Y=-16t^2+88t+12 How do I find the vertex of this equation?

The vertex is at (2.75, 133)

Step-by-step explanation:

y=-16t^2+88t+12

The x coordinate for the vertex is at the axis of symmetry

h = - b/2a

where at^2 + bt+c

a = - 16, b = 88 and c = 12

h = - 88 / (2*-16)

h = - 88/-32

h = 2.75

To find the y coordinate, we substitute this into the equation

y = - 16 (2.75) ^2 + 88 (2.75) + 12

y = - 121+242+12

y = 133