hello

[huehue]

can someone please help me with this question?

given f(x)= -5sin(2x - π/3) + 4 and g(x)= cos(x). State the transformations required to change f(x) into g(x).

thanks in advance

[Callum@1373]

Well seeing as cos(x) is also equal to sin(x + pi/2), it is in simpler terms to work out.

f(x) = -5sin(2x - pi/3) + 4
g(x) = sin(x + pi/2)

So the transformations must be:

- Translated 5pi/6 units in the negative x axis
- Translated 4 units in the negative y axis
- Reflected in the y axis
- Dilated by factor 2 from the y axis
- Dilated by factor 1/5 from the x axis

[Peanut Butter]

Well seeing as cos(x) is also equal to sin(x + pi/2)

Could you please explain this?

Because I thought cos(x) = sin (pi/2 - x)....

[Callum@1373]

Could you please explain this?

Because I thought cos(x) = sin (pi/2 - x)....

so sin(x+a) = sinxcosa + cosxsina

So sin(x + pi/2):

sinxcos(pi/2) + cosxsin(pi/2)

= 0 + cosx
= cosx

[huehue]

Well seeing as cos(x) is also equal to sin(x + pi/2), it is in simpler terms to work out.

f(x) = -5sin(2x - pi/3) + 4
g(x) = sin(x + pi/2)

So the transformations must be:

- Translated 5pi/6 units in the negative x axis
- Translated 4 units in the negative y axis
- Reflected in the y axis
- Dilated by factor 2 from the y axis
- Dilated by factor 1/5 from the x axis

thank-you!!

[Swagadaktal]

so sin(x+a) = sinxcosa + cosxsina

So sin(x + pi/2):

sinxcos(pi/2) + cosxsin(pi/2)

= 0 + cosx
= cosx

damn I don't even know about this wtf where do you learn this voodoo

[Peanut Butter]

so sin(x+a) = sinxcosa + cosxsina

So sin(x + pi/2):

sinxcos(pi/2) + cosxsin(pi/2)

= 0 + cosx
= cosx

sin(x - a) = sin(x)cos(a) - sin(a)cos(x)

sin(pi/2)cos(x) - sin(x)cos(pi/2)
= cos(x)

both our expressions work because they would be complementary angles

[Callum@1373]

sin(x - a) = sin(x)cos(a) - sin(a)cos(x)

sin(pi/2)cos(x) - sin(x)cos(pi/2)
= cos(x)

both our expressions work because they would be complementary angles

Yah

damn I don't even know about this wtf where do you learn this voodoo

You gotta be a magician to do it 

[keltingmeith]

damn I don't even know about this wtf where do you learn this voodoo

They're compound angle formulas. They were on the methods study design but never tested on - they regularly make appearances in spec exams, though.

[huehue]

YahYou gotta be a magician to do it 

one more question, is reflection in the x axis different from reflected in the x axis?

edit: hehe one more question. y=1/x dilated by a factor 1/2 from the y-axis is y=1/2x? ty!!

[huehue]

nope, yep

oh okay, thank you. two more final questions, if i reflect that in the y-axis does it become y=-1/2x? also, is translation of +3 units parallel to the x-axis essentially 3 units right?

[huehue]

okayyy, two final questions (i promise these are the last )

State the translation T_a,b required to obtain the graph with the 2^nd equation fro that with the 1^st.

a) y=¹/₃x³ ---T_a,b---> y=¹/₃x³-3x²+9x-13

b) y=²/_x ---T_a,b---> y=(-3x+14)/(x-4)

[huehue]

bump 8D!!

[qazser]

Post in 3/4 thread Mods might get upset. No clue how to do transformations from one graph to another atm but would do (x,y)->(x,y) method where u label the DRTs from the original axis and another set for the second equation then doing transformations between.

Will try to do b, might be wrong

For first equation Y=2/x
Mapping would be (x,y)->(x,2y)

For second equation y=-3x+14/x-4
Mapping would be (x,y)->(x+4,-3y+14)?<-Not sure about mapping for second one.
Then to go from (x,y)->(x,2y) to (x,y)->(x+4,-3y+14)

for x it would be translation 4 units on positive direction of x axis

for y it would be
Dilation of factor 3/2 from x axis
Reflection in x axis
Translation 14 units positive "bla bla"

Was guessing this would be the most ideal way to do it, have no clue

[HighTide]

okayyy, two final questions (i promise these are the last )

State the translation T_a,b required to obtain the graph with the 2^nd equation fro that with the 1^st.

a) y=¹/₃x³ ---T_a,b---> y=¹/₃x³-3x²+9x-13

b) y=²/_x ---T_a,b---> y=(-3x+14)/(x-4)

Hey,
I didn't spend much time on this and didn't really put it in to matrix form but if it's the transformations you require:
a) I'm pretty sure there's a much more methodical way but I did
--> divide the whole transformed side (right hand side) by 1/3.
--> you can see that the whole thing won't divide nicely so I separated the -39 into -27 and -12. That way you have a cubic expression which can be factorized into (x-3)^3 with a remainder
--> Hence the answer I got was:
y=1/3 * (x-3)^3 -4
b) Second one you can use long division. Since both degrees are the same you'll end up with a hyperbola and hence one x, (same form as the starting function) and thus you'll end up with y=2/(x-4) -3. From there the transformations are more clear.
Apologies for not posting the matrix transformation form, had little time. But if it was the transformations which you required, hope this helps.