# Thread: Cambridge Prelim MX1 Textbook Marathon/Q&A

1. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by DatAtarLyfe
Can't you just let x->infinity, making the y values ->x^2, instead of doing all the algebra above
You can

I was taught this way though ~

(I can do all the algebra in my head but I posted it for OP's sake)

2. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Crisium
You can

I was taught this way though ~

(I can do all the algebra in my head but I posted it for OP's sake)
cool, just confirming

3. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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?

4. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
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?
It's because your are trying to get rid of the pronumeral in the denominator before you let x-> infinity

5. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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.....

6. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Yes, as x tends to $\pm{\infty}$, y values will tend to $x^2+3x+3$. 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 $\lim_{x \to\infty} \dfrac{(x+1)^3}{x} = x^2+3x+3$ 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 $x^2+3x+3$ which becomes our oblique (parabolic) asymptote.

7. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
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.....
In other words, a 'parabolic asymptote'.

8. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
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.....
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

9. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

The graph of y=(x+1)^3/x in red

And it's parabolic asymptote in black.

10. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I never learnt how to do Pascals triangle, will I ever need it in 2U and or 3U in Year 12?

11. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Speed6
I never learnt how to do Pascals triangle, will I ever need it in 2U and or 3U in Year 12?
3U

12. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Drongoski
3U
Ok thanks Drongoski, one last question, is this something which I can learn independently or will I need a teacher to guide me through it? Also, is it something which can be learnt in one day?

13. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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.

14. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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.

15. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How do you do question 3 a in exercise 7D?

16. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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.

17. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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?

18. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
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?
$For large x, y\approx x^\frac{1}{2}, which would not give a parabolic shape, because this is a square root function (\emph{not} x^2 or a quadratic like that; the exponent is \frac{1}{2}, not 2). The square root function `flattens out' as x\to +\infty, which is why the curve also flattens out as x\to +\infty (we essentially have the graph of y=\sqrt{x} being an asymptote as x\to+\infty).$

19. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How do you find the focus length of the parabola x^2 = 28/5y

20. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
How do you find the focus length of the parabola x^2 = 28/5y
I take it to mean: x^2 = (28/5)y

In that case simply express in the form: x2 = 4ay

For this question: x^2 = 4* (7/5)*y

so, the focal length(not focus length) "a" is 7/5.

21. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Drongoski
I take it to mean: x^2 = (28/5)y

In that case simply express in the form: x2 = 4ay

For this question: x^2 = 4* (7/5)*y

so, the focal length "a" is 7/5.
Lol, i was right but deleted my post cause i was like "wait wtf, it's 28/5y"

22. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

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?

23. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by DatAtarLyfe
Lol, i was right but deleted my post cause i was like "wait wtf, it's 28/5y"
sorry about not making it clear

24. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For $x^2=\dfrac{28}{5}y$, vertex is $(0,0)$, focus is $(0, \dfrac{7}{5})$, hence focal length is $\dfrac{7}{5}$.

25. ## Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
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?
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

Page 5 of 61 First ... 345671555 ... Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•