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anonymous
 one year ago
NEED HELP BADLY! NEEDS TO BE DONE BY TOMORROW ~*WILL MEDAL AND FAN!*~
3. Complete Parts a, b, and c.
a. Graph the function f(x) = x + 2.
b. Translate the graph according to the rule (x,y) to (x3,y)
c. Write the function equation for the graph in Part b. Name the function g(x). *This is the part of the question I am having major problems with...*
Answer:
anonymous
 one year ago
NEED HELP BADLY! NEEDS TO BE DONE BY TOMORROW ~*WILL MEDAL AND FAN!*~ 3. Complete Parts a, b, and c. a. Graph the function f(x) = x + 2. b. Translate the graph according to the rule (x,y) to (x3,y) c. Write the function equation for the graph in Part b. Name the function g(x). *This is the part of the question I am having major problems with...* Answer:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm ok... I just really need help with this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know the part a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I already graphed it as (4,2),(3,1),(2,0),(1,1), and (0,2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it should be something like this:dw:1433437736059:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, that's what it looks like, the problem I have is that after I translate the graph by x3 I can't seem to write a function for it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now what you need to do for the part b is to graph y=(x3)+2=x1. okay?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, ok I think I see the problem, I just subtracted by x3 instead of simplifying it to x1.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in fact, part c is what we just obtained from the part b.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I see that now, thank you so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks. good luck friend!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good luck to you too! ^.^
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