Question

drag the labels to the correct locations on the image. not all labels will be used.
consider function h.
what is the range of function h?
y ∞ x -1 2 -4 -∞ -2

Explanation:

Step1: Analyze the function's graph

The function \( h(x) \) has a vertical asymptote. From the graph, we can see the behavior of the function. The range of a function is the set of all possible \( y \)-values. Looking at the graph, the function seems to have a range that excludes some values, but from the labels and the graph's behavior (the two curves and the asymptote), we can infer the range. The dashed line and the curves suggest that the function's \( y \)-values are such that \( y eq - 2 \)? Wait, no, let's re - examine. Wait, the graph has two parts. Wait, maybe the function is a rational function or a transformed function. Wait, the key is to look at the vertical asymptote and the horizontal behavior. Wait, the labels given are \( y \), \( \infty \), \( x \), \( - 1 \), \( 2 \), \( - 4 \), \( -\infty \), \( - 2 \). Wait, the range is about \( y \)-values. Let's think about the graph: the two curves, one on the left of the vertical asymptote (around \( x = 1 \) maybe? Wait, the dashed vertical line is at \( x = 1 \)? Wait, no, the grid: from the graph, the vertical dashed line is at \( x = 1 \) (since between 0 and 2). Wait, the function's graph: the left - most curve (purple) and the right - most curve (blue). Wait, the left curve passes through \( (0,4) \), \( (-2,0) \), etc. The right curve starts after \( x = 2 \). Wait, but the range: looking at the \( y \)-axis, the function's \( y \)-values. Wait, the labels for the range should be in terms of \( y \), with \( -\infty \), \( y \), and \( \infty \) or a specific value. Wait, maybe the function has a horizontal asymptote? No, the dashed line is a slant asymptote? Wait, the dashed line passes through the origin and has a slope. Wait, maybe the range is \( y eq - 2 \)? No, the labels given: \( -\infty \), \( y \), \( \infty \) but with a middle value? Wait, no, the format is \( \square<y<\square \) or \( y eq\square \)? Wait, the problem has three boxes: \( \square<y<\square \) or maybe \( -\infty < y < \infty \) except \( y=-2 \)? Wait, no, the labels include \( - 2 \), \( -\infty \), \( \infty \), \( y \). Wait, maybe the function's range is all real numbers except \( y = - 2 \), but the format here is \( -\infty < y < \infty \) with \( y eq - 2 \)? But the boxes are three: first box, then \( <y< \) second box, and third box? Wait, maybe I misread. Wait, the problem says "Drag the labels to the correct locations on the image. Not all labels will be used. What is the range of function \( h \)?" and the boxes are \( \square<\square<\square \) with the middle square being \( y \). Wait, no, the middle symbol is \( <y< \), so the first box is the lower bound, the middle is \( y \), the second box is the upper bound? Wait, no, the format is \( \square<y<\square \) or \( -\infty < y < \infty \) but with a restriction. Wait, looking at the graph, the vertical asymptote and the horizontal behavior. Wait, the left curve: as \( x\to -\infty \), \( y\to -\infty \), and as \( x\to 1^{-} \) (left of vertical asymptote), what happens to \( y \)? The left curve at \( x = 0 \) is \( y = 4 \), as \( x\to 1^{-} \), \( y\to \infty \) (since the curve goes up near the vertical asymptote). The right curve: as \( x\to 1^{+} \), \( y\to -\infty \), and as \( x\to\infty \), \( y\to \infty \). Wait, no, the right curve starts at \( x>2 \) and goes up. Wait, maybe the range is all real numbers except \( y=-2 \)? No, the labels given: \( -\infty \), \( y \), \( \infty \), \( - 2 \). Wait, maybe the correct labels are \( -\infty < y < \infty \), but with \( y eq - 2 \)? But the boxes are three: first \( -\infty \), then \( y \), then \( \infty \)? No, the format is \( \square<y<\square \) with a third box? Wait, maybe the problem has a typo, but looking at the labels, the only value that makes sense for the range's exclusion (if any) is \( - 2 \), but the format here is \( -\infty < y < \infty \) with \( y eq - 2 \), but the boxes are \( \square<y<\square \) and a third box? Wait, no, maybe the function's range is \( -\infty < y < \infty \), but that can't be. Wait, re - examining the graph: the left curve (purple) and the right curve (blue). The left curve: when \( x = 0 \), \( y = 4 \); when \( x=-2 \), \( y = 0 \); as \( x\to -\infty \), \( y\to -\infty \); as \( x\to 1^{-} \), \( y\to \infty \). The right curve: as \( x\to 1^{+} \), \( y\to -\infty \); as \( x\to\infty \), \( y\to \infty \); and it passes through \( (3, \text{some positive value}) \). So the range of \( y \) is all real numbers except \( y=-2 \)? No, the labels include \( - 2 \). Wait, maybe the vertical asymptote is at \( x = 1 \), and the slant asymptote (dashed line) has a \( y \)-intercept of 0 and slope 1, so \( y=x \). But the range: the function \( h(x) \) has a range where \( y eq - 2 \)? No, I think I made a mistake. Wait, the problem's labels: the three boxes are for the range. The first box should be \( -\infty \), the middle is \( y \), and the third is \( \infty \), but that would be \( -\infty < y < \infty \), but that's the range of a linear function. But the graph has vertical asymptotes, so the range is not all real numbers. Wait, maybe the function is \( h(x)=\frac{x + 2}{x - 1}+x \)? No, too complicated. Wait, the key is to look at the given labels. The labels are \( -\infty \), \( y \), \( \infty \), \( - 2 \). Wait, maybe the range is \( -\infty < y < \infty \), \( y eq - 2 \), but the format here is \( -\infty < y < \infty \) with the middle being \( y \) and the first \( -\infty \), third \( \infty \), but there's a third box? Wait, no, maybe the problem is mis - formatted, but based on the graph and the labels, the range is all real numbers except \( y=-2 \), but the labels given: the first box is \( -\infty \), the middle is \( y \), the third is \( \infty \), but that's not excluding \( - 2 \). Wait, maybe I'm wrong. Wait, the vertical asymptote is at \( x = 1 \), and the function's graph: the left part (purple) and right part (blue). The left part: as \( x\to 1^{-} \), \( y\to \infty \); as \( x\to -\infty \), \( y\to -\infty \). The right part: as \( x\to 1^{+} \), \( y\to -\infty \); as \( x\to \infty \), \( y\to \infty \). So the range of \( y \) is all real numbers, \( (-\infty,\infty) \), but that would mean \( -\infty < y < \infty \). But the labels include \( - 2 \), maybe it's a mistake. Alternatively, maybe the function has a horizontal asymptote at \( y=-2 \), but the graph doesn't show that. Wait, the dashed line is a slant asymptote \( y = x \), so the range is all real numbers except \( y=-2 \)? No, I think the correct labels to drag are \( -\infty \), \( y \), \( \infty \) with the understanding that the range is all real numbers, but that seems odd. Wait, maybe the problem is about the domain, but no, it's the range. Wait, the three boxes: first \( -\infty \), then \( y \), then \( \infty \), so \( -\infty < y < \infty \).

Step2: Confirm the range

From the graph of the function \( h(x) \), we can see that the function's \( y \)-values extend from negative infinity to positive infinity (as the curves go down to \( -\infty \) and up to \( \infty \) on both sides of the vertical asymptote). So the range is \( -\infty < y < \infty \).

Answer:

\( -\infty < y < \infty \) (with the labels \( -\infty \), \( y \), \( \infty \) placed in the respective boxes: first box \( -\infty \), middle box \( y \), third box \( \infty \))