Asked by A

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)=−6t2+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 to receive credit for this question.

Answers

Answered by Bot
We want to rewrite the function h(t) in the form h(t) = a(t - h)^2 + k. To do this, we complete the square:

h(t) = -6t^2 + 24t
h(t) = -6(t^2 - 4t)

To complete the square, we need to add and subtract (-4/2)^2 = 4:

h(t) = -6(t^2 - 4t + 4 - 4) (adding and subtracting 4)
h(t) = -6((t - 2)^2 - 4)

Now we can distribute the -6 and simplify:

h(t) = -6(t - 2)^2 + 24

So, a = -6, h = 2, and k = 24. The maximum height of the projectile occurs when t = 2, which is the value of h. Therefore, the maximum height is:

h(2) = -6(2 - 2)^2 + 24 = 24 feet

Answer: \boxed{h(t)=-6(t-2)^2+24}. The maximum height is 24 feet.
Answered by Tammy
in plain mathematical language: .....

h(t)=−6t2+24t
= -6(t^2 - 4t)
= -6(t^2-4t+4 - 4)
= -6( (t - 2)^2 - 4)
= -6(t - 2)^2 + 24
Answered by Bot
The function h(t) is a quadratic equation in the form ax^2 + bx + c, where a = -6, b = 24, and c = 0. To find the maximum height of the projectile, we need to rewrite h(t) in the form h(t) = a(t - h)^2 + k, where h is the horizontal shift and k is the vertical shift. To do this, we complete the square by adding and subtracting (b/2a)^2, which simplifies to (4/a)^2 = 4. Then, we factor out a from the parentheses and simplify to get the desired form. The maximum height occurs when t = h, so we plug in t = 2, which gives us k = 24, and the maximum height is 24 feet.

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