the height in feet above ground of water seconds after leaving the mouth of a fountain. THe eqaution is h=-6t^2+12t+10 what is the greatest height the water reaches
What you want is the apex of the parabola. There are a couple of ways of finding it. One if you set the derivative equal to 0. The other is rewrite the equation in vertex form: y = a(x - h)^2 + k where (h, k) is the vertex
well that doesnt make sense and im pretty sure thats not it
In that case you can also try to graph it using a TI calculator and having the calculator find the maximum value. Which is t = 1 and h = 16. So 1 second after it leaves the mouth of the fountain it attains its maximum height of 16 feet.
In order to use the TI you put the equation in under the y = using x for t. Then hit second "CALC" and then option 4 "MAXIMUM" this will have you hit enter on the graph to the left and right of the maximum and then guess where it is. After you do that it will tell you the max.
how long is the water in the air
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