If Thomas jumps in the air and his position is modeled by the equation y = 3x^2 - 12x, how long is Thomas in the air?

Question

If Thomas jumps in the air and his position is modeled by the equation y = 3x² - 12x, how long is Thomas in the air?

Answer 1

Hello Mary,

Assuming x represents the time, let's start by finding where Thomas is at the start, when x = 0

y = 3(0)^2 - 12(0) = 0

y = 0 means Thomas starts on at ground level.

Now to find how long Thomas is in the air, we have to find when Thomas will be on the ground again.

For that you replace y by 0 and solve for x: 0 = 3x² - 12x

Divide both sides by 3: 0 = x² - 4x

Factor x out: 0 = x(x-4)

So x = 0 or x =4

So Thomas is on the ground at the start (x = 0), jumps, then he's back on the ground at x =4. That means he is in the air for 4 seconds.