Answer:
Axis of symmetry: x=-2
Maximum point: (-2,8)
Step-by-step explanation:
The given function is [tex]h(x)=-\frac{1}{2}x^2-2x+6[/tex].
We can complete the square to reveal the axis of symmetry and the vertex.
We factor to get:
[tex]h(x)=-\frac{1}{2}(x^2+4x)+6[/tex]
We add the zero pairs and obtain a perfect square:
[tex]h(x)=-\frac{1}{2}(x^2+4x+4-4)+6[/tex]
[tex]h(x)=-\frac{1}{2}(x^2+4x+4)+2+6[/tex]
[tex]h(x)=-\frac{1}{2}(x+2)^2+8[/tex]
The function is now in the form [tex]h(x)=a(x-h)^2+k[/tex], where V(h,k) =(-2,8) is the vertex.
