ce 07 questions x2

The y-intercept of the graph is -2Let f(x)=ax2+bx-2(-1,0) and (4,0) are two point lie on f(x)Sub (-1,0) into f(x)0=a(-1)2-b-20=a-b-2…(1) sub(4,0) into f(x)0=a(4)2+4b-20=16a+4b-20=8a+2b-1 …(2)Sub(1x2)+(2)0=10a-5a=1/2 sub a=1/2 into (1)b=-3/2∴f(x)= (1/2)x2-(3/2)x-2=1/2(x2-3x-4)=1/2(x+1)(x-4)

13) x : y=1:2 x/y=1/2x=y/2 ∵y : z=3 : 1∴x : y : z=(3/2): 3 : 1=>3 : 6 : 2∴(x+y): (y+z) = (3+6) : (6+2) = 9:8

7. since the x-intercepts are -1 and 4, f(x) must be in the form
f(x) = k (x-4)(x+1) by factor theorem
Put f(0) = -2, k = 1/2 => Ans: A