Identifying The Function With Range { Y | Y ≤ 5 }

Understanding the range of a function is crucial in mathematics, especially when dealing with quadratic functions. The range represents all possible output values (y-values) that a function can produce. In this article, we will delve into identifying the function that has a range of {$ {y \mid y \leq 5} $}. This means we are looking for a function whose output values are less than or equal to 5. We'll explore the characteristics of quadratic functions, particularly those in vertex form, to determine the correct answer from the given options. This exploration will not only help you solve this specific problem but also enhance your understanding of how to analyze and interpret functions in general.

Understanding Range and Quadratic Functions

To identify the function with the specified range, we first need to understand the concept of range and how it relates to quadratic functions. The range of a function is the set of all possible output values (y-values) that the function can produce. For quadratic functions, which are functions of the form $f(x) = ax^2 + bx + c$, the range is closely tied to the vertex of the parabola and the direction in which the parabola opens. The vertex is the point where the parabola changes direction, and it represents either the minimum or maximum value of the function.

A quadratic function can open upwards or downwards, depending on the sign of the coefficient 'a' in the equation. If 'a' is positive, the parabola opens upwards, and the vertex represents the minimum point of the function. In this case, the range will be all y-values greater than or equal to the y-coordinate of the vertex. Conversely, if 'a' is negative, the parabola opens downwards, and the vertex represents the maximum point of the function. Here, the range will be all y-values less than or equal to the y-coordinate of the vertex. Understanding this relationship between the sign of 'a', the direction of the parabola, and the vertex is essential for determining the range of a quadratic function.

In the given problem, we are looking for a function with a range of {$ {y \mid y \leq 5} $}. This indicates that the parabola must open downwards, and the y-coordinate of the vertex should be 5. The vertex form of a quadratic function is particularly useful in this context because it directly reveals the vertex coordinates. The vertex form is given by $f(x) = a(x - h)^2 + k$, where (h, k) is the vertex of the parabola. By examining the options provided, we can determine which function satisfies these conditions: a downward-opening parabola (negative 'a') and a vertex with a y-coordinate of 5. This foundational knowledge is critical for efficiently solving the problem and comprehending the behavior of quadratic functions.

Analyzing the Given Options

Now, let's analyze each of the given options to determine which function has the range {$ {y \mid y \leq 5} $}. The options are presented in the form of quadratic functions, and we need to identify the function whose output values are less than or equal to 5. This requires a careful examination of the structure of each function, particularly its vertex form, to understand its behavior and the range of y-values it can produce.

Option A: $f(x) = (x - 4)^2 + 5$. This function is in vertex form, where the vertex is (4, 5). The coefficient of the $(x - 4)^2$ term is positive (1), which means the parabola opens upwards. Therefore, the vertex represents the minimum point of the function, and the range will be all y-values greater than or equal to 5. This does not match the desired range of {$ {y \mid y \leq 5} $}, so option A is incorrect. Understanding that a positive leading coefficient implies an upward-opening parabola is crucial in this analysis.

Option B: $f(x) = -(x - 4)^2 + 5$. This function is also in vertex form, with the vertex at (4, 5). However, the coefficient of the $(x - 4)^2$ term is negative (-1), indicating that the parabola opens downwards. This means the vertex represents the maximum point of the function. The range will be all y-values less than or equal to the y-coordinate of the vertex, which is 5. This perfectly matches the desired range of {$ {y \mid y \leq 5} $}, making option B a potential correct answer. The negative leading coefficient is the key factor that makes this option viable.

Option C: $f(x) = (x - 5)^2 + 4$. This function has a vertex at (5, 4). The coefficient of the $(x - 5)^2$ term is positive (1), so the parabola opens upwards. The range will be all y-values greater than or equal to 4, which does not align with the desired range of {$ {y \mid y \leq 5} $}. The different vertex and the upward-opening parabola disqualify this option.

Option D: $f(x) = -(x - 5)^2 + 4$. This function has a vertex at (5, 4), and the coefficient of the $(x - 5)^2$ term is negative (-1), so the parabola opens downwards. The range will be all y-values less than or equal to 4. While this parabola opens downwards, the maximum y-value is 4, which is less than 5, and thus the range does not match the required {$ {y \mid y \leq 5} $}. This option is incorrect because the y-coordinate of the vertex is not 5.

By carefully analyzing each option and considering the direction of the parabola and the vertex coordinates, we can systematically narrow down the possibilities and identify the correct answer. The process highlights the importance of understanding the relationship between the equation of a quadratic function and its graphical representation.

Identifying the Correct Function

After analyzing each option, we can now definitively identify the function that has a range of {$ {y \mid y \leq 5} $}. Our analysis focused on the vertex form of the quadratic functions, which is given by $f(x) = a(x - h)^2 + k$, where (h, k) represents the vertex of the parabola. The sign of the coefficient 'a' determines whether the parabola opens upwards (positive 'a') or downwards (negative 'a'). The y-coordinate of the vertex, 'k', is the maximum value of the function if 'a' is negative and the minimum value if 'a' is positive.

We are looking for a function whose range includes all y-values less than or equal to 5. This implies that the parabola must open downwards, and the y-coordinate of the vertex should be 5. Let's revisit the options:

Option A: $f(x) = (x - 4)^2 + 5$. This function has a vertex at (4, 5), but the positive coefficient (1) indicates that the parabola opens upwards. Thus, its range is {$ {y \mid y \geq 5} $}, which does not match our requirement.

Option B: $f(x) = -(x - 4)^2 + 5$. This function has a vertex at (4, 5), and the negative coefficient (-1) indicates that the parabola opens downwards. Therefore, its range is {$ {y \mid y \leq 5} $}, which perfectly matches the desired range. This makes option B the correct answer.

Option C: $f(x) = (x - 5)^2 + 4$. This function has a vertex at (5, 4), and the positive coefficient (1) indicates that the parabola opens upwards. Its range is {$ {y \mid y \geq 4} $}, which does not match our requirement.

Option D: $f(x) = -(x - 5)^2 + 4$. This function has a vertex at (5, 4), and the negative coefficient (-1) indicates that the parabola opens downwards. However, its range is {$ {y \mid y \leq 4} $}, which is not the range we are looking for.

Based on this analysis, option B, $f(x) = -(x - 4)^2 + 5$, is the only function that satisfies the condition of having a range of {$ {y \mid y \leq 5} $}. The negative coefficient ensures the parabola opens downwards, and the vertex (4, 5) confirms that the maximum y-value of the function is 5. This systematic approach of examining the vertex form and the sign of the leading coefficient allows us to accurately determine the range of the function and identify the correct option.

Conclusion

In conclusion, the function that has a range of {$ {y \mid y \leq 5} $} is B. $f(x) = -(x - 4)^2 + 5$. This determination was made by understanding the relationship between the vertex form of a quadratic function, the direction in which the parabola opens, and the resulting range of the function. The vertex form, $f(x) = a(x - h)^2 + k$, provides valuable information about the vertex (h, k) and the direction of the parabola (determined by the sign of 'a').

When 'a' is negative, the parabola opens downwards, and the y-coordinate of the vertex represents the maximum value of the function. In the case of option B, the negative coefficient (-1) indicates a downward-opening parabola, and the vertex (4, 5) confirms that the maximum y-value is 5. This leads to the range {$ {y \mid y \leq 5} }, which matches the given requirement. Options A and C were eliminated because they have positive coefficients, indicating upward-opening parabolas and ranges that include y-values greater than or equal to the y-coordinate of the vertex. Option D was eliminated because, although it has a negative coefficient, the y-coordinate of its vertex is 4, resulting in a range of { {y \mid y \leq 4} $}.

This problem highlights the importance of a thorough understanding of quadratic functions and their properties. By systematically analyzing the vertex form, the sign of the leading coefficient, and the vertex coordinates, we can efficiently determine the range of a function and solve related problems. This approach not only provides the correct answer but also enhances our overall comprehension of mathematical concepts and problem-solving strategies. Understanding these fundamentals is crucial for success in more advanced mathematical topics and real-world applications.