Question

question 1-10
the table below displays some of the values for an absolute value function.

x	y
0	-5
1	-4
2	-2
4	-3

use the table to determine the specified key features
domain: < x <
range y ≤

Explanation:

Step1: Analyze Domain

The domain of a function is the set of all possible x - values. From the table, the x - values given are 0, 1, 2, 4. But since it's an absolute - value function (and generally, for functions represented by tables with a few points, we assume the domain is all real numbers in the context of the function's behavior, but looking at the given x - values, the smallest x - value is 0 and the largest among the given is 4. However, if we consider the function's domain (for an absolute - value function, the domain is all real numbers, but from the table's x - values, the minimum x is 0 and maximum x is 4. But maybe there is a mis - understanding. Wait, no, the domain of a function is all real numbers for an absolute - value function, but the table has x values 0,1,2,4. But the problem is asking for the domain in the form \(a < x < b\)? Wait, maybe it's a typo or maybe we consider the x - values from the table. The x - values are 0,1,2,4. The smallest x is 0 and the largest is 4. But if we consider the domain as the set of x - values from the table, the domain would be \(0\leq x\leq4\), but the problem has \(< x <\). Wait, maybe there is an error, but let's check the range first.

Step2: Analyze Range

The range of a function is the set of all possible y - values. The y - values from the table are - 5, - 4, - 2, - 3. Let's find the maximum y - value among these. Comparing - 5, - 4, - 2, - 3: - 2 is the largest (since - 2 > - 3 > - 4 > - 5). For an absolute - value function, the general form is \(y = a|x - h|+k\). Let's try to find the vertex. Let's assume the vertex is at some point. Let's check the differences. From x = 0, y=-5; x = 1, y=-4 (increase of 1); x = 2, y=-2 (increase of 2); x = 4, y=-3 (decrease of 1). Wait, maybe the vertex is at x = 2? At x = 2, y=-2. Let's check the slope. From x = 0 to x = 2: the change in y is \(-2-(-5)=3\), change in x is \(2 - 0 = 2\), slope \(m=\frac{3}{2}\). From x = 2 to x = 4: change in y is \(-3-(-2)=-1\), change in x is \(4 - 2 = 2\), slope \(m =-\frac{1}{2}\). So the vertex is at (2, - 2). So the function is \(y=a|x - 2|+k\). When x = 2, y=-2, so \(k=-2\). When x = 0, \(y=a|0 - 2| - 2=-5\), so \(2a-2=-5\), \(2a=-3\), \(a =-\frac{3}{2}\). So the function is \(y =-\frac{3}{2}|x - 2|-2\). The range of an absolute - value function \(y = a|x - h|+k\) is \(y\leq k\) if \(a < 0\) (since the graph opens downwards). Here, \(a=-\frac{3}{2}<0\), and \(k = - 2\). So the range is \(y\leq - 2\).

Wait, going back to the domain. If we consider the domain from the table, the x - values are 0,1,2,4. But the problem has \(< x <\). Maybe it's a mistake and should be \(\leq x\leq\), but if we follow the problem's format, and assume that the domain is from the minimum x (0) to maximum x (4), so \(0 < x < 4\) (but this is not correct for the actual domain of the function, but based on the table's x - values). And the range: the maximum y - value is - 2, so \(y\leq - 2\).

Answer:

Domain: \(0 < x < 4\) (Note: This is based on the x - values in the table, though the actual domain of an absolute - value function is all real numbers. There might be a formatting error in the problem), Range: \(y\leq - 2\)

So for the domain: the first box is 0, the second box is 4; for the range: the box is - 2.