Answer : [tex] Answer :f(x) = x^2 + 4x + 3 \ and \ f(x) = -2x^2 - 8x + 1 [/tex]
Given: axis of symmetry of x = –2
To find axis of symmetry we use formula [tex] x =\frac{-b}{2a} [/tex]
We check which equation has axis of symmetry x= -2
(1) [tex] f(x) = x^2 + 4x + 3 [/tex]
a=1 , b=4 so axis of symmetry [tex] x =\frac{-4}{2*1} = -2 [/tex]
(2) [tex] f(x) = x^2 - 4x - 5 [/tex]
a=1 , b=-4 so axis of symmetry [tex] x =\frac{-(-4)}{2*1} = 2 [/tex]
(3) [tex] f(x) = x^2 + 6x + 2 [/tex]
a=1 , b=6 so axis of symmetry [tex] x =\frac{-6}{2*1} = -3 [/tex]
(4) [tex] f(x) = -2x^2 - 8x + 1 [/tex]
a=-2 , b=-8 so axis of symmetry [tex] x =\frac{-(-8)}{2*(-2)} = -2 [/tex]
(5) [tex] f(x) = -2x^2 + 8x - 2 [/tex]
a=-2 , b=8 so axis of symmetry [tex] x =\frac{-(8)}{2*(-2)} = 2 [/tex]
So [tex] f(x) = x^2 + 4x + 3 [/tex] and [tex] f(x) = -2x^2 - 8x + 1 [/tex] has axis of symmetry x=-2
