The vertical height in feet of a projectile on a planet in our solar system at a given time t in seconds is represented by the function h(t)=−4t2+24t. Re-write h(t) in the form \large h\left(t\right)=a\left(t-h\right)^2+k and determine the maximum height of the projectile. Show all work.
or look for where dh/dt = 0 (stops going up, starts down)
0 = -8 t +24
t = 3 seconds
h(3)=−4*9+24 *3 = 72-36 =36 ft
as with all quadratics, the vertex is at t = -b/2a
To rewrite the function h(t) in the form h(t) = a(t-h)^2 + k, we need to complete the square.
Given h(t) = -4t^2 + 24t, we can factor out -4 from the equation:
h(t) = -4(t^2 - 6t)
To complete the square, we need to take half of the coefficient of t, square it, and add it inside the parentheses.
The coefficient of t is -6, so half of it is -3, and the square of -3 is 9.
Adding 9 inside the parentheses:
h(t) = -4(t^2 - 6t + 9)
But since we added 9 inside the parentheses, we also need to subtract 4 * 9 to balance the equation:
h(t) = -4(t^2 - 6t + 9 - 36)
Simplifying inside the parentheses:
h(t) = -4(t^2 - 6t - 27)
Now, we need to factor the quadratic inside the parentheses:
h(t) = -4(t - 3)(t - 9)
Finally, we have rewritten h(t) in the form h(t) = a(t-h)^2 + k, where a = -4, h = 6, and k = -27.
To determine the maximum height of the projectile, we need to find the vertex of the parabolic function.
The vertex of a quadratic function in the form h(t) = a(t-h)^2 + k has coordinates (h, k). In this case, the vertex is (6, -27).
Therefore, the maximum height of the projectile is -27 feet.
To rewrite the given function, h(t) = -4t^2 + 24t, in the form h(t) = a(t - h)^2 + k, we need to complete the square.
Step 1: Begin by rearranging the terms of the function:
h(t) = -4t^2 + 24t
= -4(t^2 - 6t)
Step 2: To complete the square, we need to take half of the coefficient of the linear term (in this case, 6), square it (6^2 = 36), and add it inside the parentheses:
h(t) = -4(t^2 - 6t + 36) - 4(36)
Step 3: Simplify inside the parentheses:
h(t) = -4(t^2 - 6t + 36) - 4(36)
= -4(t - 6 + 6^2) - 4(36)
= -4(t - 6 + 36) - 4(36)
= -4(t - 6)^2 - 4(36)
= -4(t - 6)^2 - 144
Therefore, h(t) = -4(t - 6)^2 - 144.
Now, to determine the maximum height of the projectile, we observe that the maximum height occurs when the expression inside the parentheses, (t - 6)^2, is equal to zero. This means t - 6 = 0, which gives t = 6.
When t = 6, we can substitute it back into the function to find the maximum height:
h(6) = -4(6 - 6)^2 - 144
= -144
Therefore, the maximum height of the projectile is -144 feet.