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June 26, 2022, 02:50:34 pm

Author Topic: VCE Methods Question Thread!  (Read 4206479 times)  Share 

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1729

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Re: VCE Methods Question Thread!
« Reply #19320 on: March 17, 2022, 09:29:53 pm »
+4
Hi again,

Can someone please help me with q3 and q6 b.

Thanks,

beep boop
Hi beep boop  :)

Question 6
\(\mathbf{b.}\quad\) Note make sure you look back at what you started with. The the maximal domain of \(f\left(x\right)\) is \(x\in \left(0,\infty \right)\) (this is because \(\log _{10}\left(x\right)\) is only defined for \(x>0\)).
Now assuming you were able to yield to the result in part a, solve \(x=f\left(y\right)\) with consideration of the above domain.
\(\frac{y^2}{10y+10}=x\:\Longrightarrow \:y^2=x\left(10y+10\right)\Longrightarrow y^2-10xy-10=0\)
From here you can apply quadratic formula where \(a=1,\:b=-10x,\:c=-10\)
This will yield to \(y=\frac{10x+\sqrt{100x^2+40x}}{2}\:\text{or}\:\frac{10x-\sqrt{100x^2+40x}}{2}\).
However, you need to consider which solution satisifies conditons of \(f\left(x\right)\) or in other words, which solution has a range of \(\left(0,\infty \right)\). (since \(\text{dom}\:f=\text{ran}\:f^{-1}\). Testing out \(x=1\) for both of the solutions show that the first solution is the inverse function.)
Therefore, \(f^{-1}:\:\left(0,\infty \right)\rightarrow \mathbb{R},\:f^{-1}\left(x\right)=\frac{10x+\sqrt{100x^2+40x}}{2}\)
We got the domain of \(f^{-1}\) to be \(\left(0,\infty \right)\) because that is the range of the original function  :)

Question 3
\(\mathbf{a.}\quad\) For the purpose of my explanation suppose \(g\left(x\right)=f\left(x-5\right)\), then \(g\left(x+5\right)=f\left(\left(x+5\right)-5\right)=f\left(x\right)\). In other words just substitute \(x+5\) into \(f\left(x-5\right)\) to obtain original function.

\(\mathbf{b.}\quad\) Like above, lets suppose \(g\left(x\right)=2f\left(-2\left(x-\frac{1}{2}\right)\right)+5\). Therefore, \(\frac{1}{2}g\left(x\right)-\frac{5}{2}=f\left(-2\left(x-\frac{1}{2}\right)\right)\) and by solving \(-2\left(x-\frac{1}{2}\right)=x'\) you can find the horizontal transformations needed to undo it. (that is dilation of factor \(\frac{1}{2}\) from \(y\) axis, reflection in \(y\) axis and translation of \(1\) unit right.) Therefore, \(\frac{1}{2}g\left(-\frac{x-1}{2}\right)-\frac{5}{2}=f\left(x\right)\). Now substitue that into the equation. (I'm assuming this quesiton is tech-active so pop it in your CAS!)  ;D

Hope this helps :)
« Last Edit: March 17, 2022, 09:53:03 pm by 1729 »

beep boop

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Re: VCE Methods Question Thread!
« Reply #19321 on: March 18, 2022, 10:07:44 pm »
+1
Hi 1729!

Tysm. I understand more now.  8)

beep boop
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beep boop

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Re: VCE Methods Question Thread!
« Reply #19322 on: April 14, 2022, 10:08:48 pm »
0
Hi again,

Can someone please help me w/ q2 part b and q8 part e?

Much appreciated,

beep boop
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1729

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Re: VCE Methods Question Thread!
« Reply #19323 on: April 15, 2022, 09:59:54 am »
+2
Hi again,

Can someone please help me w/ q2 part b and q8 part e?

Much appreciated,

beep boop
Hi beep boop,

Question 2
\(\mathbf{b.}\quad\) Try and use the identity \(\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\). Assuming that you were able to get that \(a=\frac{1}{\sqrt{3}}\)
The equation to find the points of intersection would be \(\frac{1}{\sqrt{3}}\sin \left(x\right)=\cos \left(x\right)\) which yields to the following results
\(\Longrightarrow \frac{1}{\sqrt{3}}\frac{\sin \left(x\right)}{\cos \left(x\right)}=1\)
\(\Longrightarrow \tan \left(x\right)=\sqrt{3}\)

Question 11
\(\mathbf{e.}\quad\) Assuming you were able to get \(f'\left(x\right)=-\sin \left(\frac{x}{2}\right)\) (you could do recognize this transformation by inspection or alternatively)
Consider how the transformation maps \(f\left(x\right)\) to \(af\left(x+\pi\right)+b\) where \(af\left(x+\pi \right)+b=f'\left(x\right)\)
We know that \(af\left(x+\pi \right)+b=2a\cos \left(\frac{x+\pi }{2}\right)+b+a\pi \)
This can be simplified to \(-2a\sin\left(\frac{x}{2}\right)+b+a\pi\) using the complementary identity \(\cos \left(x+\frac{\pi }{2}\right)=-\sin \left(x\right)\).
Now, equate the coefficients \(-\sin \left(\frac{x}{2}\right)=-2a\sin \left(\frac{x}{2}\right)+b+a\pi \).
We want \(-2a=-1\) and \(b+a\pi =0\). This should get us, \(a=\frac{1}{2},\:b=-\frac{\pi }{2}\).

Hope this helps.
« Last Edit: April 18, 2022, 04:40:28 pm by 1729 »

beep boop

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Re: VCE Methods Question Thread!
« Reply #19324 on: May 16, 2022, 08:52:29 pm »
0
Hi all,

can someone please help me with this?

f(x)= 2x^3 + 3(1-k)x^2 + 6kx+ 5

given than k>-1

sketch the graph of f(x), plot the y intercept and turning points.

find the values of k such that the graph on the previous part has 3 intercepts correct to two decimal places.

ty,
beep boop
« Last Edit: May 16, 2022, 08:57:03 pm by beep boop »
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"Distance makes the heart grow fonder and proximity makes the heart want to barf."-Mr K, Never have I Ever
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1729

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Re: VCE Methods Question Thread!
« Reply #19325 on: May 16, 2022, 10:33:25 pm »
+2
I literally felt excitement in my chest when I saw a new post in the methods thread, woohoo!
f(x)= 2x^3 + 3(1-k)x^2 + 6kx+ 5

given than k>-1

sketch the graph of f(x), plot the y intercept and turning points.

I'm confused what this question is asking for because the graph can change quite a bit as \(k\) changes, are you sure there isn't anything else to the question? If not, I think it would be acceptable to sketch the graph of \(f(x)\) for any value of \(k\in \left(-1,\infty \right)\).

find the values of k such that the graph on the previous part has 3 intercepts correct to two decimal places.

To find intercepts consider turning points of the function (only way 3 intercepts are possible is if there are 2 turning points)

To find the turning points of \(f\) in terms of \(k\), solve \(f'(x)=0 \Longrightarrow 6x^2+6x\left(1-k\right)+6k=0\)
We will get two solutions (using quadratic formula) \(x=\frac{-1+k+\sqrt{k^2-6k+1}}{2},\: x=\frac{-1+k-\sqrt{k^2-6k+1}}{2}\).
To find coordinates pop these solutions into \(f(x)\).

Now let \(a\) and \(b\) be these turning points (for the purpose of keeping this post tidy). We want to solve \(f\left(a\right)<0<f\left(b\right)\). We solve this because we want the range of these turning points to include 0 so there will be 2 intercepts.

Attatched is how you would do it with CAS (I'm assuming this is CAS active)
Remember to consider the domain of \(k\) before writing down the answer!!

beep boop

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Re: VCE Methods Question Thread!
« Reply #19326 on: May 17, 2022, 11:14:07 am »
+1
Hi 1729!

I literally felt excitement in my chest when I saw a new post in the methods thread, woohoo!

I feel like I'm annoying everyone on this thread. Most of the previous posts have been me. 😭😅😂 lkfjasldkjfjffd yr 12 methods dfjlsjfsld

I'm confused what this question is asking for because the graph can change quite a bit as \(k\) changes, are you sure there isn't anything else to the question? If not, I think it would be acceptable to sketch the graph of \(f(x)\) for any value of \(k\in \left(-1,\infty \right)\).

There isn't a lot of the question. I tried to upload a photo of this last night but it didn't work for some reason. ☹ It said it didn't pass security checks even tho I've posted pics on this forum more times than I can count.

Thank so much for answering my questions from when I started posting until now! I'm so grateful for this thread. yr 12 is alksjdfasd

beep boop
« Last Edit: May 17, 2022, 02:06:15 pm by beep boop »
class of '22
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"Distance makes the heart grow fonder and proximity makes the heart want to barf."-Mr K, Never have I Ever
yr 12 stuff :)

biology1234

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Re: VCE Methods Question Thread!
« Reply #19327 on: June 05, 2022, 05:16:54 pm »
+1
ASAP hey how do you do this question? 
if the turning point of the parabola with y = x^2- bx + 21 has an x - coordinate of 4, the value of b is.
Could someone provide working out for this question, thank you much appreciated.

1729

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Re: VCE Methods Question Thread!
« Reply #19328 on: June 05, 2022, 08:05:11 pm »
+4
ASAP hey how do you do this question? 
if the turning point of the parabola with y = x^2- bx + 21 has an x - coordinate of 4, the value of b is.
Could someone provide working out for this question, thank you much appreciated.

Hi, for future reference it would be better if you showed some working out or attempt at the question so people can better help you  :)

For a quadratic, \(y=ax^2+bx+c\) the x-coordinate of the vertex is \(-\frac{b}{2a}\) (intuitively, this can be derived from completing the square on the general form of a quadratic, or finding the midpoint of x-intercepts/roots of it)

This means that \(\frac{-\left(-b\right)}{2}=4\). You can solve for \(b\) now.

biology1234

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Re: VCE Methods Question Thread!
« Reply #19329 on: June 06, 2022, 02:01:57 pm »
+2
ASAP thank you I am also a bit confused with this one.
I'm not sure how we get the y intercept.
Is it y= a( x+3) (x-3)^ 2 is where i have gotten to so far.

Commercekid2050

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Re: VCE Methods Question Thread!
« Reply #19330 on: June 06, 2022, 02:45:15 pm »
+2
ASAP thank you I am also a bit confused with this one.
I'm not sure how we get the y intercept.
Is it y= a( x+3) (x-3)^ 2 is where i have gotten to so far.

We need to find k here.

So we would use one of the two points (-3,0) or (3,0)

0= 27-27-27+k
0=-27+k
k=27

Thus y intercept is (0,27)
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PizzaMaster

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Re: VCE Methods Question Thread!
« Reply #19331 on: 2 hours ago »
0
Hi guys!!!  ;D
Could someone please help me with this
Sketch: y = (x − 1)(x^2 + 1) (Note: There is no turning point or ‘flat point’ of this cubic.)

What does "There is no turning point or ‘flat point’ of this cubic" mean here?

And also:
If we simply just have:
Sketch
y=(x+2)(x^2-x+7)

There are so many ways to sketch this. How do we do it?


Thank you!
« Last Edit: 2 hours ago by PizzaMaster »