Problem 16 - Algebra Form and Function

Question

In Exercises \(13-16,\) what happens to \(y\) when \(x\) is doubled? Here \(k\) is a positive constant. $$ \frac{y}{x^{4}}=k $$

Step-by-Step Solution

Answer

Answer: When \(x\) is doubled, the value of \(y\) becomes 16 times greater.

Step 1: Solving for y

First, let's solve the equation for \(y\) in terms of \(x\) and the constant \(k\): $$ \frac{y}{x^{4}} = k $$ To isolate \(y\), multiply both sides by \(x^4\): $$ y = kx^4 $$ Now we have an equation relating \(y\) and \(x\).

Step 2: Doubling x

Next, let's see what happens when we double the value of \(x\). We'll replace \(x\) with \((2x)\) in the equation and see how it affects \(y\): $$ y = k(2x)^4 $$

Step 3: Simplify the equation

Now, let's simplify the equation to find the relationship between \(y\) and the new value of \(x\): $$ y = k(16x^4) $$ Observe that the right-hand side of the equation has changed from \(kx^4\) to \(16kx^4\) when we doubled the value of \(x\).

Step 4: Comparing y-values

Contemplate how this affects the value of \(y\). When \(x\) is doubled, the right-hand side of the equation becomes \(16\) times greater: $$ y = 16kx^4 $$

Step 5: Conclusion

Thus, when \(x\) is doubled, the value of \(y\) becomes 16 times greater.

Key Concepts

Exponentiation

Exponentiation is like supercharging a number. It allows you to multiply it by itself a certain number of times. This is commonly represented as a base raised to an exponent, written as \( x^n \). Here, \( x \) is the base and \( n \) tells you how many times to multiply it. In the exercise, we used exponentiation with \( x^4 \), which means multiplying \( x \) by itself four times: \( x \times x \times x \times x \). When you double \( x \) and then raise it to the fourth power as \( (2x)^4 \), you multiply 2 by itself four times, resulting in 16. Hence, the term \( x^4 \) in the expression for \( y \) becomes \( 16x^4 \).

• Exponentiation helps you perform multiple multiplications quickly.
• Doubling a base before raising it to a power can greatly increase the result.
• It's a core concept helping to understand growth, scaling, and other phenomena.

Direct Variation

Direct variation describes a relationship between two variables where one is a constant multiple of the other. In simpler terms, when one variable increases, the other increases at a constant rate. In the form of the equation \( y = kx^n \), \( k \) is a constant, and the change in \( x \) directly influences \( y \). For our exercise, \( y \) varies directly with \( x^4 \), which means as \( x \) changes, \( y \) will change in a predictable way dependent on \( k \).

• Direct variation allows prediction and control of outcomes by understanding one variable influences another directly.
• Improvements in technology, science, and predictions often rely on identifying and leveraging direct variations.
• It's fundamental in both linear and nonlinear relationships.

Equation Solving

Equation solving is like figuring out a puzzle. It involves finding the value of unknowns that make the equation true. We start by isolating the unknown variable, which in our exercise was \( y \). By solving the initial equation \( \frac{y}{x^4} = k \), we multiplied both sides by \( x^4 \) to isolate \( y \), giving us \( y = kx^4 \). When we double \( x \), we substitute this new value into our equation, then simplify through exponentiation: \( y = k(2x)^4 = 16kx^4 \). This shows that doubling changes the equation's entire balance and relationship.

• Equation solving makes unknowns known by revealing relationships between variables.
• This approach is crucial in science, business, and everyday problem solving.
• Understanding the step-by-step solution process is vital for mastering algebra.