Question

Choose the equation below whose axis of symmetry is x = 0. y = x2 + 2x y = x2 − 16x + 58 y = x2 + 2 y = x2 − 4x + 2

Answer 1

y = x^2 + 2

Explanation:

find the equation below whose axis of symmetry is x = 0

to find axis of symmetry we use formula x= -b/2a

WE compare the equation with the form y =ax^2 + bx+c

y = x^2 + 2x

a= 1 and b = 2 so
`x= (-b)/(2a) = (-2)/(2) =1`

y = x^2 − 16x + 58

a= 1 and b = -16 so
`x= (-b)/(2a) = (16)/(2) =8`

y = x^2 + 2

a= 1 and b = 0 so
`x= (-b)/(2a) = (0)/(2) =0`

y = x^2 − 4x + 2

a= 1 and b = -4 so
`x= (-b)/(2a) = (4)/(2) =2`

Answer 2

If the function is symmetrical about the line x=0, it is an even function, so cannot have any odd-degree terms. The only viable choice is the 3rd one:

... y = x² + 2