A moving particle has a height given by y = 14(x-3)^2 + 3x + 9. What is the maximum height of this particle.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The height of the particle (expressed by y) is given in terms of its horizontal displacement x by `y = 14(x-3)^2 + 3x + 9` .

To determine the maximum height of the particle, find the derivative of y with respect to x.

For `y = 14(x-3)^2 + 3x +...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

The height of the particle (expressed by y) is given in terms of its horizontal displacement x by `y = 14(x-3)^2 + 3x + 9` .

To determine the maximum height of the particle, find the derivative of y with respect to x.

For `y = 14(x-3)^2 + 3x + 9` ,

`dy/dx = 14*2*(x - 3) + 3`

But `(d^2y)/(dx^2) = 28` which is positive for all x. The graph of `y = 14(x-3)^2 + 3x + 9` has an extreme value which is a minimum. The height of the particle can increase to any extent.

If the lowest height of the particle has to be determined, solve `dy/dx` = 0 for x.

`14*2*(x - 3) + 3 = 0`

=> `14*2*(x - 3) = -3 `

=> `x - 3 = -3/28`

=> `x = 3 - 3/28`

=> `x = 81/28`

At x = `81/28` , y = `14*(81/28-3)^2 + 3*81/28 + 9`

= `999/56`

The lowest height of the particle is `999/56` , there is no limit to how high it can rise.

Approved by eNotes Editorial Team