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January 19, 2020, 01:13:52 pm

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#### Jimmmy

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##### Re: VCE Methods Question Thread!
« Reply #17700 on: February 21, 2019, 09:26:08 pm »
0
Hi guys, just have a question about transformations that I'd like some help with

If a domain of a function is (-2,2)

The function then undergoes a series of transformations:

-   A translation of one unit in the negative direction of y-axis
-   A translation of one unit in the negative direction of the x-axis
-   A dilation by factor 1/3 from the x axis

What is the new domain? Does it even change?

Thanks!
Do you know what the function is? Parabola, Quartic, Inverse?  I don't think it will change the domain unless explicitly stated, but it may change the range of y-values you get in that domain.
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#### jinaede1342

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##### Re: VCE Methods Question Thread!
« Reply #17701 on: February 21, 2019, 09:29:10 pm »
0
Do you know what the function is? Parabola, Quartic, Inverse?  I don't think it will change the domain unless explicitly stated, but it may change the range of y-values you get in that domain.

It's a parabola...that's what I thought too, I guess I'll just state the domain as being (-2,2)

thanks
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#### RuiAce

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##### Re: VCE Methods Question Thread!
« Reply #17702 on: February 21, 2019, 09:31:48 pm »
+5
Do you know what the function is? Parabola, Quartic, Inverse?  I don't think it will change the domain unless explicitly stated, but it may change the range of y-values you get in that domain.

"A translation of one unit in the negative direction of the x-axis" reads like the domain (-2, 2) gets translated to the domain (-3, 1) to me.

#### Jimmmy

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##### Re: VCE Methods Question Thread!
« Reply #17703 on: February 21, 2019, 09:41:33 pm »
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"A translation of one unit in the negative direction of the x-axis" reads like the domain (-2, 2) gets translated to the domain (-3, 1) to me.
Woops

That's what happens when you let too many hours of non-Methods study happen in a day...you're spot on. For some reason, seeing a dilation warped my mind, without realising it doesn't have any relevance for domains.

Thanks for picking that up, Rui.
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#### jinaede1342

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##### Re: VCE Methods Question Thread!
« Reply #17704 on: February 21, 2019, 09:49:00 pm »
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"A translation of one unit in the negative direction of the x-axis" reads like the domain (-2, 2) gets translated to the domain (-3, 1) to me.

So dilations from the x-axis don't have an impact on the domain, but translating from the x-axis does? So one unit in the negative direction of the x-axis will result in a domain of (-3,1)? sorry, just trying to get the hang of this
« Last Edit: February 21, 2019, 09:51:35 pm by jinaede1342 »
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#### RuiAce

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##### Re: VCE Methods Question Thread!
« Reply #17705 on: February 21, 2019, 09:54:43 pm »
+5
So dilations from the x-axis don't have an impact on the domain, but translating from the x-axis does? So one unit in the negative direction of the x-axis will result in a domain of (-3,1)? sorry, just trying to get the hang of this
Dilations about the $x$-axis result in the curve being stretched vertically upwards/downwards. The $x$-coordinates stayed fixed, however the $y$-coordinates are gonna be squashed closer to the $x$-axis, i.e. the line $y=0$. (This is because your dilation factor is $\frac13$, which is less than 1.)

This has an effect on the range as pointed out above, not the domain.

Whereas translating in the negative direction of the $x$-axis literally means shift it to the left. Left/right changes obviously impact the domain whenever applicable.

#### peachxmh

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##### Re: VCE Methods Question Thread!
« Reply #17706 on: February 21, 2019, 11:05:02 pm »
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Can someone explain how to use mapping (this method specifically please!) to figure out the sequence of transformations that takes the graph of y = 2$\sqrt{4-x}$ + 3 to the graph of y= -$\sqrt{x}$ + 6? Thank you!!

Edit: fixed the formatting
« Last Edit: February 21, 2019, 11:10:24 pm by peachxmh »
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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17707 on: February 22, 2019, 12:31:49 am »
+5
Can someone explain how to use mapping (this method specifically please!) to figure out the sequence of transformations that takes the graph of y = 2$\sqrt{4-x}$ + 3 to the graph of y= -$\sqrt{x}$ + 6? Thank you!!

Edit: fixed the formatting

I'm not sure what you mean in your question about "how to use mapping" regarding transformations. Could you please elaborate? Nonetheless, I'll answer the question a few different ways.

You should note that there are often several possible sequences of transformations that will end in the same outcome. Obviously some are simpler than others, but we should keep that in mind. For example, for $y=x^2$, a dilation by factor $4$ from the $x$-axis is the same as a dilation by factor $1/2$ from the $y$-axis.

Method 1: Using dash notation

First, let's obtain expressions for the image of $x$ and $y$  ($x'$ and $y'$ respectively).

We can rewrite both equations to get  $\dfrac{y-3}{2}=\sqrt{4-x}$  and  $6-y=\sqrt{x}$.

Thus, we have  $\begin{cases}x'=4-x\\ 6-y'=\displaystyle\frac{y-3}{2}\end{cases}\implies\begin{cases}x'=-x+4\\ \displaystyle y'=-\frac{1}{2}y+\frac{15}{2}\end{cases}.$Once we have it in this form, we can just read off the transformations. Remember, convention is that we give them in the order: dilations, reflections, translations.

1) Dilation by factor $1/2$ from the $x$-axis
2) Reflection in the $x$-axis
3) Reflection in the $y$-axis
4) Translation of $4$ units in the positive $x$-direction
5) Translation of $15/2$ units in the positive $y$-direction

Method 2: Using some clever intuition

For our convenience, let's define  $f(x)=2\sqrt{4-x}+3$  and  $g(x)=-\sqrt{x}+6$.

Let's go through dilations. Clearly, there is an unwanted $2$ in the rule of $f$ in front of the square root. To get rid of that, let's apply a dilation by factor $1/2$ from the $x$-axis. That is, after the first transformation, the rule of $f$ becomes $f_1(x)=\frac12 f(x)=\frac{1}{2}\big(2\sqrt{4-x}+3\big)=\sqrt{4-x}+\frac32.$ There are no other required dilations.

Now let's consider reflections. Looking at the rule of $g$ we see that we need to implement a negative sign in front of the square root. So, we apply a reflection in the $x$-axis, giving $f_2(x)=-f_1(x)=-\left(\sqrt{4-x}+\frac32\right)=-\sqrt{4-x}-\frac32.$We also have an unwanted negative sign in front of the $x$, so we should apply a reflection in the $y$-axis too, so that $f_3(x)=f_2(-x)=-\sqrt{4-(-x)}-\frac32=-\sqrt{x+4}-\frac32.$Now, let's consider translations. The inside of the square root in the rule of $g$ is a lone $x$. So, we apply a translation of $4$ units in the positive $x$-direction: $f_4(x)=f_3(x-4)=-\sqrt{(x-4)+4}-\frac32=-\sqrt{x}-\frac32.$
Finally, we need to shift the graph up  $6-(-3/2)=15/2$  units: $f_5(x)=f_4(x)+\frac{15}{2}=\left(-\sqrt{x}-\frac32\right)+\frac{15}{2}=-\sqrt{x}+6.$All together, we end up with the same transformations. Ie: $g(x)=\frac{-1}{2} f\big(\!-\!(x-4)\big)+\frac{15}{2}.$
Either method works. Although Method 2 looks longer, it can be a lot quicker when given simpler problems involving only a few transformations.
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#### peachxmh

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##### Re: VCE Methods Question Thread!
« Reply #17708 on: February 22, 2019, 09:25:12 am »
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I'm not sure what you mean in your question about "how to use mapping" regarding transformations. Could you please elaborate?

Thank you so much for the detailed solutions (and when I said mapping I just meant dash notation (our school just calls it mapping), so sorry! )
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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17709 on: February 22, 2019, 09:22:53 pm »
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$\left(11-2x\right)\left(5-2x\right)\ge0$
$12x^2+x>6$
I know how to solve for x like first one x = 5/2 and 11/2, however i'm not sure which equality i'm supposed to use?
i'm not really fond of drawing a graph or expanding the equation out to solve this (if it's possible not to) because of time; is there a way where i can picture the graph in my head and know whether it should be greater than, less than, greater or equal to etc??
I can easily picture these graphs in my head however im not sure what the equation is solving for?
Is it like solve for when y is more than 0 for x?
i hope someone can explain this thing that has been confusing me =[
thanks!
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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17710 on: February 22, 2019, 09:44:10 pm »
+2
$\left(11-2x\right)\left(5-2x\right)\ge0$
$12x^2+x>6$
I know how to solve for x like first one x = 5/2 and 11/2, however i'm not sure which equality i'm supposed to use?
i'm not really fond of drawing a graph or expanding the equation out to solve this (if it's possible not to) because of time; is there a way where i can picture the graph in my head and know whether it should be greater than, less than, greater or equal to etc??
I can easily picture these graphs in my head however im not sure what the equation is solving for?
Is it like solve for when y is more than 0 for x?
i hope someone can explain this thing that has been confusing me =[
thanks!

Question 1
$(11-2x)(5-2x)\geq 0$
I know you said you're not really fond of drawing a graph, but you don't actually need to sketch the entire graph. Here, we only require the $x$-axis intercepts.

This first one is just a positive quadratic with $x$-axis intercepts at $x=\dfrac52$ and $x=\dfrac{11}{2}$.

From the graph, it's easy to see that the required solutions are given by  $\boxed{x\in\left(-\infty,\ \dfrac52\right]\cup\left[\dfrac{11}{2},\ \infty\right)}$.  That is, the graph of  $y=(11-2x)(5-2x)$  is above (or at) the $x$-axis when either  $x\leq \dfrac52$  or  $x\geq \dfrac{11}{2}$.

Question 2

I'll let you try this one on your own.

First, factorise  $12x^2+x-6$  and draw the graph's general shape.
« Last Edit: February 22, 2019, 09:48:03 pm by AlphaZero »
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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17711 on: February 22, 2019, 10:39:16 pm »
0
Hey,
it requires the answer to be like
-4<x<3 or like x>-4 and x<3

not that complicated xD
thanks for your answer however i do not understand how you got
$x\le\frac{5}{2}$ and $x\ge\frac{11}{2}$ with the equality sign - like how do you know which one to use?

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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17712 on: February 22, 2019, 11:09:03 pm »
+1
Hey,
it requires the answer to be like
-4<x<3 or like x>-4 and x<3

not that complicated xD
thanks for your answer however i do not understand how you got
$x\le\frac{5}{2}$ and $x\ge\frac{11}{2}$ with the equality sign - like how do you know which one to use?

The statements  "$x\leq\dfrac52$ or $x\geq\dfrac{11}{2}$"   and   "$x\in\left(-\infty,\ \dfrac52\right]\cup\left[\dfrac{11}{2},\ \infty\right)$"  mean exactly the same thing. One isn't more complicated than the other. In fact, in most questions, interval notation is preferred for clarity, especially when having to write out a union of many intervals. If the question requires you to give your answer in a certain form, then you must follow that. But if it doesn't, feel free to use any notation you like.

The graph of  $y=(5-2x)(11-2x)$  is above or at the $x$-axis in the blue parts.  Those correspond to whenever  $x\leq \dfrac52$ (the left one),  or when  $x\geq \dfrac{11}{2}$ (the right one).
« Last Edit: February 22, 2019, 11:10:58 pm by AlphaZero »
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#### Meddling

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##### Re: VCE Methods Question Thread!
« Reply #17713 on: February 22, 2019, 11:09:39 pm »
0
$\left(11-2x\right)\left(5-2x\right)\ge0$
I know how to solve for x like first one x = 5/2 and 11/2, however i'm not sure which equality i'm supposed to use?
i'm not really fond of drawing a graph or expanding the equation out to solve this (if it's possible not to) because of time; is there a way where i can picture the graph in my head and know whether it should be greater than, less than, greater or equal to etc??
I can easily picture these graphs in my head however im not sure what the equation is solving for?
Is it like solve for when y is more than 0 for x?
Hey, what you are solving for is the set of x values which corresponds to positive y values!

So, it is good that you know how to solve for the x intercepts (or when y = 0), but in this case you must provide a set of x values which corresponds to the positive y values! I.e. When the graph is 'above' the x axis (or y=0).

From your methods work, you will know that this is a positive parabola! Which makes the U shape. Hence, if it has 2 axis intercepts, the lower x-value must be relatively on the LHS and higher x-value on the RHS. As the shape of the graph is U, and you know that x intercept is 5/2 and 11/2, the 5/2 must be on the left and 11/2 on the right!.

That means, if you visualise the shape, anything to the left of the x-intercept (5/2) must be above x axis (or have a positive y value or 0 for x intercept). Similarly, anything to the right of the x-intercept (11/2) must also be above the x axis.

So you would write the solution using inequality sign.

The previous answer was perfectly done, and I am providing further explanation as you seem confused!
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#### aspiringantelope

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##### Re: VCE Methods Question Thread!
« Reply #17714 on: February 23, 2019, 10:04:12 am »
0
Hey, what you are solving for is the set of x values which corresponds to positive y values!

So, it is good that you know how to solve for the x intercepts (or when y = 0), but in this case you must provide a set of x values which corresponds to the positive y values! I.e. When the graph is 'above' the x axis (or y=0).

From your methods work, you will know that this is a positive parabola! Which makes the U shape. Hence, if it has 2 axis intercepts, the lower x-value must be relatively on the LHS and higher x-value on the RHS. As the shape of the graph is U, and you know that x intercept is 5/2 and 11/2, the 5/2 must be on the left and 11/2 on the right!.

That means, if you visualise the shape, anything to the left of the x-intercept (5/2) must be above x axis (or have a positive y value or 0 for x intercept). Similarly, anything to the right of the x-intercept (11/2) must also be above the x axis.

So you would write the solution using inequality sign.

The previous answer was perfectly done, and I am providing further explanation as you seem confused!
The statements  "$x\leq\dfrac52$ or $x\geq\dfrac{11}{2}$"   and   "$x\in\left(-\infty,\ \dfrac52\right]\cup\left[\dfrac{11}{2},\ \infty\right)$"  mean exactly the same thing. One isn't more complicated than the other. In fact, in most questions, interval notation is preferred for clarity, especially when having to write out a union of many intervals. If the question requires you to give your answer in a certain form, then you must follow that. But if it doesn't, feel free to use any notation you like.

(Image removed from quote.)

The graph of  $y=(5-2x)(11-2x)$  is above or at the $x$-axis in the blue parts.  Those correspond to whenever  $x\leq \dfrac52$ (the left one),  or when  $x\geq \dfrac{11}{2}$ (the right one).
Thanks so much both of you!!!! and dantraicos for the image! Understand now!
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