Answer:
Step-by-step explanation:
I'm not sure if this question is coming from a physics class or an algebra 2 or higher math class, but either way, the behavior of a parabola is the same in both subjects. If a parabola crosses the x axis, those 2 x values are called zeros of the polynomial. Those zeros translate to the time an object was initially launched and when it landed. The midpoint is dead center of where those x values are located. For example, if an object is launched at 0 seconds and lands on the ground 3 seconds later, it reached its max height at 2 seconds. So what we need to do is find the zeros of this particular quadratic, and the midpoint of those 2 values is where the object was at a max height of 20.06.
I used the physics equation representing parabolic motion for this, since it has an easier explanation. This equation is
[tex]x-x_{0}=v_{0}+\frac{1}{2}at^2[/tex]
where x is the max height, x₀ is the initial height, v₀ is the initial upwards velocity, t is time (our unknown as of right now), and a is the acceleration due to gravity (here, -32 ft/sec^2). Filling in our values gives us this quadratic equation:
[tex]20.06-6=30(t)+\frac{1}{2}(-32)t^2[/tex]
Simplifying that a bit gives us
[tex]14.06=30t-16t^2[/tex]
Rearranging into standard form looks like this:
[tex]0=-16t^2+30t-14.06[/tex]
If we factor that using the quadratic formula we find that the 2 times where the ball was launched and then where it came back down are
t = .925 and .95 (the ball wasn't in the air for very long!)
The midpoint occurs between those 2 t values, so we find the midpoint of those 2 values by adding them and dividing the sum in half:
[tex]\frac{.925+.95}{2}=.9375[/tex]
Therefore, the coordinates of the vertex (the max height) of this parabola are (.94, 20.06). That translates to: at a time of .94 seconds, the ball was at its max height of 20.06 feet
