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1. the range is x > -k but can someone tell me why this is so for part c find the range i thought is was - infinity to plus infinity?????
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2. There's a thread on this question somewhere, someone asked this yesterday.
4. Are you told what the equation for f(x) is, and are you given any other information in parts a and b?
5. (Original post by BrightGirl)

yeee
6. (Original post by dee321)
yeee
ahhh, i sat that paper! can't actually remember the answer to this question though, sorry
7. (Original post by ttoby)
Are you told what the equation for f(x) is, and are you given any other information in parts a and b?
here is the rest of the question
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8. (Original post by BrightGirl)
ahhh, i sat that paper! can't actually remember the answer to this question though, sorry
i know the awnser which is x> -k but dont know why?
9. (Original post by dee321)
here is the rest of the question
the range is y > -k , as it's just the y = ex graph shifted down by k (it's now asymptotic to y = -k instead of y = 0).

range is always y, the domain is always x.

edit: well in this case, you could put it as f(x) > -k (which is still basically the same as y > -k)
10. For exponential functions the range is f(x)>0 since they approach but never touch the line y=0 ( the asymptote).

However, in this question you should apply transformation. It's shifted vertically by 'K' units in the negative direction.

Therefore, the new range will be f(x)>-k
11. (Original post by FZka)
For exponential functions the range is f(x)>0 since they approach but never touch the line y=0 ( the asymptote).

However, in this question you should apply transformation. It's shifted vertically by 'K' units in the negative direction.

Therefore, the new range will be f(x)>-k
12. (Original post by BrightGirl)
the range is y > -k , as it's just the y = ex graph shifted down by k (it's now asymptotic to y = -k instead of y = 0).

range is always y, the domain is always x.

edit: well in this case, you could put it as f(x) > -k (which is still basically the same as y > -k)
13. (Original post by dee321)
No worries. Good luck.
14. (Original post by FZka)
For exponential functions the range is f(x)>0 since they approach but never touch the line y=0 ( the asymptote).

However, in this question you should apply transformation. It's shifted vertically by 'K' units in the negative direction.

Therefore, the new range will be f(x)>-k
but what about the y values below -k as -8 will have an y value less than -k is taht so??
15. (Original post by dee321)
but what about the y values below -k as -8 will have an y value less than -k is taht so??
The range of a function is the set of values which y can take.

In this question we've found out that the range is f(x)>-k i.e. the function cannot take y values below -k.

-k is the new horizontal asymptote instead of the line y=0 (the x-axis)
16. (Original post by FZka)
The range of a function is the set of values which y can take.

In this question we've found out that the range is f(x)>-k i.e. the function cannot take y values below -k.

-k is the new horizontal asymptote instead of the line y=0 (the x-axis)
okkkk fully understand now thanks

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