Maths AQA C1 revision - 14th January 2013

Hi mate, I started it I missclicked and put negative even though its correct and when you do everyone follows lmao. yeah sorry about that. your right by the way it should be 4x^2

1a) Sub in the x value of 7 from (7,k). This proves that k is indeed -4.
1b) -10/6 or -5/3
1c) Can't remember exactly, think I got -5x+3y-21=0
1d) think I got either (12,-7) or (7,-12)

2a) -1.5
2b) 0>-1.5 so the height is decreasing
2c) Ended up as 6-2, 4 for second derivative

3a) root 18 = three root 2
3b) 2/7 i.e. 2 root 2 over 7 root 2 (this is the quick method rather than the convoluted method to get the same answer)
3c) I got 5-root6, but I don't believe it's right.

4a) (x-3)^2+2
4b) 2>0, always above x-axis - no solutions.
4bii) Vertex = (3,2)

Sketch crosses y at +11, vertex at (3,2)

Transformation:

[3]
[2]

5a) Remainder is 16 i.e. -5+21
p(3)=0

p(x)=(x-3)(x-3)(x+2)

Curve crosses x at -2 and kisses x-axis without crossing it at +3

6)a Gradient = 9
y-4=9(x-1)
y=9x-5
6b) Integrate it - 2x^5-2x^3+5x+c
Forgot to actually find c. Whoops.

7) A circle with C(-3,2) has equation x^2+y^2+6x-4y=12
a). x=0 and y=-2, y=6
b). R=5.
Pythagoras - root 34
c) Got wrong answer for this, I am sure. Real answer was apparently root 9 i.e. 3 from root (34-25)=3

8a(ii) show that 4k^2-20k+9>0
b = (2k+1) a = 2 c = (3k+1)
(2k+1)^2 - 4*2*(3k+1) simplifies to 4k^2-20k+9 which factorises to...
(2k-1)(2k-9)>0

1/2, 9/2 are critical values

k<1/2 and k>9/2

For the transformation question we had to transform the curve (x-3)^2+2
to the curve x^2 so i put

[-3]
[-2]

as the vector translation, because that would move the graph -3 in x and -2 in y bringing it back to the x^2 curve.

unless you do it the other way round for some reason? it was definitely translating (x-3)^2+2 to x^2 though, not the other way round. argh i'm unsure.

Theres a similar question in the attached file I attached have a look at it, its (3,2)

Hmm interesting, are you sure that question is asking to translate from the equation of the curve given to x^2 curve? and the not the other way round? i really dont get how translating it by that vector would get it BACK to x^2?

Aaah just check this lol soz 4 quality.

yep that question is asking about going from x^2 to the equation of the curve given, so in the as exam we did today it will be the opposite way round, going from the equation of the curve given to x^2. hence

[-3]
[-2]

technically a gcse level question

Transformation:

[3]
[2]

Transformation is -3<br />
-2

For 100% UMS:

75 - January 2009
75 - June 2009
75 - January 2010
75 - June 2010
70 - January 2011
75 - June 2011
73 - January 2012
72 - June 2012

The good news is that recently they've been under 75, gives leeway for silly mistakes!

Yes, that is why I hope the exams are difficult so the boundaries are lower. I have no trouble answering difficult questions, but I have trouble being 100% perfect. I always do better on hard exams than easy exams UMS wise.

Transformation is -3
-2

Did people draw this for the graph of x squared +6x+11 with vertex (3,2)

Um that's wrong?? If the equation you've given there is right, which I think it is, then the vertex is -3,2. To work out the vertex you can use completing the square or use -b/2a. You may get a mark or two for the y=11.

No the graph was y = (x-3)^2+2, therefore solving for x would give you x = 3, so the vertex would be (3,2). but i made the silly mistake of getting the wrong transformation it was actually: [-3]
[-2]

No the graph was y = (x-3)^2+2, therefore solving for x would give you x = 3, so the vertex would be (3,2). but i made the silly mistake of getting the wrong transformation it was actually: [-3]
[-2]

There was a transformation question for that graph!?!?!?!?!?!?