VCE Methods Question Thread!

[keltingmeith]

i dont understand this?

when i enter on my cas it automatically gives me why is that and why does the cas do this?

Because that's easier. Watch closely:

[Zealous]

i dont understand this?

when i enter on my cas it automatically gives me why is that and why does the cas do this?

They are equivalent.



Edit: beaten by Eulerfan (you are too fast )

[Splash-Tackle-Flail]

I know this looks really easy but I'm probably just having a mental blank. How do you solve for n here?

n*2^n-1=192

Thanks!

[keltingmeith]

I know this looks really easy but I'm probably just having a mental blank. How do you solve for n here?

n*2^n-1=192

Thanks!

You use your calculator. This is not something you can solve with algebra (unless you get really lucky and guess the right answer. :S)

[Zealous]

I know this looks really easy but I'm probably just having a mental blank. How do you solve for n here?

n*2^n-1=192

Thanks!

Bit of algebraic manipulation can help (just play around with numbers), but it's a transcendental equation so cannot be solved algebraically.

[AirLandBus]

With regards domain and range, when graphing a square root function is the "start" point included or excluded always or is it dependent on the domain given.
For example,

F(x)= 2 + (square root) of 2
Where f:R+ -> R

Obvs, its just a standard square root function, translated up 2 units.

So the domain would be R+
And the range would be from (2, infinity).
Now because 0 is neither considered positive or negative, and as a result is not included in the domain as the domain is only + real numbers, is this why it is not inclusive in the range?

[Zealous]

With regards domain and range, when graphing a square root function is the "start" point included or excluded always or is it dependent on the domain given.
For example,

F(x)= 2 + (square root) of 2
Where f:R+ -> R

Obvs, its just a standard square root function, translated up 2 units.

So the domain would be R+
And the range would be from (2, infinity).
Now because 0 is neither considered positive or negative, and as a result is not included in the domain as the domain is only + real numbers, is this why it is not inclusive in the range?

Just be careful isn't a function of 'x' but I think you meant . But yes you are correct, we are only allowed to input 'x' values which are positive (so greater than zero). Because we are not allowed to input x=0, there is no way we can get an output (range) of 2, so therefore the range is (2,+infinity) which excludes the 2.

[AirLandBus]

Yeah, soz it was meant to be a 'x'. Thanks for your help. Add it to the mistakes/learn from book.

[AirLandBus]

What am i doing wrong in part a) Can someone elaborate.

[brightsky]

Any graph with the equation y = sqrt(ax - x^2) will pass through (0,0) so that piece of information is essentially redundant. The fact that the curve passes through (6,6), however, can be exploited.

y = sqrt(ax - x^2)
6 = sqrt(6a - 36)
36 = 6a - 36
6a = 72
a = 12

[AirLandBus]

Any graph with the equation y = sqrt(ax - x^2) will pass through (0,0) so that piece of information is essentially redundant. The fact that the curve passes through (6,6), however, can be exploited.

y = sqrt(ax - x^2)
6 = sqrt(6a - 36)
36 = 6a - 36
6a = 72
a = 12

Bingo, stupid mistake i made. Didnt make the connection between the square root over to the over side shizz.

[keltingmeith]

Mate, surely the prodigy can do it, there's nothing you cannot do!

Calm down - there are actually things in the methods curiculum it is impossible to do by hand until second year uni. And some of them are even impossible to do by hand until after a full maths degree. (mainly in the probability section, where you guys abuse your trusty CASes)

[catherine_mimi]

Hello~
I need some help determining the rule for a particular exponential graph
The question asks, 'Find the values of a and b such that the graph of y=ae^-bx goes through the points (3,50) and (6,10).
I have figured out that b= 1/3 log_e5 and that a=50e^log_e5, but how does a=250?
Thanks,
Catherine

[Orb]

Hello~
I need some help determining the rule for a particular exponential graph
The question asks, 'Find the values of a and b such that the graph of y=ae^-bx goes through the points (3,50) and (6,10).
I have figured out that b= 1/3 log_e5 and that a=50e^log_e5, but how does a=250?
Thanks,
Catherine

Hey,

So when you have e^loge, they cancel each other out, you can observe this simply by plotting e^loge(x) on your cas, the result will be y=x

After cancelling out, the result will just be 50x5 = 250 = answer

[brightsky]

So you know that the graph of y = ae^(-bx) passes through (3,50) and (6,10). Substituting these points into the equation of the exponential function yields the following equations:

50 = ae^(-3b)......(1)
10 = ae^(-6b)......(2)

(1) divided by (2):

50/10 = [ae^(-3b)]/[ae^(-6b)]
5 = e^(3b)
3b = ln(5)
b = 1/3 ln(5)

Substitute into (1):
50 = ae^(-3 * 1/3 ln(5))
50 = ae^(-ln(5))
50 = ae^(ln(1/5)) (by log laws)
50 = a (1/5) (again, by log laws)
a = 250

Note that e^(ln(BLAH)) = BLAH.