Question 130754


Since {{{f(x)=(x+2)^2}}} is in vertex form {{{y=a(x-h)^2+k}}} where a is the stretch/compression factor and (h,k) is the vertex, this means that h=-2 and k=0 (note: {{{f(x)=(x+2)^2}}} really looks like {{{f(x)=1(x-(-2))^2+0}}})



So the vertex is (-2,0)



Also, the line of symmetry is simply the equation {{{x=h}}}. So in this case the line of symmetry is {{{x=-2}}}




If we graph, we can visually verify our answer.



{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(x+2)^2),
line(-2,-12,-2,12)
)
}}} Graph of {{{f(x)=(x+2)^2}}} with a vertex of (-2,0) and the line of symmetry of {{{x=-2}}}