November 27, 2022, 11:30:05 pm

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#### Golgi Apparatus

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##### Re: VCE Methods Question Thread!
« Reply #19305 on: February 16, 2022, 10:45:00 pm »
+1
The cost of a taxi trip in a particular city is $4.00 up to and including 2 km. After 2 km the passenger pays an additional$2.00 per kilometre. Find the function f which describes this method of payment and sketch the graph of y=f(x), where x is the number of kilometres travelled. (Use a continuous model.) How do you do this?

I don’t have time to answer this fully at the moment, but I’ll give you a hint.

I think this should be a piecewise function with two linear components - one for x (0,2] and one for x (2,infinity]

Let me know if you need any more help!
VCE
2020: Biology
2021: Chemistry | English Language | Further | Methods | Psychology

2022: Bachelor of Biomedicine (UoM)

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19306 on: February 20, 2022, 03:57:41 pm »
0
Hi guys!!
I was just wondering:
How do we know when to use a calculator in the methods Cambridge 1/2 book?
Does it have a symbol anywhere or...

Thanks!

#### Rose34

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##### Re: VCE Methods Question Thread!
« Reply #19307 on: February 23, 2022, 11:52:06 am »
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Hello,

Does anyone know how to set up a stopwatch on CAS? I set up the timer accidentally but not sure how I did it so does anyone know how?

#### beep boop

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##### Re: VCE Methods Question Thread!
« Reply #19308 on: March 11, 2022, 08:11:36 pm »
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Hi all,

beep boop
class of '22
'21: viet sl [36], bio
'22: psych, methods, spesh, chem, eng lang
"Distance makes the heart grow fonder and proximity makes the heart want to barf."-Mr K, Never have I Ever
yr 12 stuff :)

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19309 on: March 11, 2022, 11:06:15 pm »
+5
Hi all,

beep boop
Hi beep boop, here are some hints and tips you should consider before looking at the solution
- Consider change of base rule, that is, $\log _ab=\frac{\log _cb}{\log _ca}$
- Consider power rule, that is, $\log _a\left(b^c\right)=c\log _ab$
- Consider the other exponent rule, $a^{\log _ab}=b$

Question 9
$\mathbf{a.}\quad$ Find the real numbers $k$ and $m$ such that $\log _9\left(x^3\right)=k\log _3\left(x\right)$ and $\log _{27}\left(512\right)=m\log _3\left(8\right)$
Solution
We know from change of base law that, $\log _9x=\frac{\log _3x}{\log _39}$.
Therefore, our equation can be simplified to $\frac{1}{\log _39}\log _3\left(x^3\right)=k\log _3x\Rightarrow \frac{3}{2}\log _3x=k\log _3x\therefore k=\frac{3}{2}$

Now for the other half of the question, $\log_38=\frac{\log_{27}8}{\log_{27}3}=3\log _{27}8$ which hopefully allows you to recognize that $3\log _{27}8=\log _{27}8^3=\log _{27}512$. From here it is obvious that the answer is $m=1$
$\mathbf{b.}\quad$ Find values of $x$ for which $\log _9x^3+\log _3x^{\frac{1}{2}}=\log _{27}512$
Solution
The above can be simplified to $3\log _9x+\frac{1}{2}\log _3x=\log _{27}512$ using logarithmic properties discussed above, now we need to change the base of these logarithims.
$\log _9\left(x\right)=\frac{\log _{27}x}{\log _{27}9}\Rightarrow \frac{3}{2}\log _27x$ and $\log _3\left(x\right)=\frac{\log _{27}x}{\log _{27}3}\Rightarrow 3\log _{27}x$.

$3\cdot \left(\frac{3}{2}\log _{27}x\right)+\frac{1}{2}\left(3\log _{27}x\right)=\log _{27}512\Rightarrow \frac{9}{2}\log _{27}x+\frac{1}{2}\left(3\log _{27}x\right)=\log _{27}512$
$\Rightarrow \log _{27}\left(x^{\frac{9}{2}}\right)+\log _{27}\left(x^{\frac{3}{2}}\right)=\log _{27}x\Rightarrow \log _{27}\left(x^{\frac{3}{2}}\cdot x^{\frac{9}{2}}\right)=\log _{27}512$
$\log _{27}\left(x^6\right)=\log _{27}512\Rightarrow x^6=512\therefore x=512^{\frac{1}{6}}$

Note: The $\log _{27}\left(x^{\frac{3}{2}}\cdot x^{\frac{9}{2}}\right)$ came from the property of $\log _ba+\log _bc=\log _bac$
Question 20
$\mathbf{a.}\quad$ Solve for $x$, $\log _2\left(x-2\right)=\log _4\left(x^2-6x+12\right)$
Solution
The above simplifies as such $\log _2\left(x-2\right)=\frac{\log _4\left(x-2\right)}{\log _42}=2\log _4\left(x-2\right)=\log _4\left[\left(x-2\right)^2\right]$ using change of base laws. Now all you have to do is solve $\left(x-2\right)^2=x^2-6x+12$.
$\mathbf{b.}\quad$ Solve for $x$, $x^{\log _ex}=e^{\left[\log _e\left(x\right)\right]^3}$
Solution
The above simplifies as such $x^{\log _ex}=\left(e^{\log _ex}\right)^3$ using the fact that $a^{bc}=\left(a^b\right)^c$.
Now $e^{\log _ex}=x$ using the property that $a^{\log _ab}=b$. Therefore, the equation simplifies to $x^3=x^{\log _ex}$. From here, we equate the exponents as such: $3=\log _e\left(x\right),\:\therefore x=e^3$.

Hope this helps, if you need anymore help don't hesitate to ask!

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19310 on: March 13, 2022, 05:46:01 pm »
0
Hi guys!!!
Question:
The graph with equation y = f(x) undergoes a series of transformations so that the equation of the image is y = −f(5x − 7). State the transformations in the correct order.
Can someone explain how to do this using the pre-image method?
I am so confused with the fractions and all.

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19311 on: March 13, 2022, 07:48:36 pm »
+2
Hi guys!!!
Question:
The graph with equation y = f(x) undergoes a series of transformations so that the equation of the image is y = −f(5x − 7). State the transformations in the correct order.
Can someone explain how to do this using the pre-image method?
I am so confused with the fractions and all.
You can use the pre-image method but it's certainly not necessary.

$y=f\left(x\right)$ maps to $y=-f\left(5x-7\right)$ which is equivalent to $y=-f\left(5\left(x-\frac{7}{5}\right)\right)$
From here, you should be able to determine it by inspection. The sequence of transformations are as follows:
1. Dilation of factor $\frac{1}{5}$ from the $y$ axis
2. Reflection in the $x$ axis
3. Translation of $\frac{7}{5}$ units in the positive direction of the $x$ axis (right)

Note: The order step 2 can be wherever (since it's the only transformation that affects the images position on the $y$ axis) but the translation must be after the dilation. This is because if you were to translate it, then dilate it you would also have to dilate the translation. This is why you should make sure you consider the order! It is very important.

If you didn't understand how I did the above by inspection, I used the fact that $f(x)$ maps to $f\left(\frac{x}{a}\right)$ after a dilation of factor $a$ from the $y$ axis and $f\left(x-h\right)$ is a translation of $h$ units right and $-f\left(x\right)$ flips the graph over the $x$ axis. I suggest you get use to this way of reading transformations as well, I have attatched the diagram which can be accessed in the methods 11 textbook in the spoiler below.

Transformation Table

If you wanted to go through the transformations using the pre-image method it would follow as such
Let the transformation map $\left(x,y\right)\mapsto \left(x',y'\right)$
$y=f\left(x\right)$ maps to $y'=-f\left(5x'-7\right)$. Rearranging the equation gives
$-y'=f\left(5x'-7\right)$. We choose to write $y=-y'$ and $x=5x'-7$, now solve for $x'$ and $y'$ and we get
$\left(x,y\right)\mapsto \left(\frac{x}{5}+\frac{7}{5},-y\right)$ which leads to the sequence of transformations.

I hope this helps
« Last Edit: March 13, 2022, 07:53:00 pm by 1729 »

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19312 on: March 13, 2022, 11:33:00 pm »
0
Thank you so much 1729!
It helps a lot!!

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19313 on: March 13, 2022, 11:42:47 pm »
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So if dilation is stated first that means it is only for that and not for the translation part, right?

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19314 on: March 13, 2022, 11:56:49 pm »
0
When we have to state the transformations that map:
y=√x  --> y=−3√2x−4   −1
Dilation by a factor of 3 from the x-axis
Dilation by a factor of 1/2 from the y-axis
Reflection over the x-axis
Translation of 2 units in the positive direction of the x-axis
Translation of 1 unit in the negative direction of the y-axis

In this case, why isn't it
"Translation of 4 units in the positive direction of the x-axis"
and why
"Translation of 2 units in the positive direction of the x-axis"
?

Thank you!

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19315 on: March 14, 2022, 11:18:58 am »
+1
So if dilation is stated first that means it is only for that and not for the translation part, right?
When we have to state the transformations that map:
y=√x  --> y=−3√2x−4   −1
Dilation by a factor of 3 from the x-axis
Dilation by a factor of 1/2 from the y-axis
Reflection over the x-axis
Translation of 2 units in the positive direction of the x-axis
Translation of 1 unit in the negative direction of the y-axis

In this case, why isn't it
"Translation of 4 units in the positive direction of the x-axis"
and why
"Translation of 2 units in the positive direction of the x-axis"
?

Thank you!
That is correct! Well done

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19316 on: March 14, 2022, 11:50:09 am »
0
Why isn't it
"Translation of 4 units in the positive direction of the x-axis"
and why
"Translation of 2 units in the positive direction of the x-axis"
?

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19317 on: March 14, 2022, 12:08:48 pm »
+1
Why isn't it
"Translation of 4 units in the positive direction of the x-axis"
and why
"Translation of 2 units in the positive direction of the x-axis"
?
Consider $f\left(2x-4\right)$ vs $f\left(2\left(x-2\right)\right)$.
Hopefully this rings a bell, from here it is a dilation of factor $\frac{1}{2}$ from $y$ axis and then a translation of $2$ units right.

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19318 on: March 14, 2022, 12:14:04 pm »
+1
Ahhhhh I get it.
Tysm!

#### beep boop

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##### Re: VCE Methods Question Thread!
« Reply #19319 on: March 17, 2022, 08:06:36 pm »
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Hi again,