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 h(t)=a(t-h)^2+k and determine the maximum height of the projectile. Show all work.
Can someone help me? Please and thank you.
-6t^2+24t
= -6(t^2-4t)
= -6(t^2+4t+4) +6(4)
= -6(t-2)^2 + 24
the vertex is at t=2, and h(2) = 24
Why did the projectile go to therapy? Because it had some major height issues! But don't worry, I'm here to help. Let's find the maximum height of the projectile.
To rewrite the function in the form h(t) = a(t-h)^2 + k, we need to complete the square.
Starting with h(t) = -6t^2 + 24t, let's factor out a common factor of -6:
h(t) = -6(t^2 - 4t)
Now, to complete the square, we take half of the coefficient of t (-4) and square it: (-4/2)^2 = 4.
To maintain the balance, we add and subtract 4 within the parentheses:
h(t) = -6(t^2 - 4t + 4 - 4)
Now, we can group the terms:
h(t) = -6((t - 2)^2 - 4)
Expanding the expression, we have:
h(t) = -6(t - 2)^2 + 24
Now we have h(t) = a(t - h)^2 + k, where a = -6, h = 2, and k = 24.
The maximum height of the projectile occurs at the vertex, which is located at the point (h, k). In this case, the vertex is (2, 24), so the maximum height of the projectile is 24 feet.
Hope this helps, and remember, even projectiles have their ups and downs!
To rewrite the function in the given form, h(t) = a(t-h)^2 + k, we need to complete the square.
Given: h(t) = -6t^2 + 24t
Step 1: Factor out -6 from the equation:
h(t) = -6(t^2 - 4t)
Step 2: To complete the square, we need to find a constant value that we can add and subtract from the equation. This value is half the coefficient of the term with t, which is -4/2 = -2.
Step 3: Add and subtract (-2)^2 = 4 inside the parentheses:
h(t) = -6(t^2 - 4t + 4 - 4)
Step 4: Rearrange the terms within the parentheses:
h(t) = -6((t^2 - 4t + 4) - 4)
Step 5: Factor the terms inside the parentheses:
h(t) = -6((t - 2)^2 - 4)
Step 6: Distribute -6:
h(t) = -6(t - 2)^2 + 24
Now, we have rewritten h(t) in the form h(t) = a(t-h)^2 + k, where a = -6, h = 2, and k = 24.
To determine the maximum height of the projectile, we look at the coefficient of the quadratic term, which is a = -6. Since the coefficient is negative, the parabola opens downward, indicating a maximum point.
In this case, the maximum height occurs when t = h = 2 seconds. Substituting this value into the equation, we get:
h(max) = -6(2 - 2)^2 + 24
= -6(0)^2 + 24
= -6(0) + 24
= 0 + 24
= 24 feet
Therefore, the maximum height of the projectile is 24 feet.
Of course! I'd be happy to help you with this question.
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) = -6t^2 + 24t, let's start by factoring out a common factor of -6 from the terms involving t: h(t) = -6(t^2 - 4t).
Next, observe the coefficient of t, which is -4. To complete the square, we need to take half of this coefficient and square it. Half of -4 is -2, and squaring it gives us 4. So, we add and subtract 4 inside the parentheses: h(t) = -6(t^2 - 4t + 4 - 4).
Now, let's group the terms inside the parentheses: h(t) = -6((t^2 - 4t + 4) - 4).
The expression inside the parentheses is a perfect square trinomial, which can be factored as (t - 2)^2: h(t) = -6((t - 2)^2 - 4).
To simplify further, we distribute the -6: h(t) = -6(t - 2)^2 + 24.
Now we have re-written h(t) in the form h(t) = a(t - h)^2 + k, where a = -6, h = 2, and k = 24.
To find the maximum height of the projectile, we can observe that the vertex of the parabola represents the highest point. The vertex of a parabola in the form h(t) = a(t - h)^2 + k is given by the coordinates (h, k).
In this case, the vertex is located at (2, 24), which means the maximum height of the projectile is 24 feet.
I hope that helps! Let me know if you have any further questions.