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=-16x^2+110x+79

## Step 1<br />The given equation is a quadratic equation in the form \(y = ax^2 + bx + c\), where \(a = -16\), \(b = 110\), and \(c = 79\).<br /><br />## Step 2<br />The maximum height of the rocket is reached at the vertex of the parabola represented by the quadratic equation. The x-coordinate of the vertex can be found using the formula \(-\frac{b}{2a}\).<br /><br />### \(x = -\frac{b}{2a} = -\frac{110}{2(-16)} = 3.4375\)<br /><br />## Step 3<br />Substitute this value of \(x\) into the equation to find the maximum height \(y\).<br /><br />### \(y = -16(3.4375)^2 + 110(3.4375) + 79 = 204.6875\)