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March 28, 2024, 11:35:03 pm

Author Topic: VCE Methods Question Thread!  (Read 4802323 times)  Share 

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deStudent

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Re: VCE Methods Question Thread!
« Reply #14400 on: January 19, 2017, 06:25:58 pm »
0
http://m.imgur.com/a/pb6vO

Is there any real method for determining the shape of these graphs? I can find out the features they will have but don't know what it'll look like altogether.

Q5) Makes a bit of sense but Q6) I knew there would be a negative point of inflection but couldn't of even guessed there would be 2 turning points like that..

Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14401 on: January 19, 2017, 09:29:03 pm »
+1
http://m.imgur.com/a/pb6vO

Is there any real method for determining the shape of these graphs? I can find out the features they will have but don't know what it'll look like altogether.

Q5) Makes a bit of sense but Q6) I knew there would be a negative point of inflection but couldn't of even guessed there would be 2 turning points like that..

You can usually find out most of the information to draw a graph like that, but the stationary points are a bit more difficult. As it's a positive function to the power of 5, it goes up like a (positive) cubic. As it's x^3(x^2-1) then you can find out the x intercepts are 0,-1,1, and since it's x^3(...) then you know there's a point of inflection at that x intercept. From there, you can plot the x intercepts, know the general shape (going up like a cubic) and you know there's a point of inflection at x=0, so from there you can deduce there are two turning points, which you find using your calculator. From left to right, it goes up to the first intercept at x=-1, then goes down the reach the second intercept at x=0 which has a point of inflection, down then back up to reach x=1 and continues upwards.
Hope this helps a little, I'm a bit rusty  :P
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deStudent

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Re: VCE Methods Question Thread!
« Reply #14402 on: January 20, 2017, 05:31:44 pm »
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Can someone help me with this http://m.imgur.com/zduz8Eq

I can't solve this with my CAS? It just says "error" when I enter the simultaneous equation. It's formatted exactly the same as the example provided.

zhen

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Re: VCE Methods Question Thread!
« Reply #14403 on: January 21, 2017, 12:52:29 am »
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Can someone help me with this http://m.imgur.com/zduz8Eq

I can't solve this with my CAS? It just says "error" when I enter the simultaneous equation. It's formatted exactly the same as the example provided.
This is how i did it

deStudent

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Re: VCE Methods Question Thread!
« Reply #14404 on: January 22, 2017, 06:15:54 pm »
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http://m.imgur.com/a/fhqHl would like some help here  :)

For Q3b) When you solve with using elimination, why do you lose one of the solutions? By eliminating you only get (0,0) but another solution is (0.5, 0.5)?

Q1a) I'm not sure why they say "/{0}", isn't that inside of the domain of k? Or is it restricted because it would cause x = undefined?

Thanks

zhen

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Re: VCE Methods Question Thread!
« Reply #14405 on: January 22, 2017, 07:02:50 pm »
+2
http://m.imgur.com/a/fhqHl would like some help here  :)

For Q3b) When you solve with using elimination, why do you lose one of the solutions? By eliminating you only get (0,0) but another solution is (0.5, 0.5)?

Q1a) I'm not sure why they say "/{0}", isn't that inside of the domain of k? Or is it restricted because it would cause x = undefined?

Thanks
For 1a), when k=0, when you plug it into the quadratic formula, the denominator is zero, which makes it undefined.

For 3b) y=2x^2 and y=x
I sub x into y, making it x=2x^2
This equals x-2x^2=0
Which equals x(1-2x)=0
This gets x=0 and x=1/2

Also, using elimination you get to the 0=x-2x^2, since you cancel out the y. From then you just do the same thing
« Last Edit: January 22, 2017, 07:05:22 pm by zhen »

lzxnl

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Re: VCE Methods Question Thread!
« Reply #14406 on: January 23, 2017, 11:43:56 am »
+3
For 1a), when k=0, when you plug it into the quadratic formula, the denominator is zero, which makes it undefined.

For 3b) y=2x^2 and y=x
I sub x into y, making it x=2x^2
This equals x-2x^2=0
Which equals x(1-2x)=0
This gets x=0 and x=1/2

Also, using elimination you get to the 0=x-2x^2, since you cancel out the y. From then you just do the same thing

More specifically, if k = 0, your equation isn't even a quadratic equation.
It's often best to not blindly substitute values into formulas and actually think about what it means to put that value in.
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Gogo14

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Re: VCE Methods Question Thread!
« Reply #14407 on: January 23, 2017, 03:15:00 pm »
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Find the Rule of the image when the graph of y=sqrt x is translated 4 units in the negative direction of the x axis, reflected in the x axis and dilated by a factor of 3 from the y axis

Solutions is y=-1/3 sqrt((x+12)/3)
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Syndicate

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Re: VCE Methods Question Thread!
« Reply #14408 on: January 23, 2017, 03:31:55 pm »
+2
Find the Rule of the image when the graph of y=sqrt x is translated 4 units in the negative direction of the x axis, reflected in the x axis and dilated by a factor of 3 from the y axis

Solutions is y=-1/3 sqrt((x+12)/3)

(x,y) -> (x-4,y) -> (x-4, -y) -> (3(x-4), -y) = (x',y')

x' = 3(x-4)   and   y' = -y
x = \( \frac{x'}{3} \) + 4  and y = -y'
\(y=  \sqrt{x} \\ -y' = \sqrt{\frac{x'+12}{3}} \\ \therefore y' = - \sqrt{\frac{x'+12}{3}} \)

Hmm... I think the answer supplied by the book is incorrect.
« Last Edit: January 23, 2017, 03:35:34 pm by Syndicate »
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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14409 on: January 23, 2017, 04:55:54 pm »
+1
Find the Rule of the image when the graph of y=sqrt x is translated 4 units in the negative direction of the x axis, reflected in the x axis and dilated by a factor of 3 from the y axis

Solutions is y=-1/3 sqrt((x+12)/3)

I think syndicate's right, don't know where the extra 1/3 at the start came from

y = √x
y = √(x+4)
y = -√(x+4)
y = -√(x/3 + 4)
y = -√((x+12)/3)

Same answer different method
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Gogo14

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Re: VCE Methods Question Thread!
« Reply #14410 on: January 23, 2017, 05:16:54 pm »
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Oh ok thanks,
Also for 3b, the answer says y=(x+1)^1/3+1
But I got y=(x+1)^1/3-1 ?
I have the same issue for 3b-d
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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14411 on: January 23, 2017, 05:28:08 pm »
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Oh ok thanks,
Also for 3b, the answer says y=(x+1)^1/3+1
But I got y=(x+1)^1/3-1 ?
I have the same issue for 3b-d

Could you post the question? :)
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Re: VCE Methods Question Thread!
« Reply #14412 on: January 24, 2017, 11:10:19 am »
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Hey,
Can someone please help me with this question:
http://imgur.com/a/SFr42
For question 8, why does it have -xy and not -2xy?

Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14413 on: January 24, 2017, 11:20:05 am »
+1
Hey,
Can someone please help me with this question:
http://imgur.com/a/SFr42
For question 8, why does it have -xy and not -2xy?


Hi,
The formula for x3 + y3 = (x+y)(x2-xy+y2)
This is the formula, and it's also a reason it can't be simplified much further. If it were -2xy then x3+y3 could be simplified to (x+y)(x-y)2
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deStudent

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Re: VCE Methods Question Thread!
« Reply #14414 on: January 24, 2017, 03:08:19 pm »
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http://m.imgur.com/a/O3hHR

I'm abit stuck on this question. I know that I need to find the gradients of the line but I'm having trouble finding the point of intersection which is the last bit of info I need.

Thanks