h(t)=-16t^2+150t+22

Answer 1

The given function is:
h(t) = -16t^2 + 150t + 22

This is a quadratic function in the form of h(t) = at^2 + bt + c, where:
a = -16
b = 150
c = 22

To find the vertex of the parabola represented by this function, we can use the formula:
t = -b / (2a)

Substituting the values of a and b, we have:
t = -150 / (2 * -16)
t = -150 / -32
t = 4.6875

The vertex of the parabola is at t = 4.6875.

To find the maximum height reached by the object, we need to substitute the value of t into the function:
h(4.6875) = -16(4.6875)^2 + 150(4.6875) + 22
h(4.6875) = -16(21.93164) + 703.125 + 22
h(4.6875) = -350.10624 + 703.125 + 22
h(4.6875) = 375.01876

The maximum height reached by the object is approximately 375.01876 units.

Answer 2

h(t)=-16t^2+150t+22 find the vertex.

Answer 3

To find the vertex of the parabola given by the equation h(t) = -16t^2 + 150t + 22, we need to find the t-coordinate of the vertex.

The t-coordinate of the vertex can be found using the formula: t = -b / (2a)

In this case, a = -16 and b = 150. Plugging these values into the formula, we get:
t = -150 / (2 * -16)
t = -150 / -32
t = 4.6875

So, the t-coordinate of the vertex is t = 4.6875.

To find the h-coordinate of the vertex, we can substitute this value of t back into the equation h(t):

h(4.6875) = -16(4.6875)^2 + 150(4.6875) + 22

Calculating this expression gives:

h(4.6875) ≈ 375.01876

Therefore, the vertex of the parabola is approximately (4.6875, 375.01876).