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Use the formula for the axis of symmetry to get the x-coordinate for the vertex. Then, use that to get the y-coordinate, which is the maximum.
Time to Maximum: 0.9375 seconds
Maximum Height: 34.125 feet
Range: 0 ≤ h ≤ 34.125
This exercise asks us to complete three tasks with the function h=-16t^2+30t+6.
In order to find maximums and minimums of a standard quadratic function, y= ax^2+ bx+c, we will need to use the equation for the axis of symmetry.
x=- b/2 a
Let's start by comparing our function to the standard quadratic.
ccc
y&=& ax^2&+& bx&+&c
↕ & & ↕ & & ↕ & & ↕
h&=& -16t^2&+& 30t&+&6
We can see that time, t, is our independent variable and that the height, h, is our dependent variable. Then we can see that a = -16, b= 30, and c=6.
Thus, takes 0.9375 seconds to reach its maximum height.
We can use the fact the maximum is at t=0.9375 how high the ball will reach.
t= 0.9375
Calculate power
Multiply
Add and subtract terms
When the ball reaches 34.125 feet, it will be at it's maximum height.
Now that we know the maximum, we know the ball will not go any higher than 34.125feet and will stop when it hits the ground at 0. Therefore, we can say that the range is 0 ≤ h ≤ 34.125.