The height of an acrobat who jumps from a trampoline 2 feet in the air is modeled by the function: y = -16x^2 + 16x + 12, where x is the time in the seconds after the acrobat jumps. 1: Find the maximum height of the acrobat. 2: Find how long it takes to reach that height. 3: Find how long the acrobat is in the air. 4: Find the acrobat's highest point.

Thanks for answering.

Maximum height: 16ft

Time to reach that height: 1 second
Time in the air: 2 seconds
Highest distance from trampoline: 4ft

SAME PERSON WHO ASKED QUESTION

change your equation to vertex form and all mysteries shall be revealed to you.

1: To find the maximum height of the acrobat, we need to find the vertex of the parabolic function. The vertex represents the highest point.

The vertex of a quadratic function in the form y = ax^2 + bx + c is given by the formula: x = -b/2a.

In this case, the quadratic function is y = -16x^2 + 16x + 12.
Comparing it with the standard form, we have a = -16, b = 16, and c = 12.

Using the formula x = -b/2a, we can find the x-coordinate of the vertex:
x = -16/(2*(-16)) = -16/(-32) = 1/2.

To find the y-coordinate of the vertex, substitute x = 1/2 into the function:
y = -16(1/2)^2 + 16(1/2) + 12
y = -16(1/4) + 8 + 12
y = -4 + 8 + 12
y = 16.

Therefore, the maximum height of the acrobat is 16 feet.

2: To find how long it takes the acrobat to reach the maximum height, we use the x-coordinate of the vertex.
In this case, the acrobat reaches the maximum height at x = 1/2.

Therefore, it takes the acrobat half a second to reach the maximum height.

3: To find how long the acrobat is in the air, we need to determine when the height is 0. In other words, we need to find the x-intercepts of the function.

Setting y = 0 in the quadratic function:
0 = -16x^2 + 16x + 12.

We can solve this equation by factoring or using the quadratic formula. Let's use the quadratic formula.

The quadratic formula is given by x = (-b ± sqrt(b^2 - 4ac))/(2a).

Substituting the values from the quadratic equation, we have:
x = (-16 ± sqrt(16^2 - 4*(-16)*12))/(2*(-16))
x = (-16 ± sqrt(256 + 768))/(-32)
x = (-16 ± sqrt(1024))/(-32)
x = (-16 ± 32)/(-32)

Simplifying further, we get:
x1 = (-16 + 32)/(-32) = 16/(-32) = -1/2
x2 = (-16 - 32)/(-32) = -48/(-32) = 3/2

Since the acrobat jumps at x = 0, we are only concerned with the positive x-value.
Therefore, the acrobat is in the air for 3/2 seconds.

4: To find the acrobat's highest point, we need the coordinates of the vertex. From the previous calculations, we found that the vertex is at (1/2, 16).

Hence, the acrobat's highest point is at (1/2, 16) with a height of 16 feet.

1) To find the maximum height of the acrobat, we need to determine the vertex of the function. The vertex represents the highest point of the parabolic path.

The given function is in the form of y = ax^2 + bx + c, where a = -16, b = 16, and c = 12.

The x-coordinate of the vertex can be found using the formula x = -b / (2a). Substituting the values, we have:
x = -16 / (2*(-16)) = -16 / -32 = 0.5.

To find the height at this x-coordinate, we substitute x = 0.5 into the equation:
y = -16(0.5)^2 + 16(0.5) + 12 = -4 + 8 + 12 = 16.

Therefore, the maximum height of the acrobat is 16 feet.

2) To find how long it takes to reach the maximum height, we already found that the time of 0.5 seconds corresponds to the maximum height.

3) To find how long the acrobat is in the air, we need to determine the interval in which the function is greater than or equal to zero. In this case, it represents the time when the acrobat is above the ground level.

Since the acrobat jumps from the trampoline, the initial height is 2 feet, which means the function must be greater than or equal to 2.

Setting y ≥ 2, we solve the quadratic equation:
-16x^2 + 16x + 12 ≥ 2.

Rearranging, we have:
-16x^2 + 16x + 10 ≥ 0.

To find the time intervals when the acrobat is in the air, we can solve this inequality by factoring or using the quadratic formula. The factors of -16x^2 + 16x + 10 are (4x + 5)(-4x + 2).

Setting each factor equal to zero, we get:
4x + 5 = 0, -4x + 2 = 0.

Solving for x, we find:
x = -5/4 and x = 1/2.

Since we are dealing with time, the negative value is not applicable. Thus, the acrobat is in the air for 0 to 0.5 seconds.

4) The highest point that the acrobat reaches is the same as the maximum height mentioned in the first part, which is 16 feet.