Question

http://imgur.com/RPAkgxB
#4. I've tried plotting points and I just get a straight line. Not sure if that's right or not.

Here are what I think the transformations are: Right 1, stretch of 3, up 1

Answer 1

Were you able to figure out the domain and range of the transformed graph?

Answer 2

can someone help me i realy need help

Answer 3

No I can't find out the domain or range because none of the points I've plotted correspond to it

Answer 4

What is the left most point on the given graph?

Answer 5

(-1, 1)

Answer 6

yes, so the left half of the domain of the original function starts with x = -1

the question is: which x value, when plugged into (1/3)*(x-1), gives a result of -1?

ie what is the solution to (1/3)*(x-1) = -1 ?

Answer 7

-2

Answer 8

good

Answer 9

this means that the left most part of the domain in the transformed function is x = -2

Answer 10

if you plugged in something smaller than -2 (say -3), then (1/3)*(x-1) produces a number that is smaller than -1 (which is outside the domain of f(x))

Answer 11

the goal is to stay in the domain of f(x) because we can't use x values that aren't defined for f(x)

Answer 12

does that make sense?

Answer 13

Not really. Why would x-values need to be defined for the original function when it's being transformed?

Answer 14

notice how we have f[ `(1/3)*(x-1)` ]

basically f[ T ] where T = (1/3)*(x-1)

Answer 15

is it possible to have f[ T ] when T is say, T = 5 ?

Answer 16

Oh, ok

Answer 17

I guess in a way, you have to think backwards

Answer 18

The lowest you can go is T = -1

So you solve for x in T = (1/3)*(x-1) and like you said, you'll get x = -2

So f[ (1/3)*(x-1) ] has the lower part of the domain be x = -2

Answer 19

My final answer is "a."

Answer 20

Hopefully I don't have to graph anything

Answer 21

You don't. They just want the domain and range of the transformed function.

Answer 22

Ok good