Identifying A Function With X-Intercepts At (1/2, 0) And (6, 0)
In the realm of mathematics, functions are the fundamental building blocks that describe relationships between variables. A crucial aspect of understanding a function is identifying its intercepts, the points where the function's graph intersects the coordinate axes. Specifically, x-intercepts, also known as roots or zeros, are the points where the graph crosses the x-axis, signifying where the function's value equals zero. In this article, we embark on a journey to decipher a function, denoted as g(x), given its x-intercepts at (1/2, 0) and (6, 0). Our mission is to pinpoint the precise function from a set of options that satisfies these conditions.
Understanding X-Intercepts: The Key to Unlocking the Function
Before diving into the options, let's solidify our understanding of x-intercepts. An x-intercept of a function is a point on the x-axis where the function's value is zero. Mathematically, if (a, 0) is an x-intercept of g(x), then g(a) = 0. This simple yet powerful concept forms the cornerstone of our approach.
In our case, we are given that g(x) has x-intercepts at (1/2, 0) and (6, 0). This translates to two crucial pieces of information:
- g(1/2) = 0
- g(6) = 0
These two equations act as filters, allowing us to sift through the given options and identify the function that satisfies both conditions. A function that doesn't satisfy either of these conditions cannot be the correct representation of g(x).
Evaluating the Options: A Step-by-Step Approach
Now, let's examine the provided options and systematically evaluate each one against our established criteria. We will substitute x = 1/2 and x = 6 into each function and check if the result is zero. If a function yields zero for both values, it is a potential candidate for g(x).
Option 1: g(x) = 2(x + 1)(x + 6)
Let's start with the first option, g(x) = 2(x + 1)(x + 6). We will substitute x = 1/2 and x = 6 into this function.
- For x = 1/2: g(1/2) = 2(1/2 + 1)(1/2 + 6) = 2(3/2)(13/2) = 39/2. This is clearly not equal to zero.
Since g(1/2) is not zero, we can immediately eliminate this option. It fails to satisfy the condition that g(1/2) = 0. There's no need to check for x = 6, as a single failure disqualifies the entire option.
Option 2: g(x) = (x - 6)(2x - 1)
Next, we consider the second option, g(x) = (x - 6)(2x - 1). We will again substitute x = 1/2 and x = 6.
- For x = 1/2: g(1/2) = (1/2 - 6)(2(1/2) - 1) = (-11/2)(1 - 1) = (-11/2)(0) = 0. This condition is satisfied.
- For x = 6: g(6) = (6 - 6)(2(6) - 1) = (0)(12 - 1) = (0)(11) = 0. This condition is also satisfied.
This option successfully passes both tests. It produces zero when x = 1/2 and when x = 6, making it a strong contender for g(x). However, we must still evaluate the remaining options to ensure we identify the most suitable answer.
Option 3: g(x) = 2(x - 2)(x - 6)
Now, let's analyze the third option, g(x) = 2(x - 2)(x - 6). We'll follow the same procedure, substituting x = 1/2 and x = 6.
- For x = 1/2: g(1/2) = 2(1/2 - 2)(1/2 - 6) = 2(-3/2)(-11/2) = 33/2. This is not equal to zero.
As with the first option, g(1/2) is not zero, so we can eliminate this option without further evaluation. It fails to meet the fundamental requirement of having an x-intercept at (1/2, 0).
Option 4: g(x) = (x + 6)(x + 2)
Finally, we examine the fourth option, g(x) = (x + 6)(x + 2). Let's substitute x = 1/2 and x = 6.
- For x = 1/2: g(1/2) = (1/2 + 6)(1/2 + 2) = (13/2)(5/2) = 65/4. This is not equal to zero.
Once again, g(1/2) is not zero, leading us to eliminate this option. It does not align with the given x-intercept at (1/2, 0).
The Verdict: Option 2 Emerges as the Solution
After meticulously evaluating all four options, we arrive at a clear conclusion. Only Option 2, g(x) = (x - 6)(2x - 1), satisfies the given conditions. It produces zero when x = 1/2 and when x = 6, accurately reflecting the x-intercepts at (1/2, 0) and (6, 0).
Therefore, the function that could be g(x) is:
g(x) = (x - 6)(2x - 1)
The Significance of Factored Form
It's worth noting that Option 2 is presented in factored form. This form provides valuable insight into the function's behavior. The factors (x - 6) and (2x - 1) directly reveal the x-intercepts. Setting each factor to zero, we can easily solve for the x-values where the function equals zero:
- x - 6 = 0 => x = 6
- 2x - 1 = 0 => x = 1/2
This direct connection between factored form and x-intercepts highlights the power of this representation in understanding a function's roots. When a quadratic function is expressed in the form a(x - r1)(x - r2), where r1 and r2 are the roots, the x-intercepts are immediately apparent.
Expanding the Horizon: Applications of Intercept Identification
Identifying intercepts is not merely an academic exercise; it has profound practical implications across various fields. In physics, intercepts can represent crucial points in a system's trajectory. In economics, they can signify break-even points where costs equal revenue. In engineering, they can indicate critical values in a system's stability. The ability to swiftly and accurately determine intercepts empowers us to model and analyze real-world phenomena.
Conclusion: Mastering the Art of Intercepts
In this exploration, we have successfully navigated the process of identifying a function based on its x-intercepts. We have honed our understanding of the fundamental relationship between x-intercepts and the function's values. Through a systematic evaluation of options, we pinpointed the function g(x) = (x - 6)(2x - 1) as the solution. This journey underscores the importance of a solid grasp of core mathematical concepts and the power of methodical problem-solving. The ability to decipher functions from their intercepts is a valuable skill that unlocks deeper insights into the world of mathematics and its applications.
By understanding the significance of x-intercepts and the factored form of functions, we empower ourselves to tackle a wide range of mathematical challenges. This exploration serves as a testament to the elegance and utility of mathematical principles in unraveling complex relationships. As we continue our mathematical pursuits, the lessons learned here will undoubtedly serve as a solid foundation for future endeavors.
