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

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.\n\n$y = -16x^{2}+228x + 110$

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.\n\n$y = -16x^{2}+228x + 110$

Answer

Answer:

$912.3$

Explanation:

Step1: Identify the coefficients

For the quadratic function $y = ax^{2}+bx + c$, here $a=-16$, $b = 228$, $c = 110$.

Step2: Find the x - value of the vertex

The x - value of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 228$ into the formula: $x=-\frac{228}{2\times(-16)}=\frac{228}{32}=\frac{57}{8}=7.125$.

Step3: Find the y - value of the vertex

Substitute $x = 7.125$ into the equation $y=-16x^{2}+228x + 110$. $y=-16\times(7.125)^{2}+228\times7.125+110$. First, calculate $-16\times(7.125)^{2}=-16\times50.765625=-812.25$. Second, calculate $228\times7.125 = 228\times\frac{57}{8}=\frac{12996}{8}=1624.5$. Then $y=-812.25+1624.5 + 110=912.25\approx912.3$.