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1. http://www.examsolutions.net/a-level...y/paper.php#Q6

how do i draw a graph of

i can drawn

but they never taught me how to draw a graph of ln something minus x how do i do it?
2. It will just be a few simple transformations (reflections, translations) so the graph will still look similar.
You may need to just re write it using latex because I can't really understand what you have put.
3. (Original post by timebent)
http://www.examsolutions.net/a-level...y/paper.php#Q6

how do i draw a graph of

i can drawn

but they never taught me how to draw a graph of ln something minus x how do i do it?
You realise ln1 is just 0 so in fact you're just drawing -x which is just a straight line through the origin with a negative gradient
4. (Original post by B_9710)
It will just be a few simple transformations (reflections, translations) so the graph will still look similar.
You may need to just re write it using latex because I can't really understand what you have put.
you want some brackets? i guess i should probably put them in, check back again after you get the notification
(Original post by target21859)
You realise ln1 is just 0 so in fact you're just drawing -x which is just a straight line through the origin with a negative gradient
yes but surely this is the transformation of
and thus i'd just move the graph right 1 unit?

but yea i want to reflect it in the y-axis if it's -x
5. (Original post by target21859)
You realise ln1 is just 0 so in fact you're just drawing -x which is just a straight line through the origin with a negative gradient
He (obviously) means .

@OP. Denote then is a reflection in the -axis of . Draw this.

Now which is a translation to the left by one unit of . Can you now draw this?
6. (Original post by timebent)
you want some brackets? i guess i should probably put them in, check back again after you get the notification

yes but surely this is the transformation of and thus i'd just move the graph right 1 unit?

but yea i want to reflect it in the y-axis if it's -x
This is just f(-x) as the a part is 0 because ln1=0
Edit: just saw Zacken's post. Didn't realise there were meant to be brackets so ignore my posts lol.
7. (Original post by Zacken)
He (obviously) means .

@OP. Denote then is a reflection in the -axis of . Draw this.

Now which is a translation to the left by one unit of . Can you now draw this?
so 1 unit to the left of the graph reflected in the y-axis?

(Original post by target21859)
This is just f(-x) as the a part is 0 because ln1=0
Edit: just saw Zacken's post. Didn't realise there were meant to be brackets.
8. (Original post by timebent)
so 1 unit to the left of the graph reflected in the y-axis?

Yes.
9. (Original post by Zacken)
Yes.
awesome

10. (Original post by timebent)
awesome

You work through it for yourself systematically first before opening the spoiler below, it's all very procedural.
Spoiler:
Show
Call . Draw it.

Now stretch this by a factor of in the direction parallel to the -axis (i.e: flatten it vertically by half). This gives you .

Now flip this in the -axis, this is . Draw it.

Now translate this upwards by units. This gives you
11. (Original post by Zacken)
You work through it for yourself systematically first before opening the spoiler below, it's all very procedural.
Spoiler:
Show
Call . Draw it.

Now stretch this by a factor of in the direction parallel to the -axis (i.e: flatten it vertically by half). This gives you .

Now flip this in the -axis, this is . Draw it.

Now translate this upwards by units. This gives you
i drew the graph like this.

y=-0.5e^x y=2-0.5e^x

i wasn't sure whether it went above the x-axis or not :/
Attached Images

12. (Original post by timebent)
i drew the graph like this.

i wasn't sure whether it went above the x-axis or not :/
1. Your attempts to use LaTeX aren't helping, could you please stick to/at least provide ascii typed mathematical expressions of the form y = e^x, y = (1/2) e^x, etc... so that we may understand what you are saying.

2. You should know basic properties of . It has an asymptote at the -axis, so . The first transformation, halving the function "changes" the asymptote to . The second transformation, reflecting the curve "changes the asymptote to . The third transformation shifts the asymptote 2 units upwards, so changes it to . So your graph is correct, indeed it is asymptotic to the line as and then diverges off to negative infinity as . Furthermore, it has a y-intercept of and an -intercept at .
13. (Original post by Zacken)
1. Your attempts to use LaTeX aren't helping, could you please stick to/at least provide ascii typed mathematical expressions of the form y = e^x, y = (1/2) e^x, etc... so that we may understand what you are saying.

2. You should know basic properties of . It has an asymptote at the -axis, so . The first transformation, halving the function "changes" the asymptote to . The second transformation, reflecting the curve "changes the asymptote to . The third transformation shifts the asymptote 2 units upwards, so changes it to . So your graph is correct, indeed it is asymptotic to the line as and then diverges off to negative infinity as . Furthermore, it has a y-intercept of and an -intercept at .
So y-intercept means x=0

so if i stick x=0 into e^x becomes 1
1x0.5=0.5
so 2-0.5=1.5 which isn't 1 or did i go wrong somewhere?

and it touches the x axis at y=0
so when y=0

where did it all go wrong?
14. (Original post by timebent)
So y-intercept means x=0

so if i stick x=0 into e^x becomes 1
1x0.5=0.5
so 2-0.5=1.5 which isn't 1 or did i go wrong somewhere?

and it touches the x axis at y=0
so when y=0

where did it all go wrong?
Yeah, that's all fine. I thought the function was whilst writing my reply up, hence the discrepancy.
15. (Original post by Zacken)
Yeah, that's all fine. I thought the function was whilst writing my reply up, hence the discrepancy.
ah ok It's all good

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