Post by: jinny1 on February 25, 2011, 09:21:04 pm
how do i find x-ints/ y-ints of modulus graphs??
lets say
y= l(x+1)^2-1l - 2
how wud i work out interecepts through working out not by looking at the drawn graph..
thank you
Post by: onur369 on February 25, 2011, 09:27:33 pm
In this case it is (x+1)^2 -1
Make it equal to zero (x+1)^2 -1=0
Then find 'x' which in this case equals to 0 and -2. (Make y=0)
To find y, (make x=0)
Post by: xZero on February 25, 2011, 09:28:27 pm
since
for y-int, let x=0
@above, you neglected the -2 outside of the modulus so your answer's wrong
Post by: nacho on February 25, 2011, 09:40:28 pm
reflect in the x-axis,
shift the reflected graph down 2 units
Post by: jinny1 on February 25, 2011, 10:18:28 pm
Draw the graph inside the modulus function,
reflect in the x-axis,
shift the reflected graph down 2 units
but when you translate the graph down 2 units.. the x and y interceptts change so you wouldnt know where in the axis the final graph will pass through. and thats where xzero's explanation comes in.
IS there any faster way to draw modulus graphs?? or is Nacho + xzero way the fastest?
Post by: pi on February 25, 2011, 10:24:47 pm
IS there any faster way to draw modulus graphs?? or is Nacho + xzero way the fastest?
It gets faster with practice, its the way I go about it too
Post by: xZero on February 25, 2011, 10:27:43 pm
IS there any faster way to draw modulus graphs?? or is Nacho + xzero way the fastest?
It gets faster with practice, its the way I go about it too
i agree, practice is the way to go
Post by: jinny1 on March 25, 2011, 09:37:35 pm
y= -2^x+3 + 1
Dilate this function by a factor of 3 from the x-axis.
I did:
y/3 = -2^x+3 + 1
y= 3(-2^x+3 + 1)
y= 3*-2^x+3 + 3
However the answer is:
y= -3*2^x+3 + 1
what did i do wrong :(
thanks in advance :P
Post by: Water on March 25, 2011, 11:04:00 pm
Dilation
Reflection
Translation
So, in any equation, you'll always dilate, then reflection if any, then translate last. With these steps, you should be able to get the right answer.
Post by: nacho on March 25, 2011, 11:51:06 pm
Post by: brightsky on March 26, 2011, 06:04:25 pm
Post by: jinny1 on March 26, 2011, 07:29:48 pm
lol
u must be laughing at my incompetence :(
Post by: brightsky on March 26, 2011, 07:36:50 pm
lol
u must be laughing at my incompetence :(
woah where did the original post go? ignore that, the post might've been deleted or something..
Post by: jinny1 on April 10, 2011, 06:52:47 pm
what is the difference between:
R- vs (-inf,0]
R+ vs [0,inf)
???
Post by: TrueTears on April 10, 2011, 06:56:20 pm
R^- means strictly negative reals
thus R^+ is equivalent to (0, infinity)
and R^- is equivalent to (-infinity, 0)
In other words, 0 is not included as it is not positive nor negative.
Post by: jinny1 on April 27, 2011, 08:36:24 pm
How would you split modulus functions into two equations to make it easier to draw absolute function graphs when the powers are greater than 1 (non-lineal)??
such as l(x-1)^2 -1l
or l(x-1)^3-1l
Thanks!!!!!
because if i do it the same way i spilt up linear functions, then i get a weird answer...
Post by: pi on April 27, 2011, 08:57:07 pm
Post by: jbebbo on April 27, 2011, 09:17:25 pm
Hey guys just a quick question here :Ppi's method for this is the best and easiest, but if you were interested, you would split |(x-1)^2-1| up like this:
How would you split modulus functions into two equations to make it easier to draw absolute function graphs when the powers are greater than 1 (non-lineal)??
such as l(x-1)^2 -1l
or l(x-1)^3-1l
(x-1)^2-1 for (x-1)^2-1>0, and
-((x-1)^2-1) for (x-1)^2-1<0
then you would solve the intersections:
(x-1)^2-1=0 => x=0 or x=2
then graph the function, and you find that y>0 for xE(-inf,0)U(2,inf) and that y<0 for xE(0,2)
therefore,
l(x-1)^2 -1l= (x-1)^2-1 for xE(-inf,0]U[2,inf)
-(x-1)^2+1 for xE(0,2)
Moderator action: removed real name, sorry for the inconvenience
Post by: jinny1 on April 27, 2011, 09:22:14 pm
Post by: Greatness on April 27, 2011, 09:28:57 pm
Moderator action: removed real name, sorry for the inconvenience
Post by: jbebbo on April 27, 2011, 09:31:20 pm
Post by: jinny1 on April 27, 2011, 09:50:20 pm
Post by: jbebbo on April 27, 2011, 10:07:20 pm
Post by: jinny1 on May 18, 2011, 10:41:32 pm
Post by: jane1234 on May 18, 2011, 10:50:43 pm
i was just wondering, in the exam when you are writing the general solutions for a sin/cos/tan equations. Do you have to put the " 'n' is an element of 'Z' " ??? Because i always forget. would you lose marks for that??I think so, as it is part of the solution (and an essential one at that, n can't be something like 2.4356754). Do enough questions and you'll start remembering to put it in :)
Post by: jinny1 on May 18, 2011, 10:53:43 pm
Post by: jane1234 on May 18, 2011, 11:00:27 pm
:P errrrr... it doesnt help that my maths teacher doesn't really care whether i put it in or not :(But the unforgiving VCAA examiners will... :)
Post by: RobM8 on May 18, 2011, 11:24:26 pm
Post by: jinny1 on May 19, 2011, 05:45:18 pm
always put the n E Z in, otherwise your answer is not a general solution.
Can i just just put it in my final answer not during working out???
Post by: pi on May 19, 2011, 06:27:25 pm
always put the n E Z in, otherwise your answer is not a general solution.
Can i just just put it in my final answer not during working out???
Best to do this when you first introduce 'n'
Post by: jinny1 on June 14, 2011, 08:57:57 pm
Post by: b^3 on June 14, 2011, 09:02:03 pm
Post by: taiga on June 14, 2011, 10:12:25 pm
(3-5)^2 = 4
:P
Post by: jinny1 on June 15, 2011, 10:12:25 am
Post by: jinny1 on July 10, 2011, 08:47:31 pm
(http://i.imgur.com/836Hu.png)
i went a little different. In the second line, i divided both sides by "x" and then squared both sides.
That would give me 4 = x+1
thus x=3 .
What did i do wrong here??? because clearly the answer is also x=0.
Post by: xZero on July 10, 2011, 08:55:55 pm
Post by: jinny1 on July 10, 2011, 09:05:54 pm
Post by: xZero on July 10, 2011, 09:13:41 pm
Post by: taiga on July 10, 2011, 09:22:09 pm
hi guys i have a problem with this.
(http://i.imgur.com/836Hu.png)
i went a little different. In the second line, i divided both sides by "x" and then squared both sides.
That would give me 4 = x+1
thus x=3 .
What did i do wrong here??? because clearly the answer is also x=0.
2x(x-3)=0
would you divide by x here? :P
Post by: jinny1 on July 10, 2011, 09:25:13 pm
hi guys i have a problem with this.
(http://i.imgur.com/836Hu.png)
i went a little different. In the second line, i divided both sides by "x" and then squared both sides.
That would give me 4 = x+1
thus x=3 .
What did i do wrong here??? because clearly the answer is also x=0.
2x(x-3)=0
would you divide by x here? :P
?? i divided by x in the second line:
2x = x(x+1)^1/2
i thought it simplied things. that becomes 2 = (x+1)^1/2
i then squared both sides.
4= x+1.
Post by: Lasercookie on July 10, 2011, 10:34:22 pm
?? i divided by x in the second line:I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.
2x = x(x+1)^1/2
i thought it simplied things. that becomes 2 = (x+1)^1/2
i then squared both sides.
4= x+1.
Post by: brightsky on July 10, 2011, 10:36:42 pm
?? i divided by x in the second line:I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.
2x = x(x+1)^1/2
i thought it simplied things. that becomes 2 = (x+1)^1/2
i then squared both sides.
4= x+1.
not quite, you're assuming that x is not 0.
Post by: moekamo on July 10, 2011, 10:38:30 pm
?? i divided by x in the second line:I think the reason why you can't simply divide both sides by x is because in doing that you make the assumption that x is positive.
2x = x(x+1)^1/2
i thought it simplied things. that becomes 2 = (x+1)^1/2
i then squared both sides.
4= x+1.
no, its because your assuming that x doesn't equal 0.
If x does equal 0(which it does) then you cannot simply divide by it
Post by: jinny1 on July 30, 2011, 07:29:55 pm
Quote
Calculating an 'area between curves' is interesting in that it does not matter whether the area is
above, below or overlapping the axes. It is worthwhile justifying to yourself why this is the case.
Could someone explain why this is the case?
thanks
Post by: b^3 on July 30, 2011, 07:52:39 pm
1. Both curves are above the x-axis
Minusing the area under the bottom curve from the area under the top curve give just the orange area, the area between the two curves.
2. One above and one below the x-axis
The signed area above the axis is postive and the signed area below the axis is negative. We minus the area of the bottom curve, that is we minus the negative, this results in the addition of the area between the axis and the bottom curve to the top curve. That is yellow - blue (blue is negative) so the areas are added.
3. Both below the x-axis
The area between the curves and the axis are both negative, but the bottom curve's area has a greater negative. When we take this greater negative away, we add a greater value and so end up with a postive result for the area that is between the two curves
4. Partly above and partly below
Here for the sections of the top curve that are above the axis, we are just doing case 2, minusing the negative so that it adds. For the part of the top graph that is under the axis, that is the green part under the axis, we can explain it with case 3. The greater negative is minused to give the postive value, the area between the two curves.
I hope this makes sense and I haven't confused you more.
Post by: moekamo on July 30, 2011, 09:27:16 pm
From Derrick Ha's bookQuoteCalculating an 'area between curves' is interesting in that it does not matter whether the area is
above, below or overlapping the axes. It is worthwhile justifying to yourself why this is the case.
Could someone explain why this is the case?
thanks
another way to think about it, you know that for an area between curves when they are both above the x axis is just the integral of 'top - bottom' between the terminals.
Now wherever they are, whether it is below and above, or only below, we can translate both curves up by the same amount by adding a constant, h, to each curve. Now when computing the integral we do 'top - bottom' and the +h from each curve cancels out. Therefore we can just do top - bottom for area between two curves for any of the situations mentioned.
Post by: jinny1 on August 21, 2011, 06:05:29 pm
So lets say on one of the spesh/methods exam 2, therre was a question asking me to factorise 2x^2+4x+3.. Can i just input that into the calculator and just copy/paste thw answer straight from the calc?? Or do i need to say first " using the factorising function in the calc it was found Blah blah...) or would i need to show working out by using the "cross method" of factorisation. The latter defeats the whole point oftech enabled exams though...as it would just be like sitting exam 1
Thanks
Post by: pi on August 21, 2011, 06:28:20 pm
Post by: jinny1 on August 21, 2011, 07:36:08 pm
and by intermediate step, do you mean word explanations or actual working out ??
thanks
Post by: pi on August 21, 2011, 08:47:40 pm
Even if it was a 2~4 mark question???
and by intermediate step, do you mean word explanations or actual working out ??
Working out (like show that you have steps in factorisation, eg. complete square, etc.), although as it is CAS, I doubt it'll be more than 1-2 marks(s) for something like that
Post by: jinny1 on August 22, 2011, 10:29:21 pm
(http://i.imgur.com/G9gBU.png)
i dont get b.
Post by: b^3 on August 22, 2011, 10:36:10 pm
so if you differentiate that you will get x2ln(x)
so you need to know the anitderivative of xln(x) so rearrange the antid you a trying to get in terms of intergal(xln(x))
so 2intergral(ln(x))=x2ln(x)-intergral(x) and go from there, then sub in terminals
Post by: brightsky on August 22, 2011, 10:39:28 pm
a) d/dx (x^2 ln(x)) = x + 2xln(x)
b) since we have the result from a), we know that x^2 ln(x) = int (x) dx + 2 int(xln(x)) dx by integrating both sides,
so int(x ln(x)) dx = 1/2(x^2 ln(x) - x^2/2)
since for the required region, we are bounded on the left by 1 and on the right by 3:
then the shaded region is:
1/2(3^2 ln(3) - (3)^2/2) - 1/2(1 ln(1) - 1/2)
=9/2 ln(3) - 9/4 + 1/4
=9/2 ln(3) - 2
hopefully no mistakes lol
Post by: jinny1 on August 22, 2011, 11:04:50 pm
so for a you would have got 2x*ln(x)+x
so if you differentiate that you will get x2ln(x)
so you need to know the anitderivative of xln(x) so rearrange the antid you a trying to get in terms of intergal(xln(x))
so 2intergral(ln(x))=x2ln(x)-intergral(x) and go from there, then sub in terminals
you meant integrate right?? :P
integration by recognition
a) d/dx (x^2 ln(x)) = x + 2xln(x)
b) since we have the result from a), we know that x^2 ln(x) = int (x) dx + 2 int(xln(x)) dx by integrating both sides,
so int(x ln(x)) dx = 1/2(x^2 ln(x) - x^2/2)
since for the required region, we are bounded on the left by 1 and on the right by 3:
then the shaded region is:
1/2(3^2 ln(3) - (3)^2/2) - 1/2(1 ln(1) - 1/2)
=9/2 ln(3) - 9/4 + 1/4
=9/2 ln(3) - 2
hopefully no mistakes lol
The point i got stuck in was when i integrated x dx.. i added a contant in and i didnt know what to do with the stupid c! So how come we dont use constants in these cases?
Thanks guys!
Post by: tony3272 on August 22, 2011, 11:22:36 pm
For example if you have
But if you take the definite integral of 2x, you would get
Post by: jinny1 on August 22, 2011, 11:33:22 pm
So without first using definitive integral, shouldnt the above equation become simpkified to : x^2ln(x) = ((int)xln(x) dx) + x^2/2 + c ???
Post by: tony3272 on August 22, 2011, 11:37:39 pm
Post by: jinny1 on August 22, 2011, 11:52:16 pm
Nvm. Silly calc error from b4. Thanks!
Post by: jinny1 on September 04, 2011, 07:07:29 pm
Post by: tony3272 on September 04, 2011, 07:20:20 pm
Post by: jinny1 on October 11, 2011, 07:50:32 pm
Cheers
Post by: Greatness on October 11, 2011, 08:03:40 pm
Yep inverse functions only exists if its 1-1.
Post by: jinny1 on October 11, 2011, 08:33:55 pm
Function: vertical and horizontal line test cross once
Yep inverse functions only exists if its 1-1.
Thanks! I always thought it was justvertical line test?? Because isnt a quadractic a function??
Post by: REBORN on October 11, 2011, 08:37:00 pm
Post by: cranberry on October 18, 2011, 08:31:59 pm
Her "8th attempt" is only dependent on her last attempt (her 7th attempt), isnt it??
Shouldnt there only be two possible orders - successful give that she missed prior x successful give that she was successful prior?
If you read the question, it clearly says her 8th attempt, which would have the same chances of success as the 3rd attempt? if not the probabilty of succes would change each attempt so the matrices would be different each attempt, no?
just rambling anyway. :-X
Post by: xZero on October 18, 2011, 08:36:12 pm
alright, this question ffrom vcaa 2008...(i know its VCAA so im probably wrong, but...)its a normal markov chain, you're over thinking it
Her "8th attempt" is only dependent on her last attempt (her 7th attempt), isnt it??
Shouldnt there only be two possible orders - successful give that she missed prior x successful give that she was successful prior?
If you read the question, it clearly says her 8th attempt, which would have the same chances of success as the 3rd attempt? if not the probabilty of succes would change each attempt so the matrices would be different each attempt, no?
just rambling anyway. :-X
Post by: jinny1 on November 05, 2011, 05:41:52 pm
Post by: Vincezor on November 05, 2011, 05:53:21 pm
Hey guys i've been wondering how to integrate an equation with given parameters. Lets say i want to integrate y'=x^2 and the questions already says y(0)=0 , then how to i make the calculator calculate the Constant as well in the process of integrating??
I'm using the classpad, and for that we use 'dsolve' (I've never used it in methods come to think of it!) but you use it heaps in spesh!
Interactive -> Advanced -> dSolve -> Type in equation
You should get
Hope that helped!
Post by: jinny1 on November 05, 2011, 05:59:23 pm
Post by: jinny1 on November 05, 2011, 06:12:03 pm
Post by: b^3 on November 05, 2011, 06:14:55 pm
nvmYou should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1
EDIT: I probably should have added inital conditions into the calculator guide.
Post by: jinny1 on November 05, 2011, 06:19:45 pm
nvmYou should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1
EDIT: I probably should have added inital conditions into the calculator guide.
Thanks buddy :)
just one last one! what would be the calculator syntax to solve this question??
(http://i.imgur.com/RVgdD.png)
Post by: b^3 on November 05, 2011, 06:26:32 pm
nvmYou should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1
EDIT: I probably should have added inital conditions into the calculator guide.
Thanks buddy :)
just one last one! what would be the calculator syntax to solve this question??
(http://i.imgur.com/RVgdD.png)
I'm not sure if you can do it in one step on the calc, you have to intergrate the stuff in front of the i,j and k, then find c and sub it back in. once you have displacement with the +c, you can start putting it in the calc and doing it the quick way (i.e. define things, solve for c, plug in t e.t.c)nvmYou should have this:
desolve(y'=x^2 and y(0)=1,x,y)
y=x^3/3 +1
EDIT: I probably should have added inital conditions into the calculator guide.
Thanks buddy :)
just one last one! what would be the calculator syntax to solve this question??
(http://i.imgur.com/RVgdD.png)
Post by: jinny1 on November 05, 2011, 06:35:41 pm
but didin't work. i dont think u can desolve vector components :(