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September 28, 2025, 01:22:33 am

Author Topic: VCE Methods Question Thread!  (Read 5701543 times)  Share 

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Toki

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Re: VCE Methods Question Thread!
« Reply #3960 on: February 09, 2014, 02:55:28 pm »
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Thanks!

P.Lipton

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Re: VCE Methods Question Thread!
« Reply #3961 on: February 09, 2014, 05:32:28 pm »
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Hey guys, I'd love some help with this question.

The functions y=x3-2x2+ax+10 and y=6+(a+b)x-4x2-x3 both have (-1,0) as an x-intercept.
Find the values of a and b.

b^3

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Re: VCE Methods Question Thread!
« Reply #3962 on: February 09, 2014, 05:38:17 pm »
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1. Substitute in and into the two curves to give two equations in terms of and .
2. Solve the two equations simultaneously for and .

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Irving4Prez

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Re: VCE Methods Question Thread!
« Reply #3963 on: February 09, 2014, 09:13:06 pm »
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Hey guys, just had a couple of questions relating to Modulus;

Is the sq root of (1 - 2x + x^2) = (x-1)^2 or |x -1|?

Would |2x - 5| be > or < 0 if it is known that 1 < x < 2.
« Last Edit: February 16, 2014, 07:15:49 pm by Irving4Prez »

Phy124

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Re: VCE Methods Question Thread!
« Reply #3964 on: February 09, 2014, 10:28:30 pm »
+2
Hey guys, just had a couple of questions relating to Modulus;

Is the sq root of (1 - 2x + x^2) = (x-1)^2 or |x -1?


Would |2x - 5| be > or < 0 if it is known that 1 < x < 2. I'm assuming it's going to be < 0 because the graph of y = 2x - 5 has negative y values between 1 and 2.


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Anchy

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Re: VCE Methods Question Thread!
« Reply #3965 on: February 10, 2014, 06:00:56 pm »
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Can i please have some help with this question? :)

6x+2y-z=1
x+y+z=2
kx+y-z=1

For what values of k, is there:
1. a unique solution?
2. no solution?

:)
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3966 on: February 10, 2014, 07:33:57 pm »
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Stuck again

2cos^2(x) + 3cos(x)  +1=0

b^3

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Re: VCE Methods Question Thread!
« Reply #3967 on: February 10, 2014, 07:40:28 pm »
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Hint: Let , factorise in terms of , substitute back in the , form the two different solutions and solve.
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Was there a domain given? If so then pick the sols that fit the domain, if not then there are a whole load of solutions, which you could write in terms of , where is an integer.
e.g.
« Last Edit: February 10, 2014, 07:43:11 pm by b^3 »
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Snorlax

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Re: VCE Methods Question Thread!
« Reply #3968 on: February 11, 2014, 12:03:41 pm »
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I have a dilation question:
With the function... y=1/x

If I want to dilate it by a factor of 2 parallel to the y-axis (aka from x-axis)
what will that look like?
If I want to dilate it by a factor of 2 parallel to the x-axis (aka from y-axis)
what will that look like?

I just need a general gist of what these look like, and whether i'm doing it correctly :/

If someone could possibly sketch it for me, that'll be appreciated! :)
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Re: VCE Methods Question Thread!
« Reply #3969 on: February 11, 2014, 12:31:41 pm »
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both graphs look the same.
if you dilate y=1/x by 2 from the xaxis, you get y=2*1/x, so y=2/x
if you dilate y=1/x by 2 from the yaxis, you get y=1/(1/2*x), which is still y=2/x
you can sketch y=1/x and y=2/x on your cas to see what they look like.

Snorlax

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Re: VCE Methods Question Thread!
« Reply #3970 on: February 11, 2014, 12:51:59 pm »
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Just focusing on the dilations from the yaxis:
If it tells me to dilate by a factor of 2 from the yaxis, I multiply the x values by 2 or 1/2?
If it tells me to dilate by a factor of 1/2 from the yaxis, I multiply the xvals by 1/2 or 2?

I think that's where i'm getting confused
Thanks
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nhmn0301

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Re: VCE Methods Question Thread!
« Reply #3971 on: February 11, 2014, 12:59:54 pm »
+1
Just focusing on the dilations from the yaxis:
If it tells me to dilate by a factor of 2 from the yaxis, I multiply the x values by 2 or 1/2?
If it tells me to dilate by a factor of 1/2 from the yaxis, I multiply the xvals by 1/2 or 2?

I think that's where i'm getting confused
Thanks
I know dilation can be a bit confusing, but it would probably help if you remember it this way.
When every there is a dilation whether from the y-axis or x-axis, I multiply the other variable by 1/k.
Eg: I have y = x^2
Dilate a factor of k from the y-axis, I multiply x by 1/k
Y = (x/k)^2
Dilate a factor of I from the x-axis, I multiply y by 1/k
Y/k= x^2
Rearrange : y= k(x^2)
Hope this helps!
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Snorlax

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Re: VCE Methods Question Thread!
« Reply #3972 on: February 11, 2014, 04:51:43 pm »
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Is reflecting in the x-axis the same as saying reflecting across the x-axis?

Damn, all this terminology...hella confusing..
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Ancora_Imparo

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Re: VCE Methods Question Thread!
« Reply #3973 on: February 11, 2014, 04:59:44 pm »
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Dilation by a factor of
From the x-axis = parallel to the y-axis
to

From the y-axis = parallel to the x-axis
to

Reflections
Over/in/across the x-axis
to

Over/in/across the y-axis
to

Translation of units
In the positive/negative x-direction = in the positive/negative direction of the x-axis = to the right/left
to

In the positive/negative y-direction = in the positive/negative direction of the y-axis = upwards/downwards
to
« Last Edit: February 11, 2014, 05:06:34 pm by Ancora_Imparo »
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Blondie21

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Re: VCE Methods Question Thread!
« Reply #3974 on: February 11, 2014, 06:34:45 pm »
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How do we find the x intercept of

5 - 4 x 31 - x = 0
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