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# C3 Graph Transformation Problem - June '08 Watch

1. Okay, just did the June 08 paper and got 92% on it, would have achieved 99% without the stupid mistakes but I lost 1% (1 mark) on one question. I just had no idea how to do it, so any help would be grateful .

http://www.edexcel.com/migrationdocu...e_20080606.pdf

Question 3) c).

For R, I set f(x) = 0, and solved. However, I tried setting x = 0 to work out Q, but got two values for y, and then I had no idea of what to do for the value of P.
2. You can redress the line's equation as f(x)=2-(-(x+1))=3+x for x+1<0, so x<-1, and f(x)=2-(x+1)=1-x for x>-1

You should be able to work out the x value of P by noting that P is where the graph goes wonky, and seeing where the equation for f(x) goes wonky (answer is above too )
3. Sorry, I don't seem to understand what you mean ><
4. All I'm doing is removing the absolute value signs, and multiplying by -1 if necessary.

If I give you a number |z|, then |z|=z if z>0, but |z|=-z if z<0. What I did was find out where x+1 was less than 0, and where it was greater than 0, and rewrite the equation accordingly.
5. You shouldn't even need to re-write the equation without the modulus for part c).

For P you just need to see that f(x) is at a maximum, which must be where f(x) = 2. Therefore (x + 1) = 0, therefore x = -1. P(-1, 2)

For Q just set x = 0, therefore y = 2 - |1| = 1. P(0, 1)

For R just set y = 0. Therefore |x + 1| = 2. Therefore x = 1 or X = -3. Clearly R lies where x = 1, so P(1, 0).

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