AdamK
  • AdamK
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
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

jim_thompson5910
  • jim_thompson5910
Were you able to figure out the domain and range of the transformed graph?
anonymous
  • anonymous
can someone help me i realy need help
AdamK
  • AdamK
No I can't find out the domain or range because none of the points I've plotted correspond to it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
What is the left most point on the given graph?
AdamK
  • AdamK
(-1, 1)
jim_thompson5910
  • jim_thompson5910
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 ?
AdamK
  • AdamK
-2
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
this means that the left most part of the domain in the transformed function is x = -2
jim_thompson5910
  • jim_thompson5910
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))
jim_thompson5910
  • jim_thompson5910
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)
jim_thompson5910
  • jim_thompson5910
does that make sense?
AdamK
  • AdamK
Not really. Why would x-values need to be defined for the original function when it's being transformed?
jim_thompson5910
  • jim_thompson5910
notice how we have f[ `(1/3)*(x-1)` ] basically f[ T ] where T = (1/3)*(x-1)
jim_thompson5910
  • jim_thompson5910
is it possible to have f[ T ] when T is say, T = 5 ?
AdamK
  • AdamK
Oh, ok
jim_thompson5910
  • jim_thompson5910
I guess in a way, you have to think backwards
jim_thompson5910
  • jim_thompson5910
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
AdamK
  • AdamK
My final answer is "a."
AdamK
  • AdamK
Hopefully I don't have to graph anything
jim_thompson5910
  • jim_thompson5910
You don't. They just want the domain and range of the transformed function.
AdamK
  • AdamK
Ok good

Looking for something else?

Not the answer you are looking for? Search for more explanations.