Which answer choice best describes $f(x)=x-12$?


AThe relationship does not show a direct linear variation.
BThe relationship shows a direct linear variation with a constant of variation of 12.
CThe relationship shows a direct linear variation with a constant of variation of -12.
DThe relationship shows a direct linear variation with a constant of variation of 1.

Verified Answer :

  • A function demonstrates direct linear variation if it can be written as \(f(x) = kx\) and passes through the origin.
  • The function \(f(x) = x - 12\) has a y-intercept of \(-12\), meaning \(f(0) eq 0\).
  • Since the function does not pass through the origin, it cannot be expressed in the form \(f(x) = kx\).
  • Final Answer: \(\boxed{\text{The relationship does not show a direct linear variation.}}\)

Explanation

  1. Analyzing the Definition of Direct Linear Variation

The problem asks to classify the nature of the function \(f(x) = x - 12\) specifically in relation to the concept of direct linear variation. A function exhibits direct linear variation if it can be expressed in the form \(f(x) = kx\), where \(k\) is a non-zero constant known as the constant of variation. In this form, the graph of the function must pass through the origin \((0,0)\), meaning that \(f(0) = 0\).

  1. Comparing the Function to the Standard Form

We examine the given function \(f(x) = x - 12\). Comparing this to the standard form of a linear function \(f(x) = mx + b\), we identify the slope \(m = 1\) and the y-intercept \(b = -12\). For direct linear variation to occur, the function must satisfy the condition that its y-intercept is zero (i.e., \(b = 0\)). Since our y-intercept is \(-12\), the condition is not met. Furthermore, evaluating the function at \(x=0\) gives \(f(0) = 0 - 12 = -12\), which is not equal to \(0\).

  1. Conclusion

Based on the analysis, the function \(f(x) = x - 12\) does not pass through the origin and cannot be written in the form \(f(x) = kx\). Therefore, it does not demonstrate direct linear variation. Among the provided options, the correct statement is that the relationship does not show a direct linear variation.

Examples

Direct linear variation is used in many real-life situations, such as calculating the total cost of an item when the price per unit is constant and there are no additional fees. For instance, if apples cost \(\$2\) per pound, the total cost \(C\) in dollars for \(w\) pounds is given by \(C = 2w\). This is a direct linear variation because doubling the weight doubles the cost, and buying zero pounds costs zero dollars. In contrast, if there was a flat shipping fee of \(\$12\), the cost function would be \(C = 2w + 12\), which is a linear function but not a direct variation because the cost at zero weight is not zero. Recognizing this distinction helps in budgeting and understanding how costs behave in different scenarios.