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 out
the time at which the rocket will reach its max, to the nearest 100th of a second.
y = -16x² + 196x + 126

The rocket will reach its maximum height after 6.13 seconds

Step-by-step explanation:

To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height

∵ y is the height of the rocket after launch, x seconds

∵ y = -16x² + 196x + 126

- Differentiate y with respect to x

∴ y' = -16(2)x + 196

∴ y' = -32x + 196

- Equate y' by 0

∴ 0 = -32x + 196

- Add 32x to both sides

∴ 32x = 196

- Divide both sides by 32

∴ x = 6.125 seconds

- Round it to the nearest hundredth

∴ x = 6.13 seconds

∴ The rocket will reach its maximum height after 6.13 seconds

There is another solution you can find the vertex point (h , k) of the graph of the quadratic equation y = ax² + bx + c, where h = and k is the value of y at x = h and k is the maximum/minimum value

∵ a = -16 , b = 196

∴

∴ h = 6.125

∵ h is the value of x at the maximum height

∴ x = 6.125 seconds

- Round it to the nearest hundredth

∴ x = 6.13 seconds

sum= when you add is how much the whole amount is

distance= when you use angles and it means how far it goes

step-by-step explanation: