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

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it