Problem 2 - A Graphical Approach to Precalculus with Limits

Question

Work Exercises \(1-6\) without pencil and paper. Do not use a calculator. If \(y\) varies directly with \(x,\) and \(y=2\) when \(x=4,\) what is the value of \(y\) when \(x=12 ?\)

Step-by-Step Solution

Answer

When \(x = 12\), \(y = 6\).

Step 1: Understand Direct Variation

When a variable \(y\) varies directly with \(x\), it means that \(y\) is directly proportional to \(x\). The relationship is given by \(y = kx\), where \(k\) is a constant.

Step 2: Find the Constant of Variation

We know that \(y = 2\) when \(x = 4\). Use these values to find \(k\). Substitute into the equation \(y = kx\) to get \(2 = k \times 4\). Solve this to find \(k = \frac{1}{2}\).

Step 3: Apply the Constant of Variation

With \(k = \frac{1}{2}\), we can now find \(y\) when \(x = 12\). Substitute \(k\) and \(x = 12\) into the equation \(y = kx\) to get \(y = \frac{1}{2} \times 12\).

Step 4: Compute the Value of y

Calculate \(y\) using the equation from Step 3: \(y = \frac{1}{2} \times 12 = 6\).

Key Concepts

Proportional Relationships

Understanding proportional relationships is essential in solving direct variation problems. When we say that two variables have a proportional relationship, it means as one variable changes, the other changes at a consistent rate. In simpler terms, when one variable increases or decreases, the other does the same in a predictable way.

In math, we express proportional relationships using a direct variation equation. For example, if \( y \) is proportional to \( x \), then \( y = kx \), where \( k \) is a constant. This constant, \( k \), tells you how much \( y \) changes based on \( x \).

This helps us understand that if we know one set of \( x \) and \( y \) values, we can find others. This means whether \( x \) doubles or triples, \( y \) will change proportionally based on the constant \( k \). So when you're faced with problems like the one above, remember it's all about how one change affects the other.

Constant of Variation

The constant of variation, often notated as \( k \) in algebraic equations, is the key element in understanding direct variation. It is the fixed ratio between the two proportional variables.

To find \( k \), you divide the known \( y \) by its corresponding \( x \). In the exercise, when \( y = 2 \) and \( x = 4 \), we find \( k \) by calculating \( k = \frac{2}{4} = \frac{1}{2} \). This \( k \) value implies that for every unit \( x \) moves, \( y \) will adjust by multiplying \( x \) with \( \frac{1}{2} \).

Remember: Finding \( k \) is crucial without using a pencil, paper, or calculator. Once \( k \) is determined, it serves as a guide to find other values of \( y \) for different values of \( x \). This allows you to solve variations quickly and efficiently, just by understanding the role of the constant in the equation.

Algebraic Equations

Algebraic equations play a vital role in representing relationships like direct variations. They are the mathematical statements that show the equality between expressions. When dealing with direct variations, the equation \( y = kx \) is a classic example.

It's fundamental because it provides a blueprint for calculating unknown values. Knowing the equation format allows students to manipulate it to find unknown variables. For instance, with a known constant of variation \( k \) and a value for \( x \), students can easily compute a new \( y \).

In the provided exercise, starting with \( y = \frac{1}{2} \times x \), students learn to substitute different values of \( x \) effortlessly. Thus, building proficiency in solving similar problems across varied scenarios.

Mastery of such equations ensures a solid understanding of broader algebraic concepts, allowing students to approach more complex mathematical challenges with confidence.