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secondary 3 | E Maths
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Date Posted: 1 year ago
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(a) y = x^2 + 2x + k
sub in x =-3 and y=0
0 = (-3)^2 + 2(-3) + k
0 = 9 - 6 + k
k = 3
sub in x =-3 and y=0
0 = (-3)^2 + 2(-3) + k
0 = 9 - 6 + k
k = 3
* k= -3
b) y = x^2 +2x - 3
Let a be x coordinate of Q.
y = (x+3)(x-a)
y = x^2 + 3x - ax - 3a
By comparing coefficients,
3-a = 2
or -3 = -3a
THEREFORE,
a=1
Q(1,0)
Let a be x coordinate of Q.
y = (x+3)(x-a)
y = x^2 + 3x - ax - 3a
By comparing coefficients,
3-a = 2
or -3 = -3a
THEREFORE,
a=1
Q(1,0)
b) when x=0,
y = x^2 + 2x - 3
y = 0^2 + 2(0) - 3
y= -3
R(0,-3)
y = x^2 + 2x - 3
y = 0^2 + 2(0) - 3
y= -3
R(0,-3)
c) By completing the square:
y = x^2 + 2x -3
y = x^2 + 2x + (2/2)^2 - (2/2)^2 - 3
y = (x+1)^2 - 4
Turning point: (-1, -4)
Since line of symmetry = to vertical line passing through the turning point, x= -1 is the line of symmetry.
y = x^2 + 2x -3
y = x^2 + 2x + (2/2)^2 - (2/2)^2 - 3
y = (x+1)^2 - 4
Turning point: (-1, -4)
Since line of symmetry = to vertical line passing through the turning point, x= -1 is the line of symmetry.
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