Expressing 6q = 3s - 9 In Function Notation

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Let's delve into the world of function notation, a fundamental concept in mathematics. Function notation is a way of representing functions algebraically, providing a concise and clear method to express the relationship between input and output values. The most common notation is f(x), where f is the name of the function, and x is the input variable. The expression f(x) represents the output value of the function for the given input x. Understanding function notation is crucial for various mathematical applications, including calculus, algebra, and data analysis. It helps us to describe how a function transforms an input into an output, making it easier to analyze and manipulate mathematical relationships. Consider this expression, 6q=3s−96q = 3s - 9, where we aim to express the equation in function notation with q as the independent variable. This means we want to isolate s in terms of q, allowing us to write s as a function of q. In essence, we are looking for a function, let's call it f, such that s = f(q). The process involves algebraic manipulation to solve the equation for s. Once we have s isolated, we can rewrite the equation in function notation, clearly showing the relationship between the independent variable q and the dependent variable s. This transformation is key to understanding how changes in q affect the value of s through the function f. Function notation not only simplifies the representation of functions but also provides a powerful tool for evaluating function values and analyzing their behavior. By expressing equations in function notation, we gain a deeper insight into the underlying mathematical relationships and can effectively use functions to model real-world phenomena.

Transforming the Equation

In this section, we will focus on transforming the given equation, 6q=3s−96q = 3s - 9, into function notation where q is the independent variable. This involves a series of algebraic steps to isolate s on one side of the equation. The initial equation presents a relationship between q and s, but it's not explicitly showing s as a function of q. Our goal is to rearrange the equation so that we have s expressed in terms of q, allowing us to write s = f(q). The first step in this transformation is to isolate the term containing s. We can do this by adding 9 to both sides of the equation. This ensures that we maintain the equality while moving the constant term away from the s term. The equation then becomes 6q+9=3s6q + 9 = 3s. Next, to completely isolate s, we need to divide both sides of the equation by 3. This step is crucial as it removes the coefficient of s, leaving us with s expressed as a function of q. Dividing both sides by 3 gives us (6q+9)/3=s(6q + 9) / 3 = s. Now, we can simplify the expression on the left side of the equation. We can divide each term in the numerator by 3, which results in 2q+3=s2q + 3 = s. This simplified equation clearly shows s as a function of q. Finally, we can rewrite this equation in function notation. Since s is now expressed in terms of q, we can write s = f(q) = 2q + 3. This function notation clearly indicates that the value of s depends on the value of q and that the function f takes q as an input and produces 2q + 3 as the output. This transformation allows us to easily evaluate s for different values of q and analyze the relationship between the two variables.

The Solution in Function Notation

Having successfully isolated s in terms of q, we can now express the solution in function notation. The equation 2q+3=s2q + 3 = s tells us that s is a function of q. To represent this in function notation, we write f(q) to denote the function of q. Therefore, we can express s as f(q) = 2q + 3. This notation is a concise way of saying that the value of s is determined by the function f applied to the input q. The function f in this case is defined by the expression 2q + 3. When we write f(q) = 2q + 3, we are defining a specific relationship between q and s. For any value of q, we can substitute it into the expression 2q + 3 to find the corresponding value of s. For example, if q = 0, then f(0) = 2(0) + 3 = 3, so s = 3. Similarly, if q = 1, then f(1) = 2(1) + 3 = 5, so s = 5. This demonstrates how function notation allows us to easily evaluate the value of the dependent variable (s) for different values of the independent variable (q). Moreover, function notation is not just a symbolic representation; it also provides a powerful tool for analyzing the behavior of functions. In this case, the function f(q) = 2q + 3 is a linear function with a slope of 2 and a y-intercept of 3. This means that for every increase of 1 in q, the value of s increases by 2. The y-intercept indicates that when q = 0, the value of s is 3. Understanding these properties is crucial for interpreting the function and its applications. In the context of the given problem, expressing the equation in function notation, f(q) = 2q + 3, provides a clear and concise representation of the relationship between q and s, making it easier to analyze and use in further mathematical operations.

Analyzing the Options

When presented with multiple options, it's crucial to carefully analyze each one to determine which correctly represents the equation 6q=3s−96q = 3s - 9 in function notation with q as the independent variable. This involves comparing each option with the derived function, f(q) = 2q + 3, and identifying any discrepancies. Let's examine each option:

  • Option A: f(q) = rac{1}{2}q - rac{3}{2}

    This option suggests that the function of q is given by rac{1}{2}q - rac{3}{2}. Comparing this with our derived function, f(q) = 2q + 3, we can see that the coefficients and constant terms are different. This indicates that Option A is not the correct representation of the given equation in function notation. The coefficient of q in Option A is rac{1}{2}, while in our derived function, it is 2. Similarly, the constant term in Option A is - rac{3}{2}, while in our function, it is 3. These differences clearly show that Option A does not match the correct function. To further verify, we can substitute a value for q into both functions and compare the results. For example, if q = 0, Option A gives f(0) = rac{1}{2}(0) - rac{3}{2} = - rac{3}{2}, while our derived function gives f(0) = 2(0) + 3 = 3. The different results confirm that Option A is incorrect.

  • Option B: f(q)=2s+3f(q) = 2s + 3

    This option presents a function of q in terms of s, which is a fundamental error. Function notation with q as the independent variable should express the function f(q) solely in terms of q, not s. The presence of s in the expression indicates that this option does not correctly represent the equation in the desired function notation. This option confuses the roles of the independent and dependent variables. In function notation f(q), q is the independent variable, and the function should express the value of the dependent variable (in this case, s) in terms of q. The expression 2s + 3 does not provide a direct relationship between f(q) and q, making this option incorrect. We are looking for a function that takes q as input and returns the corresponding value of s, but this option mixes the input and output variables.

  • Option C: f(s) = rac{1}{2}s - rac{3}{2}

    This option defines a function f of s, which is not what we are looking for. We want to express s as a function of q, not the other way around. This option represents the inverse relationship, where q would be expressed in terms of s. While this function is mathematically valid, it does not answer the question of expressing the original equation in function notation with q as the independent variable. Option C essentially solves the equation for q in terms of s, which is the opposite of what we need. To verify this, we can rearrange the original equation to solve for q. Starting with 6q=3s−96q = 3s - 9, we can divide both sides by 6 to get q = rac{1}{2}s - rac{3}{2}. This confirms that Option C correctly represents q as a function of s, but it does not represent s as a function of q.

  • Option D: f(q)=2q+3f(q) = 2q + 3

    This option matches our derived function, f(q) = 2q + 3. It correctly expresses the relationship between s and q in function notation, with q as the independent variable. The coefficient of q is 2, and the constant term is 3, both of which align with our derived function. This option accurately represents the solution to the problem. To confirm, we can substitute values for q into this function and compare the results with the original equation. For example, if q = 1, then f(1) = 2(1) + 3 = 5, which means s = 5. Substituting these values into the original equation, we get 6(1)=3(5)−96(1) = 3(5) - 9, which simplifies to 6=15−96 = 15 - 9, or 6=66 = 6. This confirms that Option D is the correct representation of the equation in function notation.

The Correct Answer

After a thorough analysis of all the options, it is evident that Option D, f(q)=2q+3f(q) = 2q + 3, is the correct answer. This function accurately represents the equation 6q=3s−96q = 3s - 9 in function notation, with q as the independent variable. It aligns perfectly with the derived function obtained by isolating s in terms of q. This process of analyzing options is a critical skill in mathematics, as it allows for a systematic approach to problem-solving. By carefully examining each option and comparing it with the derived solution, we can confidently identify the correct answer. In this case, Option D is the only option that correctly expresses s as a function of q, making it the definitive solution.

Conclusion

In conclusion, the process of converting an equation into function notation is a fundamental skill in mathematics. When given the equation 6q=3s−96q = 3s - 9, expressing it in function notation with q as the independent variable involves isolating s and rewriting the equation in the form s = f(q). Through algebraic manipulation, we determined that s=2q+3s = 2q + 3, which in function notation is written as f(q) = 2q + 3. This notation provides a clear and concise way to represent the relationship between q and s. By analyzing the provided options, we confirmed that Option D, f(q) = 2q + 3, is the correct answer. This exercise highlights the importance of understanding function notation and its applications in representing mathematical relationships. Function notation is not just a symbolic representation; it is a powerful tool for analyzing and manipulating functions, making it an essential concept in various mathematical disciplines.