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June 25, 2022, 09:23:36 pm

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#### Dominatorrr

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##### Re: VCE Methods Question Thread!
« Reply #30 on: December 16, 2011, 06:16:23 pm »
0
To dilate by factor of 2 from y-axis means you need to replace x with x/2. This should give you the answer.

Yeah I think it makes sense now.

y=(4167/x^2)-1.67
y=(4167/0.5x*0.5x)-1.67
y=(4167/0.25x^2)-1.67
y=(16668/x^2)-1.67 (CLOSE ENOUGH)

Yep, and the reason your answer was a little off is because you rounded 12500/3 up to 4167, when it should be 4166.6(recurring)

Oh I see now, cheers

#### Dominatorrr

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##### Re: VCE Methods Question Thread!
« Reply #31 on: December 17, 2011, 10:37:29 pm »
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"A light will reflect from a surface. Similarly, a function's negative portion will reflect in the x-axis when the modulus is taken. A ray of light originates from a position of (4,8) and hits the surface at an angle of 71.6 degrees."

(a) Find the equation of the light ray

I'm not sure how to use the angle to figure out the gradient?

#### TrueTears

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##### Re: VCE Methods Question Thread!
« Reply #32 on: December 17, 2011, 10:40:49 pm »
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If an angle, theta, is made with the positive x axis, then tan(theta) is the gradient (why?)
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#### Dominatorrr

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##### Re: VCE Methods Question Thread!
« Reply #33 on: December 17, 2011, 10:55:07 pm »
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Ah yes that rule, I remember now, thanks.

Yet even then, my answer doesn't correspond to that of the book which is |3x-4|

I did 90-71.6=18.4
m=tan(theta)
m=tan(18.4)=0.33
y-y1=m(x-x1)
y-8=0.33(x-4)
y-8=0.33x-1.33
y=|0.33x+6.67|

#### TrueTears

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##### Re: VCE Methods Question Thread!
« Reply #34 on: December 17, 2011, 11:00:14 pm »
+2
nah it's just tan(71.6) which is approximately 3, 71.6 is already the angle made with the positive x axis
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#### Dominatorrr

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##### Re: VCE Methods Question Thread!
« Reply #35 on: December 17, 2011, 11:07:47 pm »
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Ah cheers, I was just assuming that the angle of incidence was measured against the normal (Physics knowledge  ), hence why I used 18.4 degrees as the positive x-axis angle.

#### nisha

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##### Re: VCE Methods Question Thread!
« Reply #36 on: December 21, 2011, 05:00:31 pm »
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Hi guys

I'm kinda new to the "smartness atmosphere" in here, but I'm gonna ask the question nevertheless. I hope it doesnt sounds too kindergardenish, for all the accomplished...

Anyways, I'm having quite a bit of trouble understanding Absolute value graphs and how to draw them. I am familiar and can do the easy ones (linear, parabola and hyperbola) but I get stuck with drawing the square root function ones and some other complex ones.
If you guys have the Mathsquest 12 Methods, the questions I am stuck on are: 4(e) (f) (j) (k) and (l)
But for those who dont, this is the question:
Sketch the following:::

Q4(e):
y= 2-|x^2 -2|

Q4(f)
y= |(x+1)^2 -1| +1

Q4(j)
y= -|1/x^2 -1|

Q4(k)
y=|(sqreroot(2-x)) -2 | +3

Q4(l)
y=2- |(sqreroot(x +1)) - 8|

Looking for more of an explanation of what the hell we are supposed to do. I understand the flipping concept, just not how to obtain the certain values (points) when it has already flipped and how to do so. I hope my questions made some sense.
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#### brightsky

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##### Re: VCE Methods Question Thread!
« Reply #37 on: December 21, 2011, 05:15:08 pm »
+5
as a rule of thumb, when you get graphs of form f(|x|), then you get rid of the stuff on the left hand side of the y-axis and make a mirror image of what is on the right hand side of the y-axis. if you get graphs of |f(x)|, then reflect the stuff below the x-axis in the x-axis. For example, for q4 e, sketch y = x^2 -2 first. then reflect the stuff below the x-axis in the x-axis. then reflect it again in the x-axis (due to the negative sign). then translate the graph 2 units up.

hope this makes sense.
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#### b^3

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##### Re: VCE Methods Question Thread!
« Reply #38 on: December 21, 2011, 05:19:45 pm »
+8
Its easier to look at these from the original type of equation (which you need to identify as parabola e.t.c) and then apply the transformations.

Q4e) look at just the |x2-2| part first.
x2-2 is just x2 shifted down two units.

So from that you can apply the mod, reflecting the negative parts of the graph in the x-axis.

Now apply the reflection in the x-axis transformation. I.e. now we are graphing y=-|x2-2|

Finally shift apply the last transformation of 2 units up so that you are graphinh y=2-|x2-2|

Hope that helps, have a go at the others, but if you really get stuck ask and I'll have a look.
« Last Edit: December 21, 2011, 08:43:08 pm by b^3 »
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#### b^3

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##### Re: VCE Methods Question Thread!
« Reply #39 on: December 21, 2011, 05:26:01 pm »
+5
About finding the points that result. You can apply the transformations to the points as you go or you can find the intercepts and notable points after with a little algebra. e.g. for 4e. to find the x-intercepts you have
y=2-|x2-2|
x-intercepts, y=0
2-|x2-2|=0
|x2-2|=2
now the mod will make a negative value inside it positive and leave a positive value as positive.
i.e. x2-2=2 (when x2-2>0) and -(x2-2)=2 (when x2-2<0)
x2=4 or x2=0
x=+-2 or x=0
i.e. x-intercepts at (-2,0), (2,0) and (0,0)

Do the same for the y-intercept and use the transformations to find the turning point.

EDIT: e.g. original t.p at (0,-2)
Apply the mod, all the negative y-values will become positive as it is flipped in the x-axis.
(0,-2)->(0,2)
Then apply the whole reflection of the graph in the x-axis.
(0,2)->(0,-2)
Then apply the shift of two units up.
(0,-2)->(0,0)
« Last Edit: December 21, 2011, 05:29:44 pm by b^3 »
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#### nisha

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##### Re: VCE Methods Question Thread!
« Reply #40 on: December 22, 2011, 11:31:03 am »
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So the only way to do absolute values is by the steps?

I understand it now. Man, if teachers actually explained things properly then maths would be so much easier.

Anyway, I have another question (I'm just going to keep asking them):

Was doing Transformations with matrices and found this question:
If f(x)= -g(2(x+1)) +1 and g(x)= sqroot (x), fing f(x) in terms of x only without using a calculator?

Tried using algebra, but kind of failed.
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#### b^3

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##### Re: VCE Methods Question Thread!
« Reply #41 on: December 22, 2011, 12:02:48 pm »
+2
if $g(x)=\sqrt{x}$ then $g(2(x+1))=\sqrt{2(x+1)}$ (you replace the x in g(x) with whateve is in the brackets of g(   ), i.e. here we replace it with 2(x+1))
So then $f(x)=-\sqrt{2(x+1)}+1$
-g(2(x+1)      1

i.e.$f(x)=1-\sqrt{2(x+1)}$
« Last Edit: December 22, 2011, 12:04:31 pm by b^3 »
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#### nisha

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##### Re: VCE Methods Question Thread!
« Reply #42 on: December 22, 2011, 12:13:41 pm »
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thankyou!
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#### Insa

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##### Re: VCE Methods Question Thread!
« Reply #43 on: December 22, 2011, 09:27:59 pm »
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Hey guys, I've begun flicking through pages of Maths Quest 12 methods textbook and started doing some questions. I'm having trouble with this question and need some help with it. The question is in the attachment. Thanks for any replies

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#### brightsky

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##### Re: VCE Methods Question Thread!
« Reply #44 on: December 22, 2011, 09:36:03 pm »
+2
(2/x + 3x)^5 = 5C0*(2/x)^5(3x)^0 + 5C1*(2/x)^4(3x)^1 + 5C2*(2/x)^3(3x)^2 + ... 5C5*(2/x)^0(3x)^5
so which one of these terms has a degree of 2?
none of them!
so the coefficient would be 0.
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2014 - 2016: Bachelor of Biomedicine, The University of Melbourne
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Currently selling copies of the VCE Chinese Exam Revision Book and UMEP Maths Exam Revision Book, and accepting students for Maths Methods and Specialist Maths Tutoring in 2020!