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Projectile motion describes the movement of an object that is launched into the air and allowed to travel freely under the influence of gravity. In this context, we consider a simple situation where a projectile is fired vertically upward. Understanding the motion requires considering the forces acting on it.
The functional equation given in the original exercise, \(f(x) = 96x - 16x^2\), represents this concept of projectile motion without any air resistance. Here are some key points to keep in mind:

• Gravity negatively affects the projectile's motion, pulling it back down to the ground.
• The equation is expressed as a quadratic function, reflecting how projectile motion forms a curved path or parabola.
• The positive coefficient for \(x\) indicates upward motion initially, while the negative coefficient for \(x^2\) implies it will eventually downturn due to gravity.

When solving problems about projectile motion, understanding this quadratic representation simplifies finding significant points such as the maximum height or the time taken for that motion.

The vertex of a parabola is a key point in any quadratic function. It represents either the maximum or minimum point of the curve, depending on the direction in which the parabola opens. In this scenario, because the coefficient \(a\) in \(ax^2 + bx + c\) is negative, the parabola opens downward.
The vertex thus indicates the highest point reached by the projectile, which is crucial for calculating the maximum height. Let's delve into the basics of locating the vertex in a quadratic function:

• The standard form of a quadratic function is \(ax^2 + bx + c\).
• The vertex \((x, f(x))\) provides the peak of the parabola, and the formula to find the \(x\)-value of the vertex is \(x = -\frac{b}{2a}\).
• Once we know \(x\), we can determine \(f(x)\) by substituting \(x\) back into the original function.

In our example, substituting the given \(b = 96\) and \(a = -16\) into the formula yields an \(x\)-value of 3 seconds, confirming that 3 seconds is when the projectile reaches its peak.

Finding the maximum height of the projectile directly involves substituting the vertex's \(x\)-value back into the quadratic function. This calculation gives the precise height reached at the highest point of the projectile trajectory.
Let's break down how you proceed from knowing the vertex to calculating maximum height:

• Using the \(x\)-coordinate derived from the vertex formula \(x = -\frac{b}{2a}\), substitute that back into the original equation \(f(x) = ax^2 + bx + c\).
• In the exercise example, after determining \(x = 3\), continue by finding \(f(3)\).
• Perform the computation: \(f(3) = 96(3) - 16(3)^2 = 288 - 144 = 144\).

The result, 144 feet, represents the peak reached by the projectile. Understanding this final step in solving for maximum height is essential because it not only involves basic algebraic substitution but also ties back to the behavior of quadratic functions.