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# C3 Questionnnnnn Helppppppppppppppp watch

1. how would you draw ln(3-x) - would you reflect ln(x) in the y-axis and then move it to the left by 3 or would you work out separetly where it cuts the x and y axis and then draw it
2. I'd go about it by getting the general graph in my head by doing what you said first, it's reflected in Y, then moved 3 to the RIGHT (asymptote at x=3). Then work out where it crosses the axis so you can draw it well
3. You can do either tbh.
4. (Original post by sheligalo)
I'd go about it by getting the general graph in my head by doing what you said first, it's reflected in Y, then moved 3 to the RIGHT (asymptote at x=3). Then work out where it crosses the axis so you can draw it well
why would you move it to the RIGHT? shouldn't it be to the LEFT
5. what paper is this????
6. (Original post by Remarqable M)
what paper is this????
its from my edexcel maths book
7. You definitely move it to the left along the x-axis, by -3.
new asymtope will b at x=-3.
it will cross y-axis at (0,ln3)
8. after reflection it moves 3 to the right
9. (Original post by infinity123321)
after reflection it moves 3 to the right
how how can it move to the right? the equation is f(x)=ln(-x+3)
10. your sketch has to be wrong
11. i used my graphical calculator and thats what it showed
12. (Original post by morebritishthanindian)
your sketch has to be wrong
oooh i seee what i got wrong anyways here is new sketch.
13. shouldnt it be :
Attached Images

14. The person who sketched their graph above (Now they deleted it) is wrong. lnx is a reflection of y=e^x in y=x not y=0.

So the graph would be a translation of 3 to the left and then a reflection in x y axis.
15. It is the graph of y=ln(-x), but shifted along the x axis by +3.

Here is a diagram, but where the graph just seems to stop at y=3, that's where the curve is tending to -infinity.
http://tinyurl.com/yfhlwtb
16. (Original post by Remarqable M)
oooh i seee what i got wrong anyways here is new sketch.
thats right sorryi did e^x
17. (Original post by JTeighty)
The person who sketched their graph above (Now they deleted it) is wrong. lnx is a reflection of y=e^x in y=x not y=0.

So the graph would be a translation of 3 to the left and then a reflection in x y axis.
i just realised it now! thank goodness i came on here
18. (Original post by Remarqable M)
oooh i seee what i got wrong anyways here is new sketch.
Correct
19. (Original post by JTeighty)
Correct
20. (Original post by Remarqable M)
Draw curve of y= 2^x (exponential curve passing through (0,1)

Then reflect this in the y axis.
Then reflect this in the x axis.

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