can some1 plz explain jun 2007 p4 q 2f and 4 c-e 
take 2 sets of c-ordinates (2,1) (3,3)
(1 0) * (2 3) = (2 3)
(-1 1) (1 3) (-1 0)
do the same with the other co-ordinate
(2,1) --> (2,-1)
(3,3) --> (3,0)
(5,1) --> (5,-4)
ii)
https://studentforums.biz/index.php/topic,5575.msg167872.html#msg167872 check that..
if u see carefully, its the matrix which represents..shear of shear factor = -1 and y invariant.
iii) inverse of the transformation matrix which mapped ABC to A3B3C3
that is inverse of (1 0) = (1 0)
(-1 1) (1 1)
4c) this means that find two values for 'y' for which there are 2 values of x for each of them
for this u have to use the graph, check the diagram..
put straight lines where u think the line will intersect the graph twice..
(red lines)
ull notice, that wen y = 3...x has 2 values..
also, wen y = -9, x has two values
4e) f(x) + x -1 =0
as f(x)=y
y+x-1 = 0
y = 1-x
draw the line y = 1-x in the graph
eii) it cuts the graph at 3 points...meaning it has 3 solutions
