question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time…

question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y = -16x² + 166x + 74 answer attempt 2 out of 2

question a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y = -16x² + 166x + 74 answer attempt 2 out of 2

Answer

Explanation:

Step1: Identify the coefficients

The quadratic equation is $y = -16x^{2}+166x + 74$, where $a=-16$, $b = 166$, $c = 74$.

Step2: Find the x - coordinate of the vertex

The formula for the x - coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 166$ into the formula: $x=-\frac{166}{2\times(-16)}=\frac{166}{32}=\frac{83}{16}=5.1875$.

Step3: Find the maximum height

Substitute $x = 5.1875$ into the equation $y=-16x^{2}+166x + 74$. $y=-16\times(5.1875)^{2}+166\times5.1875 + 74$. First, calculate $(5.1875)^{2}=26.912109375$. Then $-16\times(5.1875)^{2}=-16\times26.912109375=-430.59375$. $166\times5.1875 = 861.125$. $y=-430.59375+861.125 + 74$. $y=504.53125\approx504.5$.

Answer:

$504.5$