Y=-16X^2+100X+6 where x is the time in seconds and Y is the height in feet.
find the zeros of the function?
What do the zeros represent and are they realistic? and about how high does the rocket fly before hitting the ground?
solve
16x^2 - 100x - 6 = 0
x = (100 ± √(100^2-4(16)(-6))/32
= 6.31 or -0.06
Only the positive time makes sense.
the vertex will be the midway time between the 2 above, which is 3.13
sub that into y = ..
to get the maximum height.
To find the zeros of the function, we need to solve the equation -16X^2 + 100X + 6 = 0. This can be done by using the quadratic formula.
The quadratic formula states that for an equation of the form Ax^2 + Bx + C = 0, the solutions for x can be found using the formula:
x = (-B ± √(B^2 - 4AC)) / (2A)
In this case, A = -16, B = 100, and C = 6. Substituting these values into the formula, we get:
x = (-(100) ± √((100)^2 - 4(-16)(6))) / (2(-16))
Simplifying further:
x = (-100 ± √(10000 + 384)) / (-32)
x = (-100 ± √(10384)) / (-32)
x = (-100 ± 101.890) / (-32)
Now, we have two possible solutions:
x = (-100 + 101.890) / (-32) ≈ 0.047
x = (-100 - 101.890) / (-32) ≈ 6.347
So, the zeros of the function are approximately x = 0.047 and x = 6.347.
The zeros of the function represent the time at which the height is zero, or when the object is at ground level. In this case, they represent the times in seconds when the rocket hits the ground.
However, it is important to note that the second zero (x = 6.347) is not physically realistic because it implies that the rocket hits the ground 6.347 seconds after being launched, which is not possible given the provided equation. The realistic zero is x = 0.047, which corresponds to approximately 0.047 seconds after the rocket is launched.
To find the maximum height the rocket reaches before hitting the ground, we can plug in the x-coordinate of the vertex into the equation. The x-coordinate of the vertex is given by:
x = -B / 2A
Substituting the given values, we have:
x = -100 / 2(-16)
x = -100 / -32
x ≈ 3.125
So, the rocket reaches its peak height after approximately 3.125 seconds. To find the height at this time, we substitute x = 3.125 into the equation:
y = -16(3.125)^2 + 100(3.125) + 6
y ≈ -156.25 + 312.5 + 6
y ≈ 162.25
Therefore, the rocket reaches a height of approximately 162.25 feet before hitting the ground.
To find the zeros of the function, we need to solve the equation Y = -16X^2 + 100X + 6 for X when Y is equal to zero. In other words, we need to find the values of X where the height of the rocket (Y) is equal to zero.
1. Set Y = 0 in the equation:
0 = -16X^2 + 100X + 6
2. Rearrange the equation:
16X^2 - 100X - 6 = 0
Now, we can solve this quadratic equation using the quadratic formula.
The quadratic formula is: X = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 16, b = -100, and c = -6. Plugging these values into the quadratic formula, we have:
X = (-(-100) ± √((-100)^2 - 4 * 16 * (-6))) / (2 * 16)
X = (100 ± √(10000 + 384)) / 32
X = (100 ± √10384) / 32
Now, we can simplify this expression further:
X = (100 ± 101.888) / 32
Thus, the zeros of the function are approximately:
X = (100 + 101.888) / 32 = 6.025
X = (100 - 101.888) / 32 = -0.069
The zeros of the function represent the times in seconds when the rocket hits the ground. In this case, we have two zeros: X ≈ 6.025 and X ≈ -0.069.
A negative time value (-0.069) does not make physical sense in this context because time cannot be negative. Therefore, we can conclude that the realistic zero of the function is X ≈ 6.025 seconds.
To find the maximum height reached by the rocket, we can use the equation of the vertex of the parabolic function. The vertex formula for a quadratic function in standard form (Y = ax^2 + bx + c) is X = -b / (2a).
In our case, a = -16 and b = 100. Plugging these values into the formula, we have:
X = -100 / (2 * -16)
X = -100 / -32
X ≈ 3.125
So, at approximately 3.125 seconds, the rocket reaches its maximum height. To find the height, substitute this time value into the original equation:
Y = -16 * (3.125)^2 + 100 * 3.125 + 6
Calculating this, we get:
Y ≈ 193.75
Therefore, the rocket flies to a height of approximately 193.75 feet before hitting the ground.