1. ## Transformations

The question is

For each of the following transformations find the rule and the invariant points (if they exist)

One of them is: a dilation from the x axis of factor 4

I know the rule is $(x,y) \longrightarrow (x,4y)$

But for the invariant points, I know that x isn't getting moved so that is the point but the answers say $\{(x,0):x\in R\}$

Can someone explain to me what this means?

2. Originally Posted by jgv115
The question is

For each of the following transformations find the rule and the invariant points (if they exist)

One of them is: a dilation from the x axis of factor 4

I know the rule is $(x,y) \longrightarrow (x,4y)$

But for the invariant points, I know that x isn't getting moved so that is the point but the answers say $\{(x,0):x\in R\}$

Can someone explain to me what this means?
If the point is invariant then you require y = 4y ....

3. Originally Posted by jgv115
The question is

For each of the following transformations find the rule and the invariant points (if they exist)

One of them is: a dilation from the x axis of factor 4

I know the rule is $(x,y) \longrightarrow (x,4y)$

But for the invariant points, I know that x isn't getting moved so that is the point but the answers say $\{(x,0):x\in R\}$

Can someone explain to me what this means?
Only if $y=0$ will the vertical co-ordinate of the point also not move under $(x,y)\rightarrow\ (x,4y)$

since $4y=y\Rightarrow\ 4y-y=0\Rightarrow\ 3y=0\Rightarrow\ y=0$

This means that any point on the graph that lies on the x-axis
remains on the x-axis.

(think of the graph of a line... by applying the dilation you multiply the slope by 4,
but it still crosses the x-axis at the same point,
or think of a sinewave that crosses the x-axis...
after the dilation, the sinewave has 4 times the original amplitude
but crosses the x-axis at the same points as before).

4. mm.. sorry I still don't fully understand...

I understand that only if y=0 will the vertical coordinate of the point also not move under $(x,y)\rightarrow\ (x,4y)$ because yea...

But what's that got to do with $\{(x,0):x\in R\}$

What does $x\in R$ mean?

Sorry if I'm really stupid or something

5. Originally Posted by jgv115
mm.. sorry I still don't fully understand...

I understand that only if y=0 will the vertical coordinate of the point also not move under $(x,y)\rightarrow\ (x,4y)$ because yea...

But what's that got to do with $\{(x,0):x\in R\}$

What does $x\in R$ mean?
Every point on the x-axis is invariant under the dilation.

The x-axis is the line $y=0$

so $\{(x,0):x\in R\}$ is referring to all points on the x-axis.

(x may be any real value)

6. Thanks for the answers! I think I understand

A transformation has rule $(x,y) \longrightarrow (3-x,2y+1)$ Find the coordinates of the point which is invariant under this transformation.

So if the point is invariant then you require $3-x=x$ and $2y+1 = y$

So that means the invariant points are $\displaystyle (\frac{3}{2}, -1 )$

Is this right or wrong...

Now I'm doing composition of transformations and there's only question I don't understand.

Find the rule for a reflection in the x axis followed by a reflection in the line x=2.

So I know the first part is $(x,y)\rightarrow (x,-y)$ but the next bit reflection in the line x=2. I don't understand how to do this... Can someone help me out?

7. Originally Posted by jgv115
Thanks for the answers! I think I understand

A transformation has rule $(x,y) \longrightarrow (3-x,2y+1)$ Find the coordinates of the point which is invariant under this transformation.

So if the point is invariant then you require $3-x=x$ and $2y+1 = y$

So that means the invariant points are $\displaystyle (\frac{3}{2}, -1 )$ point, not points

Is this right or wrong...

Yes, that is the invariant point. A single point has a pair of x and y co-ordinates.

Now I'm doing composition of transformations and there's only question I don't understand.

Find the rule for a reflection in the x axis followed by a reflection in the line x=2.

So I know the first part is $(x,y)\rightarrow (x,-y)$ but the next bit reflection in the line x=2. I don't understand how to do this... Can someone help me out?
Will the y co-ordinate change ?

The x co-ordinate has a certain distance to the line x=2.
How far away will the new x co-ordinate be from the point $(x,-y)\;\;?$

8. mm..

The y coordinate won't change...

But sorry I just can't get my head around this

I'm thinking if x=0 what would the rule be and then altering it to x=2, maybe +2 or -2 somewhere?

Am I getting somewhere? More clues?

If the x point is 2 units on the right of the line x=2 that means x' will be 2 units on the left of x=2 therefore it will be x-4... but if the x point is 1 unit on the right of x=2 then everything changes.

I don't know

9. Geometrically, to "reflect in a line", you draw the line perpendicular to the given line and mark a point on that perpendicular, on the other side of the line, at the same distance from the line.

Let (x, y) be the original point and (x', y') the reflected point. Then the point on the line, (2, y), must be midway between (x, y) and (x', y'). That means that (x+ x")/2= 2 and (y+ y')/2= y. From the first equation, x+ x'= 4 so x'= 4- x. From the second equation, y+ y'= 2y so y'= 2y- y= y. (x, y) is reflected into (4- x, y).

10. Originally Posted by jgv115
mm..

The y coordinate won't change...

But sorry I just can't get my head around this

I'm thinking if x=0 what would the rule be and then altering it to x=2, maybe +2 or -2 somewhere?

If a point's horizontal co-ordinate is x=0,
then the vertical line x=2 would be halfway between x=0 and the new x, so the new x co-ordinate will be 4.
If the horizontal co-ordinate of a point is x=7, then the vertical line x=2 is halfway between x=7 and the new x,
so the new x is x=-3

Am I getting somewhere? More clues?

If the x point is 2 units on the right of the line x=2 that means x' will be 2 units on the left of x=2 therefore it will be x-4... but if the x point is 1 unit on the right of x=2 then everything changes.

But it will also be 4-x, and that will give you the image of a point with x co-ordinate 3 also

I don't know
As HallsofIvy showed, start with "the average of the x co-ordinate and it's image is 2".

11. Originally Posted by jgv115
mm..

The y coordinate won't change...

But sorry I just can't get my head around this

I'm thinking if x=0 what would the rule be and then altering it to x=2, maybe +2 or -2 somewhere?

Am I getting somewhere? More clues?

If the x point is 2 units on the right of the line x=2 that means x' will be 2 units on the left of x=2 therefore it will be x-4... but if the x point is 1 unit on the right of x=2 then everything changes.

I don't know
Do you understand what invariant means? If so, I don't see why the very first replies (including my own) do not provide a more than satisfactory explanation.

12. Ok I got it.

I'm up to Determining Transformations and the book does a horrible horrible job of explaining. HORRIBLE (well at least for me)

For example

Find the single transformation which maps $y=x^2$ to $y = 2(x + 3)^2-4$

The book tells me to assume the composition which maps $(x,y) \rightarrow (x',y')$

But then it says "Rearrange to make the transformation from $y=x^2$ more obvious"...

Then it somehow gets $y=\frac{y'+4}{1}$ and $x=(x'+3)$

Where did the square on the (x+3) go?

Ahh I don't get it

13. Originally Posted by jgv115
Ok I got it.

I'm up to Determining Transformations and the book does a horrible horrible job of explaining. HORRIBLE (well at least for me)

For example

Find the single transformation which maps $y=x^2$ to $y = 2(x + 3)^2-4$

The book tells me to assume the composition which maps $(x,y) \rightarrow (x',y')$

But then it says "Rearrange to make the transformation from $y=x^2$ more obvious"...

Then it somehow gets $y=\frac{y'+4}{1}$ and $x=(x'+3)$

Where did the square on the (x+3) go?

Ahh I don't get it
Firstly, think of the point labelled $(x,y)$ as being on the curve $y=x^2.$

Any point on $y=x^2$ is mapped to a corresponding image point on $y=2(x+3)^2-4,$ a different curve.

To distinguish between points on the two curves, the points on $y=2(x+3)^2-4$

may be labelled $\left(x',y'\right).$

We want to find transformation equations that map $x\rightarrow\ x'$ and $y\rightarrow\ y'.$

So we can begin, knowing that the points labelled $\left(x',y'\right)$ are on the image curve.

Hence write $y'=\left2(x'+3\right)^2-4$

We want to arrive at $x'=g(x,y),\;\;\;y'=h(x,y)$

$y'=2\left(x'+3\right)^2-4\Rightarrow\ y'+4=2\left(x'+3\right)^2$

$\displaystyle\frac{y'+4}{2}=\left(x'+3\right)^2$

This is of the form $y=x^2$ if $y=\displaystyle\frac{y'+4}{2}$ and if $x=x'+3$

Hence, the transformation equations are $y'=2y-4$ and $x'=x-3$

You can check this....

$y=x^2$

$y'=2y-4,\;\;x'=x-3$

$\displaystyle\ y=\frac{y'+4}{2},\;\;x=x'+3$

$\displaystyle\frac{y'+4}{2}=\left(x'+3\right)^2\Ri ghtarrow\ y'+4=2\left(x'+3\right)^2\Rightarrow\ y'=2\left(x'+3\right)^2-4$

and as the "dashes" are only used to distinguish between point co-ordinates on both curves,

the image curve is $y=2(x+3)^2-4$

14. Okok I understand. I think of it as basically making x/y equal to something in the new equation then subbing it in, sort of like reducible equations.

I have a few that I'm confused about, these ones have more than one step:

$y= x^2$ to $y=2x^2-3$

This is what I've done...

$y'+3 = 2(x')^2$

so...

$y= y' +3$ and $x=2x'$

So... $\displaystyle \frac{x}{2} = x'$ and $y+3 = y'$

The answers say: A dilation factor of 2 from the x axis followed by a translation determined by the vector (0,-3) (this should be vertical).

How can it be a dilation factor of 2 from the x axis... that is $(x,y)\rightarrow (x,2y)$ How come I get +3 and it says -3?

Also..

$\displaystyle y= \frac{1}{x}$ to $\displaystyle y=\frac{2}{x-3}$

I don't rearrange the equation because I think it is fine... so..

$y=y'$ and $\displaystyle x = \frac{2}{x'-3}$

$\displaystyle x' = \frac{2}{x}+3$

Is this correct? What do I do from here

Also I just want to say how much I appreciate the help from you guys, I love you guys soo much!!

15. Originally Posted by jgv115
Okok I understand. I think of it as basically making x/y equal to something in the new equation then subbing it in, sort of like reducible equations.

I have a few that I'm confused about, these ones have more than one step:

$y= x^2$ to $y=2x^2-3$

This is what I've done...

$y'+3 = 2(x')^2$

so...

$y= y' +3$ and $x=2x'$

no, it's not ready to classify the image points yet

So... $\displaystyle \frac{x}{2} = x'$ and $y+3 = y'$

The answers say: A dilation factor of 2 from the x axis followed by a translation determined by the vector (0,-3) (this should be vertical).

How can it be a dilation factor of 2 from the x axis... that is $(x,y)\rightarrow (x,2y)$ How come I get +3 and it says -3?

Also..

$\displaystyle y= \frac{1}{x}$ to $\displaystyle y=\frac{2}{x-3}$

I don't rearrange the equation because I think it is fine... so..

no, it still needs work!

$y=y'$ and $\displaystyle x = \frac{2}{x'-3}$

$\displaystyle x' = \frac{2}{x}+3$

Is this correct? What do I do from here

Also I just want to say how much I appreciate the help from you guys, I love you guys soo much!!
You're almost there...

For the first one....

$y=x^2$

$\displaystyle\ y'=2\left(x'\right)^2-3\Rightarrow\ y'+3=2\left(x'\right)\Rightarrow\frac{y'+3}{2}=\le ft(x'\right)^2$

Now, we have the form $\displaystyle\ y=x^2$

Hence $\displaystyle\frac{y'+3}{2}=y\Rightarrow\ 2y=y'+3\Rightarrow\ y'=2y-3$

$x'=x$

Check:

$\displaystyle\ y=x^2\Rightarrow\frac{y'+3}{2}=\left(x'\right)^2$

$y'+3=2\left(x'\right)^2\Rightarrow\ y'=2\left(x'\right)^2-3$

For the second one....

$\displaystyle\ y=\frac{1}{x}$

$\displaystyle\ y'=\frac{2}{x'-3}\Rightarrow\frac{y'}{2}=\frac{1}{x'-3}$

Now it's in the form $\displaystyle\ y=\frac{1}{x}$

$\displaystyle\ y=\frac{y'}{2}\Rightarrow\ y'=2y.$

$\displaystyle\ x=x'-3\Rightarrow\ x'=x+3$

Check:

$\displaystyle\ y=\frac{1}{x}\Rightarrow\frac{y'}{2}=\frac{1}{x'-3}\Rightarrow\ y'=\frac{2}{x'-3}$

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