A baseball is thrown in the air at 5 ft. with an initial velocity of 42 ft per second. let t represent time and seconds. after the baseball is thrown let h represent height of the baseball. quadratic function h(t)= -16t squared + 42 t + 5

Find domain and range

h(t) = -16t^2 + 42t + 5

find the vertex of this parabola using the method you learned

your domain is obviously t ≥ 0 to the value of t when h(t) = 0
since your h(t) is clearly above the ground, h(t) ≥ 0 up to the max of h(t), obtained at the vertex.

I got 32.5 for vertex...the rest doesn't make since to me

Here is a picture of what you have:

https://www.wolframalpha.com/input/?i=h%28t%29+%3D+-16t%5E2+%2B+42t+%2B+5

The t of your vertex is 42/32 = 21/16
that is the time it takes to reach your maximum
sub that back into the original equation, h(21/16) = 521/16 = appr 32.56
so your vertex is (21/16, 521/16)
You had the height of the vertex correct,
so for the range: 0 ≤ h(t) ≤ 32.56

to find the right side of your domain:
-16t^2 + 42 + 5 = 0
so your domain would be all values of t ≥0 up to the positive answer of your equation.

To find the domain and range of the quadratic function h(t) = -16t^2 + 42t + 5, you need to consider the restrictions on the input variable t (time) and the output variable h (height).

1. Domain:
The domain of a quadratic function depends on the physical constraints or time period under consideration. In this case, since we are considering the height of a baseball that is thrown in the air, the domain would depend on how long the ball remains in the air before hitting the ground.

To determine the domain, we need to find the values of t for which the ball is in the air. The ball will hit the ground when h(t) = 0. So let's set h(t) = 0 and solve for t:

-16t^2 + 42t + 5 = 0

To solve this quadratic equation, you can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = -16, b = 42, and c = 5. Plugging these values into the quadratic formula, you can find the roots of the equation. The roots will give you the values of t when the ball hits the ground.

2. Range:
The range of a quadratic function is the set of all possible output values. In the case of the given quadratic function, h(t), the range will depend on the vertex of the parabola.

The vertex of a quadratic equation in the form h(t) = at^2 + bt + c is given by the formula:

t = -b / 2a

In our case, a = -16 and b = 42. Plugging these values into the formula, you can find the time at which the height is maximized (vertex).

Once you have the time at which the height is maximized (t value at the vertex), substitute it back into the equation h(t) = -16t^2 + 42t + 5 to find the maximum height reached by the ball.

Therefore, the range of the function h(t) will be a single value, which is the maximum height the ball reaches.

By finding the domain and range of the quadratic function h(t) = -16t^2 + 42t + 5 using the steps described above, you can determine the possible values of t and the corresponding heights, respectively.