Hi! I was just wondering if some one could be nice enough and help me solve a maths question?
Paper 4, Oct/Nov 2004, Question 7. I'm realllly bad at this topic...I sloved the first part....but got stuck after that. 
Help, Please?
Nov 2004 Q7)a)i)
1x(4)^2+(-2)(4)+(-3)
= 5
ii)Solve the Equation (x^2-2x-3)
Answer = (x-3)(x+1)
X=3 or X=-1
So cordinates for K (-1,0) and for L it is (3,0)
III) -b divided by 2a
-(-2)/(2x1)
X=1 Subsitite X in the Equation and find Y = -4
Hence Cordinates is (1,-4)
B)it will become upside down simple enough
ii)if there is no Y-intercept then the GRAPH SHOULD PASS THROUGH THE ORIGIN
C)C=0
II)(3,0) and (4,8) Equation is Y=ax^2+Bx
0=A(3^2)+3B
8=A(4^2)+4b THEN SOLVE SITMIELOUSLY
A=2
B=-6
