Problem 25 - College Algebra with Corequisite Support

Question

For the following exercises, use the given information to find the unknown value. \(y\) varies directly as the square of \(x .\) When \(x=2,\) then \(y=16\). Find \(y\) when \(x=8\).

Step-by-Step Solution

Answer

When \(x=8\), \(y=256\).

Step 1: Identify the direct variation equation

Since \(y\) varies directly as \(x^2\), we can express this relationship as \(y = kx^2\), where \(k\) is a constant that we need to find.

Step 2: Use given values to find the constant

Substitute \(x = 2\) and \(y = 16\) into the equation \(y = kx^2\). This gives us \(16 = k(2)^2\). Simplifying gives \(16 = 4k\). Solve for \(k\) by dividing both sides by 4, yielding \(k = 4\).

Step 3: Substitute the known constant to find the unknown value of y

Now that we have \(k = 4\), substitute \(x = 8\) into the equation \(y = 4x^2\). This gives \(y = 4(8)^2\).

Step 4: Calculate \(y\) for \(x = 8\)

Calculate \(8^2\) which is 64. Then multiply this by 4: \(y = 4 \times 64 = 256\).

Key Concepts

Understanding Algebra Through Direct Variation

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating those symbols. It is all about finding the unknown values. In the context of direct variation, algebra helps us to express the relationship between two variables.
In our exercise, we have variables \( y \) and \( x \), where \( y \) varies directly as the square of \( x \). This means that any change in \( x \) directly affects \( y \), following a specific rule or pattern. Understanding how to set up and simplify algebraic expressions with direct variation is a key skill. This involves steps like substituting known values and solving for unknown variables, which are foundational in algebra.

Decoding Constants in Direct Variation

In any direct variation, the constant is a crucial part of the formula. It serves as the fixed ratio that connects the dependent variable to the independent variable. In this exercise, the constant \( k \) is found by using given values of the variables.
Here's how we determine the constant: Given \( y = kx^2 \), if \( y = 16 \) when \( x = 2 \), we substitute these values into our equation to solve for \( k \). This gives us \( 16 = k(2)^2 \) or \( 16 = 4k \). Dividing both sides by 4 reveals \( k = 4 \). Understanding constants helps in forming relationships between variable quantities in equations.

Crafting Equations with Known Constants

An equation is like a statement that shows two expressions are equal. In algebra, we often use equations to describe relationships and solve for unknown variables.
For direct variation, once we determine the constant \( k \), the generalized equation becomes a powerful tool: \( y = kx^2 \). Using this equation, we can substitute any value of \( x \) to find the corresponding \( y \). By substituting \( k = 4 \) and \( x = 8 \) into the equation, we find \( y = 4(8)^2 \), which simplifies to \( y = 256 \). Crafting equations correctly ensures that we can predict and understand how variables interact.

Exploring Functions and Their Application

Functions are a fundamental concept in mathematics, describing a relationship between inputs and outputs. In this exercise, the relationship between \( x \) and \( y \) is modeled as a function. Here, \( y \) depends on \( x \) squared, scaled by the constant \( k \).
Functions provide a way to easily adapt and calculate different scenarios by changing the input variables. This is illustrated when we change \( x \) from 2 to 8. With the function \( y = 4x^2 \), we simply substitute 8 for \( x \) and recalculate to find the new value of \( y \), demonstrating flexibility and utility of functions in solving mathematical problems.