Why is it the highest power of x in the denominator , i thought it was just the highest power in either the numerator or denominator?
For 5)i
y = (x+1)^3/ x
i know there is a vertical asymptote of x = 0
But is there an horizontal asymptote.
I expanded the numerator and did lim x ---> infinity
and get x^2 + 3x + 3.....
Yes, as x tends to , y values will tend to . Note that what we are trying to find is a curve which 'bounds' the graph of f. This is only just for graphical purposes. It is not correct to say since the expression on the LHS still means that y tends on infinity as you increase x. Graphically, however, it will be 'bounded' off by which becomes our oblique (parabolic) asymptote.
Last edited by VBN2470; 8 Jul 2015 at 6:54 PM.
Just keep in mind these general rules when your trying to identify what kind of asymptotes exist:
1/ If the highest power of x in the denominator = the highest power of x in the numerator, then you have a horizontal asymptote
2/ If the highest power of x in the denominator < the highest power of x in the numerator (BY ONE), then you have an oblique asymptote
3/ If the highest power of x in the denominator < the highest power of x in the numerator (BY TWO OR MORE), then you have another graph as an asymptote
So in your case, the highest power in numerator was 3 and highest power in denominator was 1, thus you have a parabolic asymptote
I never learnt how to do Pascals triangle, will I ever need it in 2U and or 3U in Year 12?
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The Pascal Triangle itself is dead easy; nothing to it. But if you are talking about the many interesting properties associated with it, that requires a bit more algebra.
Last edited by Drongoski; 8 Jul 2015 at 11:33 PM.
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Pascal's triangle is more of a reference point to begin the binomial theorem topic. It's not a major part in the syllabus.
1 (this is called the '0th' row)
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
etc.
For the binomial theorem, what's interesting is that the expansion of (1+x)^n, the coefficients match up the nth row of Pascal's triangle. Also, Pascal's triangle can be written in combinations. It might also be worth mentioning that the sum of the coefficients on each row is 2^n. Besides that you don't really need much more for MX1.
How do you do question 3 a in exercise 7D?
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so the question is y = x^2 1/2
so make it an improper fraction
y = x^5/2
then it simply:
y' = 5/2 x^ 3/2 which = 5/2 x^ 1 1/2
you bring down the 5/2 and minus 1 from the power.
for question 6a from 10F
y = x^(1/2) + 1/x^(1/2)
i don't understand why the curve flattens out as x ---> + infinity?
why is it not a parabola shape?
How do you find the focus length of the parabola x^2 = 28/5y
Last edited by Drongoski; 9 Jul 2015 at 2:56 PM.
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i though the focus length was the distance from the vertex and the focus.
I got the focus was (0,7/5)
and the vertex ( 14/5, 7/5)
so I got the focus length = 14/5
Where did i go wrong?
For , vertex is , focus is , hence focal length is .
Your vertex is incorrect. If you look at your original equation, x^2=28/5y, your vertex is actually (0,0)
To determine the vertex from your equation, you use the standard equation (x-h)^2=4a(y-k), where (h,k) is your vertex. So in your equation, h=0, k=0 and 4a=28/5
Hence your vertex is (0,0), your focal length is 7/5 and thus your focus is (0,7/5)
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