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