Problem 1 - Algebra Form and Function

Question

In Exercises \(1-4\), write a formula for \(y\) in terms of \(x\) if \(y\) satisfies the given conditions. Proportional to the \(5^{\text {th }}\) power of \(x,\) and \(y=744\) when \(x=2\)

Step-by-Step Solution

Answer

Answer: \(y=23.25x^5\)

Step 1: Write a proportionality equation

Since \(y\) is proportional to the \(5^{\text{th}}\) power of \(x\), we can write the proportionality equation as: \(y = k x^5\) where \(k\) is the constant of proportionality.

Step 2: Substitute the given values and find the formula for \(y\)

Now, we're given that \(y=744\) when \(x=2\). Let's substitute these values into our equation to solve for \(k\): \(744 = k(2)^5\) \(744 = k(32)\) To find the value of \(k\), we simply divide both sides of the equation by 32: \(k=\frac{744}{32}\) \(k=23.25\) Now that we have the value of \(k\), we can write the final formula for \(y\) in terms of \(x\): \(y=23.25x^5\)

Key Concepts

Constant of Proportionality

The constant of proportionality is a key concept that simplifies relationships between variables by expressing them in proportional terms. When something is described as proportional to another variable, it means they change at a consistent rate relative to each other. In algebra, when we say "New York is proportional to the square of Los Angeles", it typically manifests as an equation like \( y = k x^2\). Here, \(k\) is the constant of proportionality. It represents how much \(y\) changes when \(x\) changes.
To determine \(k\), specific values of \(x\) and \(y\) are usually provided. As in the given exercise, knowing that \(y = 744\) when \(x = 2\) allows us to calculate \(k\) by substituting these values into the proportionality equation and solving for \(k\). This constant makes it possible to predict the value of \(y\) for any other \(x\).
In the problem discussed, \(k\) was calculated to be 23.25. This means that for any value of \(x\), multiplying it by \(x^5\) and then by 23.25 gives us \(y\). The constant of proportionality bridges the specific relationship between the variables, revealing the predictive power of the equation.

Power Functions

In algebra, power functions take the form \( y = ax^b \) where \(a\) and \(b\) are constants. Power functions are powerful tools because they allow for modeling relationships where one variable is dependent on another raised to a power. For example, in the context of the exercise, the given power function is \( y = kx^5 \), which indicates that \(y\) depends on \(x\) raised to the fifth power.
Such functions are useful for representing real-world phenomena where the relationship isn't linear. For instance, physical laws like Newton's law of gravity or the formula for the area of a circle can be expressed using power functions. The constant \(k\) affects the steepness of the curve, while the exponent determines the rate of increase or decrease. For each power, the graph develops a unique shape, offering insights into the dynamics of change between the variables.
In this exercise, the final formula \( y = 23.25x^5 \) derived is a classic example of a power function, where the constant of proportionality \(23.25\) scales the power relationship between \(y\) and \(x\).

Solving Equations

Solving equations is at the heart of algebra, enabling us to find unknown values by manipulating mathematical expressions. When given an equation with one or more variables, the goal is to determine the value of these variables. In solving, we apply operations like addition, subtraction, multiplication, or division to isolate the unknown.
For the original problem, we started with the equation \(744 = k(2)^5\). To solve for \(k\), we used division, calculated the power separately, and then derived \(k\) by dividing 744 by 32. This allowed us to discover the constant of proportionality, which was then utilized to predict \(y\) for other values of \(x\).
It's important to follow each step carefully in equation solving, checking the work as you go. The validations ensure that solutions are correct. Solving equations accurately is a valuable skill because it unlocks solutions to real-world problems that can be modeled mathematically.