Answer:
[tex](C)y = x^2 +2[/tex]
Step-by-step explanation:
Given the general form of quadratic equation [tex]ax^2+bc+c=0[/tex], the axis of symmetry is derived using the equation [tex]x=-\frac{b}{2a}[/tex]
Given the equation
[tex]y = x^2 + 2x\\a=1, b=2\\Axis of Symmetry, x=-\frac{2}{2(1)}=-1[/tex]
Next,
[tex]y = x^2 -16x+58\\a=1, b=-16\\$Axis of Symmetry$, x=-\frac{-16}{2(1)}=8[/tex]
Next,
[tex]y = x^2 +2\\a=1, b=0\\$Axis of Symmetry$, x=-\frac{0}{2(1)}=0[/tex]
Lastly,
[tex]y = x^2 -4x+2\\a=1, b=-4\\$Axis of Symmetry$, x=-\frac{-4}{2(1)}=2[/tex]
Clearly, [tex]y = x^2 +2[/tex] is the equation whose axis of symmetry is x=0.
