# Today P2 Identity question

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#21

(Original post by

Can anyone show me what the graph of ln|3x-6| looks like?

**imasillynarb**)Can anyone show me what the graph of ln|3x-6| looks like?

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#22

(Original post by

If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie

**crana**)If you can imagine it this is how I drew it:

Asymptote at x = 2

line coming up from the asymptote on the right crossing the x axis at 7/3 (i think) then the curve slowed down but y continued to increase with x

then this was sort of reflected in the asymptote crossing x axis at 5/3 and y axis at ln 6

rosie

coz dy/dx = 3/(3x-6) = 0 hence x is 2

but i couldnt get a corresponding y value ... ill upload an image from my graphic calc

it was wierd in a way ...

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#23

(Original post by

this is how i drew it:

**mockel**)this is how i drew it:

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#24

(Original post by

i got a dip at x=2

coz dy/dx = 3/(3x-6) = 0 hence x is 2

but i couldnt get a corresponding y value ... ill upload an image from my graphic calc

it was wierd in a way ...

**infekt**)i got a dip at x=2

coz dy/dx = 3/(3x-6) = 0 hence x is 2

but i couldnt get a corresponding y value ... ill upload an image from my graphic calc

it was wierd in a way ...

Im no great shakes at maths but if there is no "turning point" as such then the dy/dx = 0 could be that at x =2 the gradient *is* 0, because there is no graph there?

If that makes sense.

I dont have a graphics calculator but I am fairly sure that it is undefined ie asymptote at x=2

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#26

(Original post by

hey guys, was ques 5 the binomials right with [x^3-1/2x]^12? and then 6 was that iteration stuff (zzz) and 7 was that trap rule stuff?

you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?

**BossLady**)hey guys, was ques 5 the binomials right with [x^3-1/2x]^12? and then 6 was that iteration stuff (zzz) and 7 was that trap rule stuff?

you know for the iteration, how did u show it was 1.70 or whatever, did u just choose 1.65 and 1.75, stick it in and show sign change ya/?

For the "independent of x" term in the binomial question, it's possible without expanding the whole expression. Since the first term has postive index (x^3) and the second term has negative index (-0.5x^-1), you need to find the stage in the binomial expansion that makes both x terms have the same index number, and then they will cancel each other, leaving only a number and no x-term. So you need the term when (-0.5x^-1) is to the power of 9, and therefore (x^3) will be to power of 3. (Total powers must equal 12, because that is n.)

So you then get 12C9 . (x^3)^3 . (-0.5x^-1)^9

the x terms cancel out and you're left with a fraction which cancels down to

-55/128

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#27

(Original post by

Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

**satin**)Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

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#28

Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

spot on dude!

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#29

btw, what was the one for.. I think it was the iteration one and what value of x0 is this not valid for?

I think I put as it was something like root (blah blah /x+1) that it was when x=-1 so it is over 0 and hence undefinined..

I think I put as it was something like root (blah blah /x+1) that it was when x=-1 so it is over 0 and hence undefinined..

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#30

(Original post by

spot on my friend ... thts wat i did ... but i made the curve come very close to x=2 so tht it seemed tht they were joined

**infekt**)spot on my friend ... thts wat i did ... but i made the curve come very close to x=2 so tht it seemed tht they were joined

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#31

hey ppl, u no wat sum1 jus told me?? dat last nyt a copy of da p2 edexcel got stolen+sum yoke put it online sumwer askin 4 help wif ans bt den da site ppl contacted edexcel! so wat y'all tink will happen?? c if we afta do it again....

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could the graph go below the x axis??? i thought it would reflect in x=2 but i tohught that it wouldnt cross the axis

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#33

(Original post by

yeah i did that in the exam, but my mouse control isn't very good

**mockel**)yeah i did that in the exam, but my mouse control isn't very good

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#34

(Original post by

could the graph go below the x axis??? i thought it would reflect in x=2 but i tohught that it wouldnt cross the axis

**lexazver203**)could the graph go below the x axis??? i thought it would reflect in x=2 but i tohught that it wouldnt cross the axis

for example: x = 2.01

y = ln |3x - 6| = ln (0.03) = -3.506..

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#35

(Original post by

What I mean is, anyone have some sort of graph sketching program and can like save the picture and post it for me?

**imasillynarb**)What I mean is, anyone have some sort of graph sketching program and can like save the picture and post it for me?

http://www.coolmath.com/graphit/

however it does seem to get a tad confused with modulus's, you can't really put them in.

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#36

**satin**)

Thats the right method for iteration, but because it said "correct to 2dp" I think you had to use 1.695 and 1.705, can anyone confirm?

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