Transformation

[Sophie Atherden]

y=2x-1
- translations of unit in the positive direction of x-axis, and 2 units in the positive y direction
- reflection in the y axis
- dilation of 2 from the x-axis
- reflection in the line y=x
find the values of x and y in terms of x' y'

[bridgetkelly123]

Yo,
Wasn't it  - ind the equations of the image of x' and y', in terms of x and y

[abeybaby]

There is a chapter that explains this beautifully in Connect's methods Unit 3 notes. This is a really good example of a difficult question. BAsic principle is: Always draw a diagram. And if you want the equation of the line under those transformations, just substitute the x and y we found on the side back into the equation

[EmmaQuinn]

There is a chapter that explains this beautifully in Connect's methods Unit 3 notes. This is a really good example of a difficult question. BAsic principle is: Always draw a diagram. And if you want the equation of the line under those transformations, just substitute the x and y we found on the side back into the equation

(Image removed from quote.)

Hey thanks for this, that was really helpful! So If I wanted to find the image of a point from the original graph, I would just substitute the new x and y values into the corresponding coordinates.
EG: find image of (5,-3), I would go ->  5=-y'-1 and -3=X'/2 - 2 

[abeybaby]

That's okay!

So remember, the whole principle is that you take a point (x,y), and after applying a sequence of transformations, wherever you end up is called (x', y').

So in our example, we're saying if x=5 and y=-3, what are x'and y'?

x' = 2(y+2)
x' = 2*(-3+2)
x' = 2*-1 = -2

y' = -(x+1)
y' = -(5+1)
y' = -6

so under this sequence of transformations, (5, -3) becomes (-2, -6).

If you wanted to find the equation of the line y=2x -1, you would say substitute in the x' and y'

y = 2x-1
(x'-4)/2 = 2*(-y'-1) -1
x'-4=-4y'-4-1
(x'-4+5)/-4=y'
y = -(x+1)/4 is the equation of the image