What is the vertex (x, y)? h(t)=2x^2+4x+7
pls show work

Hack
for y=ax^2+bx+c
the xvalue of the vertex is -b/2a
and the y value is found by subsituting that for x

so

-b/2a

h(t)=2x^2+4x+7
a=2
b=4
-b/2a=-4/(2*2)=-4/4=-1
the y value
h(-1)=2(-1)^2+4(-1)+7
h(-1)=2(1)-4+7
h(1)=2+3
h(1)=5

(-1,5) is da vertex