Problem 30 - Algebra Form and Function

Question

Put the functions in the form \(Q=k t\) and state the value of \(k\). $$ Q=(t-3)(t+3)-(t+9)(t-1) $$

Step-by-Step Solution

Answer

Answer: -8

Step 1: Expand the given expression

We start by expanding both the products in the given expression. To do this, we'll use the distributive property, which states that \(A(B + C) = AB + AC\). In this case, we have: $$ (t - 3)(t + 3) = t^2 - 3t + 3t - 9 = t^2 - 9 $$ and $$ (t + 9)(t - 1) = t^2 - t + 9t - 9 = t^2 + 8t - 9 $$ Now, we can rewrite the original expression as: $$ Q = (t^2 - 9) - (t^2 + 8t - 9) $$

Step 2: Combine like terms

Now let's combine the like terms in the expression. We have two quadratic terms, \(t^2\), two linear terms (multiplied by \(t\)), and two constant terms. $$ Q = (t^2 - t^2) - 8t - (-9 + 9) $$ Notice that the constant terms and quadratic terms cancel out. This leaves us with: $$ Q = -8t $$

Step 3: Identify the value of k

Now, our expression is in the form of \(Q = kt\). In this case, k is the constant factor multiplied by \(t\), which is -8. So, we have: $$ Q = -8t $$ And the value of \(k\) is -8.

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that helps us simplify expressions and solve equations. It states that multiplying a number by a sum is the same as multiplying each addend separately and then adding the products together. Mathematically, it can be expressed as: \( A(B + C) = AB + AC \). In our given exercise, we applied the distributive property to expand expressions such as \((t - 3)(t + 3)\). Here’s how the distributive property helps:

• First, multiply the terms \(t\) and \(t\), giving you \(t^2\).
• Then, multiply \(t\) by \(3\) and \(-3\) by \(t\), which cancel out as \(3t\) and \(-3t\).
• Finally, multiply \(-3\) by \(3\) to get \(-9\).

This process simplifies expressions and aids in making algebraic manipulation more manageable. It’s a critical step for breaking down complex problems into simpler parts.

Combining Like Terms

Combining like terms is a process used in algebra to simplify expressions or equations, which involves summing coefficients of terms that have the same variable and exponent. Essentially, it means putting together all the terms that look alike.In the example provided, after expanding the expressions using the distributive property, we obtain terms that include: - Quadratic terms (\(t^2\) in this case)- Linear terms (terms with just \(t\))- Constant terms (just numbers)When we write out our expanded expression for \(Q\) as \((t^2 - 9) - (t^2 + 8t - 9)\), you can combine like terms:

• The \(t^2\) terms cancel each other out because \(t^2 - t^2 = 0\).
• The numerical constants \(-9\) and \(9\) also cancel each other, leaving us with \(-8t\).
• Thus, only the linear term \(-8t\) remains.

This simplification is crucial when you want to identify the main part of an expression, especially when you need to write it in a specific form like \(Q = kt\). It is essential to work through each term carefully to ensure accuracy in your solutions.

Linear Functions

Linear functions form the foundation of algebraic equations involving a constant rate of change. They are represented by expressions such as \(Q = kt\), where \(Q\) and \(t\) are variables, and \(k\) is a constant. Understanding linear functions is vital as they describe a straight line when graphed and imply a relationship that changes at a constant rate.Within the context of the exercise given, once you have simplified the expression to \(Q = -8t\), you have identified a linear function:

• Here, the coefficient \(-8\) represents the constant \(k\), which shows the rate at which \(Q\) changes concerning \(t\).
• The linear relationship means that for every unit increase in \(t\), \(Q\) decreases by \(8\) units, reflecting a negative slope in the graph of the function.
• The simplicity of linear functions allows easy calculation and prediction of outcomes, which is particularly useful in solving real-world problems where relationships can be modeled in a linear way.

Recognizing the form of the equation and the value of \(k\) not only helps in mathematical problem-solving but also builds intuition for more complex functions encountered in advanced mathematics.