Identifying Functions With A Specific Range Y ≤ 5
Understanding Range in Functions
When delving into the world of mathematical functions, understanding the range is crucial. The range of a function represents the set of all possible output values (y-values) that the function can produce. In simpler terms, it's the span of values you get after plugging in all possible input values (x-values) into the function. We are particularly interested in identifying which function, among a given set of options, has a range limited to all y-values less than or equal to 5, represented mathematically as {y | y ≤ 5}. This constraint means the function's output will never exceed 5. To effectively determine which function fits this criterion, we need to analyze each option and its graphical representation. The shape and orientation of the graph of a function provide valuable insights into its range. For instance, quadratic functions, which graph as parabolas, are particularly insightful. A parabola that opens upwards has a minimum value, while a parabola that opens downwards has a maximum value. These extreme points directly influence the range of the function. Functions with ranges limited to y ≤ 5 usually have a parabolic shape opening downwards, with the vertex (the highest point) at y = 5 or lower. Understanding this key concept, we can narrow down our choices by examining the coefficients and constants in the function's equation. The coefficient of the squared term (the term with x²) determines the direction of the parabola's opening: a positive coefficient indicates an upward opening, and a negative coefficient indicates a downward opening. The constant term and other transformations to the function influence the vertical position of the graph, hence affecting the range. By carefully considering these factors, we can systematically evaluate each option and identify the function whose range falls within the specified limit. The process involves not only understanding the algebraic representation of the function but also visualizing its graphical behavior, thereby enhancing our comprehension of the relationship between a function's equation and its range.
Analyzing Option A: f(x) = (x - 4)² + 5
Let's dive into analyzing the first option, f(x) = (x - 4)² + 5. This is a quadratic function, recognizable by the squared term involving x. The general form of a quadratic function is f(x) = a(x - h)² + k, where (h, k) represents the vertex of the parabola, and a determines whether the parabola opens upwards or downwards. In our case, a = 1, h = 4, and k = 5. Since a is positive (1), the parabola opens upwards. This is a crucial piece of information because it tells us that the vertex will be the minimum point of the graph. The vertex of this parabola is at the point (4, 5). This means the lowest y-value the function can achieve is 5. As the parabola opens upwards, the y-values will only increase from this minimum point. Consequently, the range of this function will be all y-values greater than or equal to 5, written as {y | y ≥ 5}. To further illustrate this, consider what happens as x moves away from 4. Whether x is less than or greater than 4, the term (x - 4)² will always be positive or zero (since squaring any number results in a non-negative value). Adding 5 to this non-negative value ensures that f(x) will always be 5 or greater. For example, if x = 3, f(x) = (3 - 4)² + 5 = 1 + 5 = 6. If x = 5, f(x) = (5 - 4)² + 5 = 1 + 5 = 6. No matter the value of x, f(x) will never be less than 5. Therefore, this function does not satisfy the condition that the range must be {y | y ≤ 5}. The parabola's upward-opening nature, coupled with the vertex at (4, 5), makes it clear that the function's output values will only go up from 5, not down.
Examining Option B: f(x) = -(x - 5)² + 4
Now, let's turn our attention to option B: f(x) = -(x - 5)² + 4. This function is also a quadratic, but a critical difference lies in the negative sign in front of the squared term. The general form remains f(x) = a(x - h)² + k, and in this case, a = -1, h = 5, and k = 4. The negative value of a is a game-changer because it indicates that the parabola opens downwards. This immediately suggests that the vertex will represent the maximum point of the graph, not the minimum. The vertex of this parabola is located at (5, 4). This means the highest y-value the function can achieve is 4. As the parabola opens downwards, all other y-values will be less than 4. Consequently, the range of this function will be all y-values less than or equal to 4, expressed as {y | y ≤ 4}. This range is different from the target range of {y | y ≤ 5}, but it's crucial to understand why. The term -(x - 5)² will always be negative or zero, because squaring any number results in a non-negative value, and the negative sign in front flips the sign. Adding 4 to this negative or zero value means that f(x) will never be greater than 4. To illustrate this, let's consider a few examples. If x = 5, f(x) = -(5 - 5)² + 4 = 0 + 4 = 4. If x = 4, f(x) = -(4 - 5)² + 4 = -1 + 4 = 3. If x = 6, f(x) = -(6 - 5)² + 4 = -1 + 4 = 3. In all cases, f(x) remains at or below 4. Therefore, while this function has a limited range, it is not the one we are looking for, as its range is {y | y ≤ 4} rather than {y | y ≤ 5}. The downward-opening parabola with a vertex at (5, 4) confirms that the function's output values will only go down from 4, not up to 5.
Evaluating Option C: f(x) = -(x - 4)² + 5
Let's now meticulously evaluate option C: f(x) = -(x - 4)² + 5. This quadratic function bears a striking resemblance to option B, but with a subtle yet significant difference in the vertex. Once again, we identify the key parameters in the general form f(x) = a(x - h)² + k. Here, a = -1, h = 4, and k = 5. The negative value of a (-1) is a critical indicator that the parabola opens downwards. This implies that the vertex will be the maximum point of the graph, and all other points on the parabola will have y-values less than or equal to the y-coordinate of the vertex. The vertex of this parabola is located at (4, 5). This is a crucial finding, as it means the highest y-value the function can achieve is 5. Given that the parabola opens downwards, all other y-values will be less than or equal to 5. Therefore, the range of this function is precisely {y | y ≤ 5}, which is the range we are looking for. To understand why this function meets the criteria, consider the term -(x - 4)². As we discussed earlier, squaring any number yields a non-negative value. However, the negative sign preceding the squared term flips the sign, making the entire term negative or zero. When we add 5 to this negative or zero value, the result will always be 5 or less. For instance, if x = 4, f(x) = -(4 - 4)² + 5 = 0 + 5 = 5. If x = 3, f(x) = -(3 - 4)² + 5 = -1 + 5 = 4. If x = 5, f(x) = -(5 - 4)² + 5 = -1 + 5 = 4. Regardless of the x-value, f(x) never exceeds 5. This confirms that option C is indeed the function with the desired range of {y | y ≤ 5}. The downward-opening parabola, combined with the vertex at (4, 5), definitively establishes that the function's output values will only go down from 5, satisfying the specified range constraint. Therefore, after careful analysis, option C emerges as the correct answer.
Analyzing Option D: f(x) = (x - 5)² + 4
Finally, let's examine option D: f(x) = (x - 5)² + 4. This quadratic function, like option A, features a positive coefficient for the squared term, which will influence the direction of the parabola. We again identify the key parameters in the general form f(x) = a(x - h)² + k. In this case, a = 1, h = 5, and k = 4. The positive value of a (1) indicates that the parabola opens upwards. This crucial characteristic means that the vertex will represent the minimum point of the graph. The vertex of this parabola is located at (5, 4). This implies that the lowest y-value the function can achieve is 4. Since the parabola opens upwards, all other y-values will be greater than or equal to 4. Consequently, the range of this function is all y-values greater than or equal to 4, expressed as {y | y ≥ 4}. This range does not match our target range of {y | y ≤ 5}. The term (x - 5)² will always be positive or zero, as squaring any real number results in a non-negative value. Adding 4 to this non-negative value ensures that f(x) will always be 4 or greater. To illustrate this, let's consider a few examples. If x = 5, f(x) = (5 - 5)² + 4 = 0 + 4 = 4. If x = 4, f(x) = (4 - 5)² + 4 = 1 + 4 = 5. If x = 6, f(x) = (6 - 5)² + 4 = 1 + 4 = 5. No matter the value of x, f(x) will never be less than 4. Therefore, this function does not satisfy the condition that the range must be {y | y ≤ 5}. The parabola's upward-opening nature, combined with the vertex at (5, 4), makes it clear that the function's output values will only go up from 4, not down to include values less than or equal to 5. Thus, option D is not the correct answer.
Conclusion: Identifying the Function with Range y ≤ 5
In conclusion, after a thorough analysis of all the given options, we have successfully identified the function with a range of y | y ≤ 5}. By carefully examining the quadratic functions and their properties, we determined that the correct answer is **option C. Similarly, option B, f(x) = -(x - 5)² + 4, was incorrect because, despite having a downward-opening parabola, its vertex at (5, 4) gives it a range of {y | y ≤ 4}, which is less restrictive than {y | y ≤ 5}. Finally, option D, f(x) = (x - 5)² + 4, was also eliminated due to its upward-opening parabola and vertex at (5, 4), leading to a range of {y | y ≥ 4}. This methodical approach, involving understanding the relationship between a quadratic function's equation and its graphical representation, allowed us to confidently pinpoint the function that satisfies the given range constraint. The process underscores the importance of analyzing the sign of the leading coefficient and the coordinates of the vertex when determining the range of a quadratic function. Through this comprehensive analysis, we have not only found the correct answer but also reinforced our understanding of functions and their ranges.
