Apaper airplane is thrown out of a third-floor window. the quadratic expression -2x2 + 5x + 33 models the height of the paper airplane, where x is the time in seconds.

1) Since you have not provided the full question, I will work the quadratic expression that models the height of the paper airplane to find as much information as it gets.

2) Firstly, note that the quadratic function -2x^2 + 5x + 33 has these characteristics:

i) It is a parabola

ii) Since, the coefficient of x^2 is negative (-2) it opens downward and has a maximum, which indicates the maximum height of the airplane

iii) The, y-intercept, i.e. the y-value for x = 0, is the initial height of the ariplane, the height from which it was launched, and it is - 2 (0) + 5(0) + 33 = 33.

Then, the airplane was launched from a height of 33 units.

3) The vertex of the parabola is the maximum and it tells both the time to reach the maximum height and the value of that maximum.

4) You can easily find the vertex coordinates by completeing squares. This is how:

Start: -2x^2 + 5x + 33

Factor - 2 from the first two terms: - 2 (x^2 - 5/2x) + 33

Add and subtract the square of the half of x's coefficient:

- 2 ( x^2 - 5/2x + 25/16) + 25/8 + 33

Form the perfect square binomial: - 2 (x - 5/4)^2 + 289/8

By comparission with the vertex form of the equation of the parabola: A(x - h)² + k, the vertex is:

(h,k) = (5/4, 289/4) = (1.25, 36.125).

5) Then, the maximum height is 36.125 units, when the time is 1.25 seconds.

6) You can also find the time when the airplane lands on the ground by making -2x^2 + 5x + 33 = 0

For that you can factor the expression -2x^2 + 5x + 33

-2x^2 + 5x + 33 = - (x + 3) (2x - 11)

Equal to zero: - (x + 3)(2x - 11) = 0 ⇒ x = - 3 and x = 11/2 = 5.5.

That means that the airplane will land on the ground at 5.5 seconds.

7) You can also find the heights at different times, by just pluging in different values of x in the expression -2x^2 + 5x + 33.

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step-by-step explanation: