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[Jim Groovester]

Okay, so the y-int is 0, easy enough. X-int is when y=0, so 2x^3=12x^2. I've only come up with 0 and 6 as possible values of x for that, if I had batteries for my gfx calculator I'd be more certain.

That's not how you're supposed to do that. Do this instead:

0 = 12x² - 2x³
0 = x²(12 - 2x)
x = 0 (mulitplicity two), x = 6

You still found the right intercepts, but you'll lose solutions if you're not careful when you set x terms equal to each other and do something like dividing out the xs. (The reason this is bad is because you could possibly be dividing by zero.) If you set one side of an equation equal to zero and then factor, you won't have that problem.

Now, as to the rest of your problem....

when 6x^2=24x

There you go doing that again. Avoid doing that in the future.

My question is thus: With my x-intercepts at x=0 and x=6, does it work to have a turning point between those points but *not* in the centre (i.e. x=3), or have I done something wrong?

That's perfectly fine. There's no issue if your math checks out, which I think it does.

Question for other people who know aught of cack-yool-uss:

Let's say we have a graph that's formula is (1)/(X-2). The graph is discontinuous at X=2, because you divide by zero. Why is the limit as X=>2 'no limit' instead of 'limit is undefined'? What is a limit?

As x -> 2⁺ (which is to say, the limit from the positive side), the limit approaches positive infinity.
As x -> 2^- (which is to say, the limit from the negative side), the limit approaches negative infinity.

Because the left hand limit and the right hand limit do not match, the limit does not exist.

If the left hand limit and the right hand limit were both the same when, for example, both sides went to positive infinity, then the limit would exist and it would said to be undefined.

[Vector]

Because the left hand limit and the right hand limit do not match, the limit does not exist.

If the left hand limit and the right hand limit were both the same when, for example, both sides went to positive infinity, then the limit would exist and it would said to be undefined.

Crap, I was wrong on this one.  Thanks for the correction.

Notation will be the death of me... I automatically assumed it was approaching from the negative side.

[Jim Groovester]

But as x -> 2^- the function approaches negative infinity.

VECTOR CORRECTION COMBO x2!

I am the man.

[Vector]

Meh.  It's all the same to me by now, dude =/

This is why I left engineering.  "Negative infinity?  Positive infinity?  Same difference, so long as it converges!*"

*For certain definitions of convergence.

But yeah, good catches!  I can tell I need to go review calculus again, since it's gotten mighty rusty.

[Blargityblarg]

My question is thus: With my x-intercepts at x=0 and x=6, does it work to have a turning point between those points but *not* in the centre (i.e. x=3), or have I done something wrong?

That's perfectly fine. There's no issue if your math checks out, which I think it does.

Righto. I was confused, 'cause with parabolas the turning point is always the mean of the x-intercepts. Cheers.

[Heron TSG]

Thank you, comrades.

[eerr]

Question for other people who know aught of cack-yool-uss:

Let's say we have a graph that's formula is (1)/(X-2). The graph is discontinuous at X=2, because you divide by zero. Why is the limit as X=>2 'no limit' instead of 'limit is undefined'? What is a limit?

The place the line heads to as it runs away.

for 1/x-2, it heads to infinity in  one direction, and negative infinity in the other direction.

Limit as x approaches 2 from the right, positive, is positive infinity.
Limit as x approaches 2 from the left, negative, is negative infinity.

limit as x approaches 2, is undefined because lim+ and lim- don't meet.

[Heron TSG]

I 'know' it should be undefined, I just don't know why that would be marked wrong, with 'no limit' as the correct answer.

[Jim Groovester]

That's because 'undefined' and 'no limit' have two distinct meanings. It's incorrect to use them interchangeably.

Undefined => limit approaches infinity
no limit => left hand limit and right hand limit do not match, meaning the limit does not exist

[ed boy]

These have been causing me great trouble:

Spoiler: first problem (click to show/hide)

Prove that for n lines in a plane, no two of which a parallel and no three of which meet in a point, that the number of areas the plane is divided into is:
(n²+n+2)/2

Spoiler: second problem (click to show/hide)

let p=a^b-1. If p is prime, prove that a=2 and b is a prime number (except for the case of a=3 and b=1)

[Virex]

For the first one, are there only situations in which 2 lines cross, or is just m = 3 excluded, but m < 3 not?


For the second one, are you supposed to prove that for real numbers or for integers?

[ed boy]

for the first, the only time lines intersect is when exactly two lines intersect (that is, at any given point there will be either no lines, one line, or two lines intersecting).

for the second, it is for positive integers.

[Vector]

for the second, it is for positive integers.

You might be having trouble with it because the statement is false.

[ed boy]

That would explain a lot. Thanks.

[Vector]

That would explain a lot. Thanks.

Hey, no problem.

I remember having significant difficulty with the first problem when I encountered it, but I think a little induction should do the trick.