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

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Step-by-step explanation: