Question

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

time after launch, x in seconds, by the given equation. Using this equation, find the
maximum height reached by the rocket, to the nearest tenth of a foot.
y = -16x2 + 119x + 73

Answer 1

The equation we have is:

`y = { - 16x}^(2 ) + 119x + 73`

This is equation of a parabola that opens down.

To find the maximum point or the maximum height, we need to find the vertex of the parabola.

First, let's define variables:

A will be the coefficient of the x squared.

b is the coefficient of the x and

C is the independent term:

`a = - 16 \\ b = 119 \\ c = 73`

The vertex of will have the following x coordinate:

`x = - (b)/(2a)`

Substituting the values of a and b:

`x = - (119)/(2( - 16))`

Solving the operations:

`x = - (119)/( - 32 ) \\ x = 3.71875`

The next step is to substitute this x value into our equation to find the maximum height y:

`y = { - 16x}^(2) + 119x + 73 \\ \\ substituting \: x = 3.71875 \\ \\ y = - 16(3.71875)^(2) + 119(3.71875) + 73`

Solving the operations:

`y = - 16(13.8291) + 442.49555 + 73 \\ \\ y = 294.23`

Rounding to the nearest tenth (1 decimal place)

y=294.2

Answer: 294.2 feet.

Answer 2

• 294.3 ft

Explanation:

Given function

• y = -16x^2 + 119x + 73

This is quadratic function

It reaches the maximum value at vertex, which is calculated by the formula:

• x = -b/2a

Substituting values

• x = -119/2*(-16) = 3.71875

y-value of same point is

• y = -16*3.71875^2 + 119*3.71875 + 73 = 294.265625 ≈ 294.3 ft

So maximum height is 294.3 ft