Questions: 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=-16 x^2+198 x+77

Solution Steps

The given equation is a quadratic equation of the form:

\[ y = -16x^2 + 198x + 77 \]

This is a standard form of a quadratic equation, \( ax^2 + bx + c \), where \( a = -16 \), \( b = 198 \), and \( c = 77 \).

The maximum height of the rocket corresponds to the vertex of the parabola, since the parabola opens downwards (as indicated by the negative coefficient of \( x^2 \)).

The \( x \)-coordinate of the vertex can be found using the formula:

\[ x = -\frac{b}{2a} \]

Substituting the values of \( a \) and \( b \):

\[ x = -\frac{198}{2 \times -16} = \frac{198}{32} = 6.1875 \]

Substitute \( x = 6.1875 \) back into the original equation to find the maximum height \( y \):

\[ y = -16(6.1875)^2 + 198(6.1875) + 77 \]

Calculate \( (6.1875)^2 \):

\[ (6.1875)^2 = 38.2852 \]

Substitute back into the equation:

\[ y = -16 \times 38.2852 + 198 \times 6.1875 + 77 \]

Calculate each term:

\[ y = -612.5632 + 1225.3125 + 77 \]

Add the terms:

\[ y = 689.7493 \]

Final Answer