Problem 103 - College Algebra with Modeling and Visualization

Question

Assume that the constant of proportionality is positive. Suppose \(y\) is directly proportional to the second power of \(x .\) If \(x\) is halved, what happens to \(y ?\) $$ \begin{array}{ccccc} x & 2 & 3 & 4 & 5 \\ \hline y & 2 & 4.5 & 8 & 12.5 \end{array} $$

Step-by-Step Solution

Answer

When \( x \) is halved, \( y \) becomes one fourth of its original value.

Step 1: Understanding Proportionality

Since \( y \) is directly proportional to the square of \( x \), we can express this relationship with the equation \( y = kx^2 \), where \( k \) is the constant of proportionality.

Step 2: Finding the Constant of Proportionality

Use one pair of values \( (x, y) = (2, 2) \) to find \( k \). Substitute into the equation: \( 2 = k \times 2^2 \). Simplifying gives \( 2 = 4k \) so \( k = \frac{1}{2} \).

Step 3: Expressing the Relationship

Now that we know \( k = \frac{1}{2} \), the relationship between \( x \) and \( y \) is given by \( y = \frac{1}{2} x^2 \).

Step 4: Effect of Halving \( x \)

To determine the effect on \( y \) when \( x \) is halved, substitute \( x = x/2 \) into the relationship: \( y = \frac{1}{2} \left(\frac{x}{2}\right)^2 = \frac{1}{2} \times \frac{x^2}{4} = \frac{1}{8} x^2 \).

Step 5: Conclusion on the Effect

The new expression \( y_{new} = \frac{1}{8} x^2 \) shows that the new value of \( y \) is one fourth the original value (since \( y = \frac{1}{2} x^2 \) originally). Therefore, when \( x \) is halved, \( y \) becomes one fourth of its original value.

Key Concepts

Proportional Relationships

Direct variation is a specific type of proportional relationship. In the context of mathematics, if two quantities vary directly, it means that they increase or decrease in the same ratio. This kind of relationship is described by the equation \( y = kx \), where \( y \) and \( x \) are variables, and \( k \) is a constant.Generally, in a direct variation, when one variable changes, the other changes in a predictable way, maintaining a constant ratio. For example:

• If \( x \) doubles, then \( y \) also doubles.
• If \( x \) is halved, \( y \) is halved too.

In our exercise, we see this concept expanded to involve powers of \( x \), such as squares, resulting in a quadratic relationship, which we'll explore next.

Quadratic Functions

Quadratic functions are polynomial equations of degree 2, typically expressed in the form \( ax^2 + bx + c = 0 \). While they may appear complex, they describe a wide range of real-world phenomena, from the trajectories of projectiles to economic models.In our direct variation exercise, the quadratic aspect comes into play because \( y \) is proportional to the square of \( x \). Specifically, the equation becomes \( y = kx^2 \). This means:

• If \( x \) is doubled, \( y \) increases by a factor of four.
• Conversely, if \( x \) is halved, \( y \) decreases to one fourth of its original value.

Quadratic functions often graph as parabolas when plotted, exhibiting symmetry and a singular turning point. These characteristics can help in visualizing how changes in \( x \) affect \( y \).

Constant of Proportionality

The constant of proportionality, often denoted as \( k \), is a crucial part of understanding direct variation. It determines the specific rate at which one variable changes relative to another. In the equation \( y = kx^2 \), it signifies how many times more or less \( y \) is than \( x^2 \).Finding \( k \) involves using known values of \( x \) and \( y \) to solve the equation. From the exercise, using the pair \((x, y) = (2, 2)\), we solved:

• \( k = \frac{1}{2} \), meaning \( y \) is half of \( x^2 \).

This constant allows for predictions and calculations about \( y \) when \( x \) changes. It helps bridge the relationship between the two variables, maintaining the proportional nature of their connection, no matter their values.