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A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)
y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1
eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1
Answer :A--> R,K
B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10
answer B--> I,≈W if it is written -1/2*x² (square has been forgotten)
C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J
D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L
E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!
y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.
y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]
Step-by-step explanation:
i) 
ii) 
iii) 
Answer:
y = 
Step-by-step explanation:
Let the equation of the line is,
y = mx + b
Here m = slope of the line
b = y-intercept
Slope of a line passing through two points 
and 
is,
m = 
From the graph attached,
Since, the given line passes through (0, -1) and (3, 4),
Slope 'm' = 
m = 
y - intercept 'b' = -1
Therefore, equation of the line will be,
y =
