Maths AQA C1 revision - 14th January 2013

I got that one too! What did people do for 6b? Was it integration as I didn't know what to do.

You're probably right. I didn't know how to work out the vertex. I should've revised more. How did you work out the vertex?

I'm sure the answer is 3 to the right and 2 up. After you have completed the square, the vertex is (3,2). The vertx of the y=x^2 is (0,0) therefore the graph has been translated by the vector [3,2]

Should it be positive 3 and positive 2?

From what ive heard it was an easy paper in general and i dont believe i dropped any marks. They cannot however make the A boundary above 67/75 due to the gap they must leave for students getting above 90% aiming for A* in maths and the grade boundaries are normally 6 or 7 marks away from eachother so im assuming

62-64 for an A
55-57 for a B
47-49 for a C
Etc

The question asked for the geometric transformation of y = (x-3)^2 + 2 onto y = x^2

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Should it be positive 3 and positive 2?

I got something like<br />
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2x^5 + 2x^3 + something I can't remember -1<br />
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You had to integrate to find the original equation of the curve and then subsititue the coordinates of P to find out the value of C

I think I read the question wrong now<img src="images/smilies/angry.gif" border="0" alt="" title="" smilieid="12" class="inlineimg" />

The completed square of the equation was (x-3)^2 + 2

Therefore to make the bracket = 0, you have to put a positive 3 in so 3 is the x value and the 2 outside the bracket is the y value of the vertex.

Ah nvm, thats a mark ive lost. However i found this paper fairly decent, anyone else get the c on the integration question to be -1?

no, as it was asking for the translation from the graph to y=x2, so it was -3, -2, dw, i think a lot of people got that one wrong, its confusing as they normally ask for it the other way round

How many marks was the translation one?

Wasn't it (x-3)^2 - 9 + 11 or something like that?

How many marks was the translation one?

-9+11 = 2