what are the zeros of this parabola. h(t) = -16t^2 + 16t + 12

Answer 1

Simplify by removing 4:

4*(-4t^2 +4t =3)

Then factor the remaining expression.

(-4t^2 +4t =3) = (-2t + 3)*(2t + 1)

The two points where the parabola reach 0 are:

-2t + 3 = 0; t = -0.5, and

2t + 1 = 0; t = 1.5

Answer 2

To find the zeros of the parabola means we are looking for the x values when y is 0.

So we have to factor this quadratic. We could possibly have 2 roots because the x is squared indicating the number of roots.

1. -16 t ^2 + 16 t + 12 =0
2. -4 ( 4 t^2 - 4 t -3) =0
3. -4 ( 2 t + 1) (2 t - 3) =0
4. t= -1/2 and t = 1.5
5. I hope this helps.