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

    how do i draw a graph of

    ln(1-x)?

    i can drawn ln(1+x)\ or\ ln(x-1)\ or\ ln(2x)\ or\ ln(x^2)

    but they never taught me how to draw a graph of ln something minus x how do i do it?
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    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.
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    (Original post by timebent)
    http://www.examsolutions.net/a-level...y/paper.php#Q6

    how do i draw a graph of

    ln1-x?

    i can drawn ln1+x\ or\ lnx-1\ or\ ln2x\ or\ lnx^2

    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
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    (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
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    (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 \ln (1-x).

    @OP. Denote f(x) = \ln x then g(x) = f(-x) = \ln (-x) is a reflection in the y-axis of f(x). Draw this.

    Now h(x) = g(x+1) = \ln(-x +1) = \ln(1-x) which is a translation to the left by one unit of g(x). Can you now draw this?
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    (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.
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    (Original post by Zacken)
    He (obviously) means \ln (1-x).

    @OP. Denote f(x) = \ln x then g(x) = f(-x) = \ln (-x) is a reflection in the y-axis of f(x). Draw this.

    Now h(x) = g(x+1) = \ln(-x +1) = \ln(1-x) which is a translation to the left by one unit of g(x). Can you now draw this?
    so 1 unit to the left of the graph y=\ln x 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.
    sorry about that
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    (Original post by timebent)
    so 1 unit to the left of the graph y=\ln x reflected in the y-axis?

    Yes.
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    (Original post by Zacken)
    Yes.
    awesome

    so what about the flipside

    y=2-\dfrac{1}{2} e^x

    what about one of these?
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    (Original post by timebent)
    awesome

    so what about the flipside

    y=2-\dfrac{1}{2} e^x

    what about one of these?
    You work through it for yourself systematically first before opening the spoiler below, it's all very procedural.
    Spoiler:
    Show
    Call f(x) = e^x. Draw it.

    Now stretch this by a factor of \frac{1}{2} in the direction parallel to the y-axis (i.e: flatten it vertically by half). This gives you g(x) = \frac{1}{2}f(x) = \frac{1}{2}e^x.

    Now flip this in the x-axis, this is h(x) = -g(x) = -\frac{1}{2}e^x. Draw it.

    Now translate this upwards by 2 units. This gives you m(x) = 2 + h(x) = 2 + -\frac{1}{2}e^x = 2 - \frac{1}{2}e^x
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    (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 f(x) = e^x. Draw it.

    Now stretch this by a factor of \frac{1}{2} in the direction parallel to the y-axis (i.e: flatten it vertically by half). This gives you g(x) = \frac{1}{2}f(x) = \frac{1}{2}e^x.

    Now flip this in the x-axis, this is h(x) = -g(x) = -\frac{1}{2}e^x. Draw it.

    Now translate this upwards by 2 units. This gives you m(x) = 2 + h(x) = 2 + -\frac{1}{2}e^x = 2 - \frac{1}{2}e^x
    i drew the graph like this.
    y=e^x\ y=\frac{1}{2}e^x
    y=-0.5e^x y=2-0.5e^x


    i wasn't sure whether it went above the x-axis or not :/
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    (Original post by timebent)
    i drew the graph like this.
    y=e^x\ y=\frac{1}{2}e^x

y=-\frac{1}{2}e^x\ y=2-\frac{1}{2}e^x

    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 \exp. It has an asymptote at the x-axis, so y=0. The first transformation, halving the function "changes" the asymptote to y = \frac{0}{2} = 0. The second transformation, reflecting the curve "changes the asymptote to y = -\frac{0}{2} = 0. The third transformation shifts the asymptote 2 units upwards, so changes it to y = -\frac{0}{2} + 2 = 2. So your graph is correct, indeed it is asymptotic to the line y = 2 as x \to -\infty and then diverges off to negative infinity as x \to \infty. Furthermore, it has a y-intercept of (0, 1) and an x-intercept at (\ln 2, 0).
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    (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 \exp. It has an asymptote at the x-axis, so y=0. The first transformation, halving the function "changes" the asymptote to y = \frac{0}{2} = 0. The second transformation, reflecting the curve "changes the asymptote to y = -\frac{0}{2} = 0. The third transformation shifts the asymptote 2 units upwards, so changes it to y = -\frac{0}{2} + 2 = 2. So your graph is correct, indeed it is asymptotic to the line y = 2 as x \to -\infty and then diverges off to negative infinity as x \to \infty. Furthermore, it has a y-intercept of (0, 1) and an x-intercept at (\ln 2, 0).
    So y-intercept means x=0

    so if i stick x=0 into y=2-\frac{1}{2} e^x 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
    0=2-\frac{1}{2} e^x

    2=\frac{1}{2} e^x

    4=e^x

    ln4=x

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

    so if i stick x=0 into y=2-\frac{1}{2} e^x 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
    0=2-\frac{1}{2} e^x

    2=\frac{1}{2} e^x

    4=e^x

    ln4=x

    where did it all go wrong?
    Yeah, that's all fine. I thought the function was 2 - e^x whilst writing my reply up, hence the discrepancy.
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    (Original post by Zacken)
    Yeah, that's all fine. I thought the function was 2 - e^x whilst writing my reply up, hence the discrepancy.
    ah ok It's all good
 
 
 
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