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.
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.
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
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.
To rewrite the function h(t) in the form h(t) = a(t - h)^2 + k, we need to complete the square.
Step 1: Start with the given function h(t) = -6t^2 + 24t.
Step 2: Divide the entire equation by the coefficient of t^2 (in this case, -6) to simplify the equation. This step will make the coefficient of t^2 equal to 1.
h(t) = (-6t^2 + 24t)/(-6)
= t^2 - 4t
Step 3: Take half of the coefficient of t (-4), square it, and add the resulting value to the equation. This step allows us to complete the square.
h(t) = t^2 - 4t + (-4/2)^2
= t^2 - 4t + 4
Step 4: Rearrange the equation to match the form a(t - h)^2 + k. In this case, a = 1, h = 2, and k = 4.
h(t) = 1(t - 2)^2 + 4
Therefore, the function h(t) can be rewritten as h(t) = (t - 2)^2 + 4.
To find the maximum height of the projectile, we can look at the vertex of the parabolic function. In the form a(t - h)^2 + k, the vertex has coordinates (h, k). Hence, the maximum height of the projectile is 4 units.