Problem 45 - Algebra A Combined Function

Question

An object is thrown upward from the ground with an initial velocity of 64 feet per second. Neglecting air resistance, the height \(h(t)\) in feet of the object after \(t\) seconds is given by the polynomial function \(h(t)=-16 t^{2}+64 t\) a. Write an equivalent factored expression for the function \(h(t)\) by factoring \(-16 t^{2}+64 t\). b. Find \(h(1)\) by using $$ h(t)=-16 t^{2}+64 t $$ and then by using the factored form of \(h(t)\). c. Explain why the values found in part (b) are the same.

Step-by-Step Solution

Answer

The factored form of \(h(t)\) is \(-16t(t-4)\). Both methods give \(h(1) = 48\) since they are equivalent expressions.

Step 1: Factor the Expression

To factor the polynomial \(-16t^2 + 64t\), start by factoring out the greatest common factor from both terms, which is \(16t\). \(-16t^2 + 64t = 16t(-t + 4)\). Thus, the factored form of the function is \(h(t) = -16t(t - 4)\).

Step 2: Evaluate Using the Original Function

Substitute \(t = 1\) into the original equation \(h(t) = -16t^2 + 64t\). This gives \(h(1) = -16(1)^2 + 64(1) = -16 + 64 = 48\).

Step 3: Evaluate Using the Factored Form

Substitute \(t = 1\) into the factored form of the equation \(h(t) = -16t(t - 4)\). This results in \(h(1) = -16(1)((1) - 4) = -16 \times 1 \times -3 = 48\).

Step 4: Justify the Consistency of Results

Both methods of evaluating \(h(1)\) give the same result, 48, because they are using equivalent expressions. Factoring is merely a different representation of the original polynomial, and substitution applies the same mathematical operations regardless of form.

Key Concepts

Factoring Polynomials

Factoring polynomials is an essential skill in algebra that allows us to simplify polynomial expressions. This involves breaking down a complex expression into simpler terms, which are multiplied together to obtain the original expression.
In the given problem, the polynomial expression \(-16t^2 + 64t\) was factored. The first step in factoring is identifying the greatest common factor (GCF) that can be extracted from all terms. Here, the GCF is \(16t\), which is common to both terms of the polynomial.
Factoring \(16t\) out, we get:

• First term \(-16t^2\) becomes \(-16t\times t\)
• Second term \(64t\) becomes \(16t\times 4\)

Thus, the factored form of the polynomial is \(-16t(t - 4)\). Factoring provides a useful way to understand and solve polynomial equations more easily, especially when dealing with quadratic functions.

Quadratic Functions

Quadratic functions are a specific type of polynomial that are characterized by the highest power of the variable being squared. The general form of a quadratic function is \(ax^2 + bx + c\).
In the exercise, the function \(h(t) = -16t^2 + 64t\) is a quadratic function because the highest degree is 2 (i.e., \(t^2\)). It's represented in the form \(at^2 + bt\), where there is no constant term. Quadratics are commonly expressed with a parabolic shape when graphed. The negative sign before the \(16t^2\) indicates the parabola opens downwards.
Understanding how to manipulate quadratic functions, including factoring them, assists in solving problems related to motion, such as the height of an object thrown upward. These functions allow you to predict and analyze the trajectory of objects in physical contexts, simplifying their dynamics in time.

Function Evaluation

Function evaluation involves substituting a specific value for the variable in a function to find the corresponding output. This process is essential for understanding how functions behave with different inputs.
In the problem, we evaluated the function at \(t = 1\) for both the original and factored forms of the function \(h(t)\).
Let's break it down:

• Using the original: Substitute \(t = 1\) into \(-16t^2 + 64t\), resulting in \(-16(1)^2 + 64(1)\), simplifying to \(-16 + 64 = 48\).
• Using the factored form: Substitute \(t = 1\) into \(-16t(t - 4)\), leading to \(-16(1)(1 - 4)\), which simplifies to \(-16 \times 1 \times (-3) = 48\).

This confirms that despite the difference in appearance between the original and factored forms, the values after evaluation remain consistent. Function evaluation is pivotal for verifying solutions and ensuring that different forms of the same function yield equivalent results.