VCE Methods Question Thread!

[Zues]

Hey Guys, so me and my friend where talking and he came up with this new way of evaluating answers, heres what he said.

"i was doing 1.b) which involves the equation y=x^2-1, and i noticed that when you put in an x value, for example 7, the answer will be the product of the values above and below the number. i.e. 7^2-1=48, and 6X8=48. so in other words if you want to find the square of a number, which is hard with big numbers by hand, you use the rule y=(x-1)(x+1)+1...for instance if you can't figure out the square of 23 and you aren't allowed a calculator, you go 22X24+1 which is doable by hand and you get the answer..the plus one is because the product of the numbers above and below equals 1 less than the number squared, so you have to add the 1...ie. 7^2=49 and 6X8=48..same as 12^2=144 and 11X13=143.

i also figured out one for cubes, you use y=x(x-1)(x+1)+x, i.e 7^3=343, so you go 6X7X8+7 which equals 343"

so ?

[pi]

No offence to your mate, but 23x23 seems like less effort to me rather than 22x24+1.

[Zues]

none the less, its a bit interesting?

[grannysmith]

Hey Guys, so me and my friend where talking and he came up with this new way of evaluating answers, heres what he said.

"i was doing 1.b) which involves the equation y=x^2-1, and i noticed that when you put in an x value, for example 7, the answer will be the product of the values above and below the number. i.e. 7^2-1=48, and 6X8=48. so in other words if you want to find the square of a number, which is hard with big numbers by hand, you use the rule y=(x-1)(x+1)+1...for instance if you can't figure out the square of 23 and you aren't allowed a calculator, you go 22X24+1 which is doable by hand and you get the answer..the plus one is because the product of the numbers above and below equals 1 less than the number squared, so you have to add the 1...ie. 7^2=49 and 6X8=48..same as 12^2=144 and 11X13=143.

i also figured out one for cubes, you use y=x(x-1)(x+1)+x, i.e 7^3=343, so you go 6X7X8+7 which equals 343"

so ?

When trying to mentally calculate e.g. I just break it up and go 23x20 + 3x20 + 3x3 = 529.

[keltingmeith]

none the less, its a bit interesting?

When trying to mentally calculate e.g. I just break it up and go 23x20 + 3x20 + 3x3 = 529.

Look, I reckon both of these are weird... But if you think it cool/you find it easier to calculate cubes/squares these ways, don't let anybody else tell you other ways! There are wrong ways to do things, but there is no one right way to do them either.

[catherine_mimi]

Hello,
someone please help me graph the equation y= e^2x-1/2
. It confuses me because there is a power of x. I would know how to graph y= e^2x. I know its something of an exponential graph, but I cannot seem to find the asymptotes or the shape correctly.
Please help me

[knightrider]

Hello i have a question about the update for the casio classpad 400.

In the picture attached it says now with sliders etc etc. How do you actually get the sliders to appear on the graph as shown in the photo. So from this it looks like you can change the a,h and k values.How do you actually get this feature to appear on the graphs as shown in the photo.

Thanks

[Zues]

Hello,
someone please help me graph the equation y= e^2x-1/2
. It confuses me because there is a power of x. I would know how to graph y= e^2x. I know its something of an exponential graph, but I cannot seem to find the asymptotes or the shape correctly.
Please help me

if you mean
y=e^(2x-0.5)

then you change it into "transformation form" which is
y=e^2(x-0.25)

- exponential graph
- horizontal asy at y=0
-dilated from the x axis by a factor of 1 and from y by a factor of 1/2
- horizontal translation of 1/4 units

if you mean
y=e^2x^-1/2 i dont know if this is possible but my calc produces a graph which i am not familiar with (the shape yes, the equation no)

[keltingmeith]

Hello,
someone please help me graph the equation y= e^2x-1/2
. It confuses me because there is a power of x. I would know how to graph y= e^2x. I know its something of an exponential graph, but I cannot seem to find the asymptotes or the shape correctly.
Please help me

You take comfort in knowing that VCAA will never ask this on a tech-free exam, then you turn on your calculator and use that to graph it. Where did you pull this question from?

[Zues]

You take comfort in knowing that VCAA will never ask this on a tech-free exam, then you turn on your calculator and use that to graph it. Where did you pull this question from?

how would you even graph it?

[keltingmeith]

how would you even graph it?

You go to the graphing page, and after the "f(x)=" you put the function in.

[knightrider]

i have a question relating to the binomial theorem.

In my books the notes say something i don't understand.Could someone explain what this means?

Thanks

[Zues]

You go to the graphing page, and after the "f(x)=" you put the function in.

By hand

[AirLandBus]

i have a question relating to the binomial theorem.

In my books the notes say something i don't understand.Could someone explain what this means?

Thanks

Check out this. this is what I followed when learning it.

https://www.khanacademy.org/math/algebra2/polynomial_and_rational/binomial_theorem/v/binomial-theorem

[keltingmeith]

By hand

You're not expected to know this for methods.

Indeed, there are a lot of ways that you could sketch a graph having never seen it before (namely through the use of calculus), however you are not expected to know of or use these methods in either methods or specialist maths.