Question

The distance of an object thrown upward is modeled by the equation h=-2x^2+3t+4, where h is the height of the object in feet at any given time and t is the time in seconds. The object is visible whenever it is at h (greater than or equal to) 0. For how long is the object visible?

Answer 1

So for this, we will have to find the zeros (x-intercepts) of the equation. In this case, we will be using the quadratic formula, which is $x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a}$ (a = x^2 coefficient, b = x coefficient, c = constant). Using our info, our equation is such: $x=\frac{-3+\sqrt{3^2-4*(-16)*4}}{2*(-16)},\frac{-3-\sqrt{3^2-4*(-16)*4}}{2*(-16)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-3+\sqrt{9+256}}{-32},\frac{-3-\sqrt{9+256}}{-32}$

Next, do the addition: $x=\frac{-3+\sqrt{265}}{-32},\frac{-3-\sqrt{265}}{-32}$

Next, plug in the equations into the calculator and your answer will be: $x=-0.41,0.60$

Since we can't have negative time in this situation, it can't be -0.41 seconds. Which means that the object is visible for about 0.60 seconds, or C.