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MATHS study thread. (1 Viewer)

YBK

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Horizontal asymptote is at 0

i think that's because bottom power is one higher than the top and has coefficient 1

so it's 0/1 = 0
 

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Horizontal asymptote is at 0

From the limit as x -> infinity, the curve approaches 0 as it goes to infinity, yes, but the curve is actually 0 when x is -1 or -3, so it's not an asymptote per se. Unless you want to start debating about the definition of "asymptote" (stupid maths teacher...)


I_F
 

YBK

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Riviet said:
Haha same here, didn't have time to find the pof's, we only had 45 mins and this question was 1 of 2... the other was a hard parametrics question.
u could simply sketch f'x = 0
 

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He could, but that usually means finding f''(x) anyway to sketch f'(x)


I_F
 

YBK

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When above x-axis then gradient is +ve and below gradient is -ve at that particular point.
 

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Haha nah, i wouldn't have had enough time for it but thx for the idea.
 

kido_1

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who is doing integration of implicit and parametric functions?
 
P

pLuvia

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kido_1 said:
who is doing integration of implicit and parametric functions?
Me neither, having even touched on parametrics

Slow down kido :p
 

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For anyone that's interested:

Find the gradient of the tangent to the curve y=x24-98x13 at x=2

[Yes i know the numbers are a little exaggerated. :D]
 

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Riviet said:
For anyone that's interested:

Find the gradient of the tangent to the curve y=x24-98x13 at x=2

[Yes i know the numbers are a little exaggerated. :D]
haha

dy/dx=24x23-98*13x12

replace x=2 into derrivative ...

and ... tada!!!

:D
 

YBK

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here, i have a harder one (harder... not too hard though :D)... from our end of yr MX 1 assessment

The equation x^3 + kx + 2 = 0 has 2 equal roots. Find all three roots and the value of k.
 

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If no one wants to do it, then i will. :D

I have little experience with this type of question, but i think i've got it.

Let the equal root be "a" and the other root be "b".

Then (x-a)2(x-b) = x3+kx+2

Expanding and then factorising LHS into the form px3+qx2+rx+s we obtain the equality:

x3-(b+2a)x2+(2ab+a2)x-a2b = x3+0x2+kx+2

Equating coefficients of q, r, and s (x3's cancel), we obtain 3 equations:

-(b+2a) = 0 *

2ab+a2 = k **

-a2b = 2 ***

From *,
b = -2a
sub -> ***
=> 2a3 = 2
a3 = 1
a = 1

.: b = -2

.: k = 2ab+a2
= -4+1
= -3

Therfore, the equal root is 1, the other root is -2, and k = -3
 
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YBK

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Riviet said:
If no one wants to do it, then i will. :D

I have little experience with this type of question, but i think i've got it.

Let the equal root be "a" and the other root be "b".

Then (x-a)2(x-b) = x3+kx+2

Expanding and then factorising LHS into the form px3+qx2+rx+s we obtain the equality:

x3-(b+2a)x2+(2ab+a2)x-a2b = x3+0x2+kx+2

Equating coefficients of q, r, and s (x3's cancel), we obtain 3 equations:

-(b+2a) = 0 *

2ab+a2 = k **

-a2b = 2 ***

From *,
b = -2a
sub -> ***
=> 2a3 = 2
a3 = 1
a = 1

.: b = -2

.: k = 2ab+a2
= -4+1
= -3

Therfore, the equal root is 1, the other root is -2, and k = -3
Correct! :) btw, in my exam, i did it in a different way to yours but it was still right.

What I did was find the sum of roots, and product of roots. Made one of them an equation in alpha and then substituted in the other...

I think it's shorter than your method since there really is no need to expand and factorise... :)
 

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what do the /'s mean?

Integral sign. By the way, the Windows Character Map (Start Menu, All Programs, Accessories, System Tools) has a nice integral sign - ∫ :)


I_F
 

YBK

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insert-username said:
what do the /'s mean?

Integral sign. By the way, the Windows Character Map (Start Menu, All Programs, Accessories, System Tools) has a nice integral sign - ∫ :)


I_F
ahh, cool thanks.

We haven't done integration yet . :D
 

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if u expand it....

∫x(x^2+1)^3 dx

=∫x(x^6+3x^4+3x^2+1)dx

=∫(x^7+3x^5+3x^3+x)dx

=1/8(x^8)+3/6(x^6)+3/4(x^4)+1/2(x^2)+c

=1/8(x^8)+1/2(x^6)+3/4(x^4)+1/2(x^2)+c

i think there is a quicker and more easier method...but this is one way =P
 

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