Transformations of Graphs: y = 4^(x+3) to y = (2^x)/5

[Monkeyseat]

1) Describe geometrically how the curve y=4^(x+3) can be transformed into the curve y=(2^x)/5 by a sequence of stretches.

Okay so I rewrote it as:

y = (4^x) * 4^3
y = (2^2x) * 64

I'm not sure now though about the stretches. Moving outwards from x this is what I did:

Horizontal stretch, scale factor 2 gives y = (2^x) * 64.
Vertical stretch, scale factor 1/64 gives y = 2^x.
Verical stretch, scale factor 1/5 gives y = (2^x)/5.

Is that correct? If not, where have I gone wrong?

2) The graph of y = f(x) is reflected in the x-axis and then the y-axis to produce the graph with equation y = g(x).

a) Find g(x) in terms of f and x.

Okay so reflection in the x-axis gives y = -f(x) and then a reflection in the y axis gives y = -f(-x).

b) Describe geometrically the single transformation that maps the graph of y = f(x) onto the graph of y = g(x).

Is it a reflection in a line? Sorry, I don't know, I've drawn the graph but I can't see it:

Any help?

Thanks very much.

 

[stapel]

1) Describe geometrically how the curve y=4^(x+3) can be transformed into the curve y=(2^x)/5 by a sequence of stretches.

You have the originating function:

. . .y = 4[sup:1fie10wn]x + 3[/sup:1fie10wn]
. . . . .= 4[sup:1fie10wn]x[/sup:1fie10wn] 4[sup:1fie10wn]3[/sup:1fie10wn]
. . . . .= (2[sup:1fie10wn]2[/sup:1fie10wn])[sup:1fie10wn]x[/sup:1fie10wn]) (64)
. . . . .= (64) (2[sup:1fie10wn](2x)[/sup:1fie10wn])

You have the target function:

. . .y = (2[sup:1fie10wn]x[/sup:1fie10wn]) / 5
. . . . .= (1/5) (2[sup:1fie10wn]x[/sup:1fie10wn])
. . . . .= (1/5) (2[sup:1fie10wn](1/2) (2x)[/sup:1fie10wn])
. . . . .= (1/320) (64) (2[sup:1fie10wn](1/2) (2x)[/sup:1fie10wn])

I'm not sure if the scale factor, in going from 2[sup:1fie10wn]2x[/sup:1fie10wn] to 2[sup:1fie10wn](1/2)(2x)[/sup:1fie10wn] is "2" (as you have written) or "1/2"...? Or if the overall vertical "stretch" is "320" or "1/320"...?

2) The graph of y = f(x) is reflected in the x-axis and then the y-axis to produce the graph with equation y = g(x). a) Find g(x) in terms of f and x.

Okay so reflection in the x-axis gives y = -f(x) and then a reflection in the y axis gives y = -f(-x).

Sounds good to me. But, if you need convincing, try drawing a few graphs, making the changes both by reflecting and by sign-changing.

b) Describe geometrically the single transformation that maps the graph of y = f(x) onto the graph of y = g(x).

Is it a reflection in a line? Sorry, I don't know, I've drawn the graph but....

Yes, it's a line. To "find" this line, draw lines between corresponding points on the two graphs, and draw dots at these segments' midpoints. Eventually you'll have enough dots (points) to draw the line. It's a very simple one. :wink:

Eliz.

 

[Monkeyseat]

Re:

Thanks for the reply.

1) Describe geometrically how the curve y=4^(x+3) can be transformed into the curve y=(2^x)/5 by a sequence of stretches.

You have the originating function:

. . .y = 4[sup:3uqsagk4]x + 3[/sup:3uqsagk4]
. . . . .= 4[sup:3uqsagk4]x[/sup:3uqsagk4] 4[sup:3uqsagk4]3[/sup:3uqsagk4]
. . . . .= (2[sup:3uqsagk4]2[/sup:3uqsagk4])[sup:3uqsagk4]x[/sup:3uqsagk4]) (64)
. . . . .= (64) (2[sup:3uqsagk4](2x)[/sup:3uqsagk4])

You have the target function:

. . .y = (2[sup:3uqsagk4]x[/sup:3uqsagk4]) / 5
. . . . .= (1/5) (2[sup:3uqsagk4]x[/sup:3uqsagk4])
. . . . .= (1/5) (2[sup:3uqsagk4](1/2) (2x)[/sup:3uqsagk4])
. . . . .= (1/320) (64) (2[sup:3uqsagk4](1/2) (2x)[/sup:3uqsagk4])

I'm not sure if the scale factor, in going from 2[sup:3uqsagk4]2x[/sup:3uqsagk4] to 2[sup:3uqsagk4](1/2)(2x)[/sup:3uqsagk4] is "2" (as you have written) or "1/2"...? Or if the overall vertical "stretch" is "320" or "1/320"...?

I don't know, that's why I was asking. :lol: I got a different outcome to you (or did I just do the vertical stretches separately?) - I don't really understand it! Can someone else please clarify how to do this then or who is right/wrong?

b) Describe geometrically the single transformation that maps the graph of y = f(x) onto the graph of y = g(x).

Is it a reflection in a line? Sorry, I don't know, I've drawn the graph but....

Yes, it's a line. To "find" this line, draw lines between corresponding points on the two graphs, and draw dots at these segments' midpoints. Eventually you'll have enough dots (points) to draw the line. It's a very simple one. :wink:

Sounds simple enough. I did the dot thing but I didn't know which side of the curve corresponded to one another and it just became a mess. Is it y = x? :?

 

[stapel]

I'm not sure if the scale factor, in going from 2[sup:3spcl15m]2x[/sup:3spcl15m] to 2[sup:3spcl15m](1/2)(2x)[/sup:3spcl15m] is "2" (as you have written) or "1/2"...? Or if the overall vertical "stretch" is "320" or "1/320"...?

I don't know, that's why I was asking.

Well, check your book. If the "scale factor" always a whole number or can the scale factor be a fraction or decimal? You need to provide us with your book's naming convention. :idea:

I got a different outcome to you (or did I just do the vertical stretches separately?)

You divided by 64, and then by 5. I divided by (64)(5) = 320. Whether or not your book requires that the division (in going from the originating function to the final function) must, for some reason, stop at a whole number in between (in this case, "1") is something we can't "see" from here.

To "find" this line....

I did the dot thing but I didn't know which side of the curve corresponded to one another....

Um... If you don't know how the one line relates the other, then how did you draw the other...? :shock:

Eliz.

 

[Monkeyseat]

Re:

I'm not sure if the scale factor, in going from 2[sup:1fi5c2f6]2x[/sup:1fi5c2f6] to 2[sup:1fi5c2f6](1/2)(2x)[/sup:1fi5c2f6] is "2" (as you have written) or "1/2"...? Or if the overall vertical "stretch" is "320" or "1/320"...?

I don't know, that's why I was asking.

Well, check your book. If the "scale factor" always a whole number or can the scale factor be a fraction or decimal? You need to provide us with your book's naming convention. :idea:

I've not got a book at the moment only a photocopy, but in previous questions the scale factor has been either - in some it has been a whole number and some a fraction.

To "find" this line....

I did the dot thing but I didn't know which side of the curve corresponded to one another....

Um... If you don't know how the one line relates the other, then how did you draw the other...? :shock:

Eliz.

I used a graph drawing tool. I just drew lines between the corresponding points on both graphs (image in the first post) but it wasn't very clear:

That was using y = (x+2)^2 and y = -(-x+2)^2. It didn't appear very conclusive!

 

[Monkeyseat]

So is part b a reflection in y = x?

For some reason it seems to work for some but not for others:

Using y = (x-5)+1 and y = -((-x-5)+1)
Therefore, y = x works.

Using y = (x+5)^2 and y = -((-x+5)^2)
Doesn't seem to work - I can't see any line of reflection here. I've drawn the graph multiple times and I know I'm probably wrong but I can't see it. I did the dot thing and got a mass of lines - the mid-points all appeared to be at the origin. It made me think it might be y = 1/(-x) from what I saw.

Sorry, I don't want it to seem like you're giving me the answer but I don't know where to go from here. Thanks.

 

[Monkeyseat]

Sorry to do this but I'm just bumping it up the page as it's in for tomorrow and mainly I still don't know part (2b), although I'm not 100% certain about part (1) either.

Someone said it was a rotation but they're not even on the syllabus - it's just translations, rotations and reflections. Please could you give me some more info? I've drawn numerous graphs for part (b) but I can't see any obvious/clear line of reflection, sorry! I can't get one single transformation that maps the graph of y = f(x) onto the graph of y = g(x) - those 2 graphs could be anything and from what I've seen what works for one pair may not work for another pair. Now there is obviously an answer but I have tried for a long time and can't see it. Please, could you just assist me in finishing this?

Thanks for helping, much appreciated, it is just sometimes harder to understand things over the internet than face to face.

 

[stapel]

Using y = (x+5)^2 and y = -((-x+5)^2)
Doesn't seem to work - I can't see any line of reflection here.

For my previous reply, I imagined pictures in my head, but didn't draw things out. If you do the thing with the dots, you should find that the midpoints of segments joining corresponding parts (vertices, going in and up/down, and going out and down/up) keep landing in the same spot. (See below.) So this is actually reflection through a point.

Sorry!

Eliz.

 

[Monkeyseat]

Re:

Using y = (x+5)^2 and y = -((-x+5)^2)
Doesn't seem to work - I can't see any line of reflection here.

For my previous reply, I imagined pictures in my head, but didn't draw things out. If you do the thing with the dots, you should find that the midpoints of segments joining corresponding parts (vertices, going in and up/down, and going out and down/up) keep landing in the same spot. (See below.) So this is actually reflection through a point.

Sorry!

Eliz.

No problem stapel, thanks for helping. I had a diagram exactly like yours and I was wondering "where's the line???" :lol: Someone had directed me to this website shortly before your reply which showed a similar problem where the answer was a reflection in the origin (I didn't even know you could have that). You've helped confirm it - so I take it that it's a reflection in the origin then?

Many thanks.