Saxandthecity
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Hey guys, I was just going over the June 2014 paper again and I don't really understand the mark scheme for question 7iii) (http://static1.squarespace.com/stati...+June+2014.pdf).

The mark scheme is found here:

http://static1.squarespace.com/stati...ark+scheme.pdf

The mark scheme mentions 'end-points' that exist in the 1st and 3rd quadrant at (-2, -3) and (2,1), but I was under the impression that the 1st and 3rd quadrants were the top right and the bottom left quadrants respectively. Also, with a negative quadratic, isn't the 'end-point' at an infinite magnitude y-value? Thus, I don't understand how they can say that end points exist at (-2,-3) and (2,1)

Any help is greatly appreciated!
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Hasufel
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haven`t looked at the shceme but....don`t forget, for your t values: for t = -Pi/2, x=-2, y= -3 and for t= Pi/2, x=2, y=1 are the end points (think of this as only PART of the quadratic, since the restricted t values also restrict the x and y values. Think of this as like a "piecewise" curve which stops at the given points.

(if you even wished to take say the t range as -100Pi to 100Pi, all you would be doing is travelling back and forth along the quadratic from left to right from the indicated points (the (x,y) ones) since the trig functions involved are periodic and certain multiples would give the same x and y values.
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