to use the CAS quickly (more instinctive than looking/checking for buttons etc) Also remember in the actual exam you'll probably end up going a lot slower.

This. I used to do spesh exam 2s in an hour and come the day of the exam, I got so paranoid about checking over my answers that it took me 100 minutes. I ended up not having the time to check through my entire exam and ironically lost marks for careless errors because I was too careful (didn't have time to check)

Thanks again sine
Do you know how to change y = (4x-7)/(x-2) into the standard form to work out the asymptotes?

If you just want to find asymptotes, you don't even have to divide. One way is this. Your vertical asymptote is obvious; x = 2. Your horizontal asymptote is the y value you get when x becomes large. Now, if x becomes large (positive or negative), 4x - 7 is close to 4x. Try it for x = 1000000 etc. Similarly, x - 2 is close to x. Therefore you would expect (4x - 7)/(x - 2) to be very close to 4 as x gets large.