Question

Parallel Movement of Functions Tutorial!!!

Answer 1

Let's understand how graths of functions y=ƒ(x) and y=ƒ(x)+n are connected. For this case, let n=2.
Let show that y=ƒ(x)+2 can be derived from y=ƒ(x) applying the vertical parallel movement.
Any point M(x0, y0) will translate into point M1(x0, y0+2).
If any point M(x0, y0) is on grath y=ƒ(x) then y0=ƒ(x0) and
y0+2=ƒ(x0)+2. Thus point M1(x0, y0+2) is on grath y=ƒ(x)+2.
Consequently apyyling the parallel movement 2 places by y-asis any point of grath y=ƒ(x) will translate to corresponding point on y=ƒ(x)+2. If another point K(x1, y1) is on y=ƒ(x)+2 then y1=ƒ(x)+2. y1-2=ƒ(x1). Thus K(x1, y1-2) is on grath y=ƒ(x).

We can make a conclusion that:
- Grath of function y=ƒ(x)+n can be derived from
grath y=ƒ(x) applying the traslation n units by y-asis if n>0 or |n| units if n<0. -

On this grawing you see y=x² and y=x²-4. Parabila y=x²-4 derived from y=x².

Now lets see how y=ƒ(x) and y=ƒ(x-m) are connected.
Let m=3. Using our previous conclusin we can say
that y=ƒ(x-3) can be derived from y=ƒ(x) by traslating it 3 units by x-asis to the right. Thus any point M(x0, y0) it traslates to point M1(x0+3, y0).
Grath y=ƒ(x-m) can be derived from grath y=ƒ(x) appyling the translation m units by x-asis to the right if m>0 or |m| to the left if m<0.
In parabolas the point of parabola and line if simmetry interssect is called the vertex. In parabola y=x²+2 the line of simmetry is y-asis and in parabola y=(x-3)² the symmetry is x=3.
The final conclusion is that grath of function y=ƒ(x-m)+n may be derived from grath y=ƒ(x) as the result of traslation m units by x-asis and and traslation of grath y=(x-m)² by y asis on n units.

Answer 2

@jagr2713

Answer 3

Great job

Answer 4

thx

Answer 5

@sammixboo @TheSmartOne @thomaster @pooja195 @bingtamekia

Answer 6

Amazing!

Answer 7

Good Job! :)

Answer 8

$$\Large nice~ work ~:)$$