What is the maximum or minimum of h(t)=2x^2+4x+7 show work for points

That is the vertex
since the leading coefient (the term in front of the the x^2 term)

get into form
y=a(x-h)^2+k
te vertex is (h,k)
max value is k which occurs at x=h

complete the square
h(t)=(2x^2+4x)+7
h(t)=2(x^2+2x)+7
h(t)=2(x^2+2x+1-1)+7
h(t)=2((x+1)^2-1)+7
h(t)=2(x+1)^2-2+7
h(t)=2(x+1)^2+5
vertex is at (-1,5)
the max value is 5