Question

decide whether the function has extrema: extremum (0, -1) relative minimum relative maximum absolute minimum absolute maximum select a statement that describes the function: add extremum no more extrema identify the increasing or decreasing intervals. interval increasing or decreasing describe the end behavior: as x → -∞, y → as x → ∞, y → ing key characteristics of polynomial functions

Explanation:

Step1: Analyze End Behavior (As \( x \to -\infty \))

Observe the left - hand side of the graph (as \( x \) values become very small, i.e., \( x \to -\infty \)). The graph seems to approach a horizontal line (the horizontal asymptote). From the visual, as \( x \to -\infty \), the \( y \) - value of the function approaches \( 2 \)? Wait, no, looking at the graph, the left - most part of the blue curve is at \( y = 2 \) (approximately) and as \( x \) goes to \( -\infty \), it stays at that level? Wait, no, maybe I misread. Wait, the graph has a horizontal asymptote? Wait, no, the red dashed line is \( y = 0 \)? Wait, no, the blue curve: on the left, when \( x \to -\infty \), the \( y \) - value is approaching \( 2 \)? Wait, no, the user's previous input had some pre - filled values as \( -\infty \), but let's re - examine. Wait, the graph: the blue curve, as \( x \to -\infty \), the \( y \) - coordinate seems to approach \( 2 \)? But the pre - filled box was \( -\infty \), maybe that's wrong. Wait, no, maybe it's a rational function? Wait, the vertical dashed line is at \( x=-1 \), so it's a function with a vertical asymptote at \( x = - 1 \). For the end behavior as \( x\to-\infty \): we look at the part of the graph where \( x < - 1 \) (left of the vertical asymptote). The blue curve on the left ( \( x < - 1 \)) is a horizontal line? Wait, no, the blue curve on the left (for \( x < - 1 \)) is at \( y = 2 \) (from the grid, each square is 1 unit). So as \( x\to-\infty \), \( y\to 2 \)? But the pre - filled option was \( -\infty \), maybe that's an error. Wait, maybe the function is a rational function. Let's think again.

Wait, the problem is about describing end behavior. Let's assume that maybe the graph is of a function where as \( x\to-\infty \) and \( x\to\infty \), the function approaches a horizontal line. But the pre - filled boxes were \( -\infty \), but maybe that's incorrect. Wait, no, maybe the graph is of a function like \( y=\frac{something}{(x + 1)}+...\). Wait, the vertical asymptote is at \( x=-1 \). For the right - hand side ( \( x\to\infty \) ), the blue curve approaches \( y = 0 \) (the red dashed line). For the left - hand side ( \( x\to-\infty \) ), the blue curve approaches \( y = 2 \)? But the user's pre - filled boxes are \( -\infty \), maybe there's a mistake in the pre - filled values. Wait, maybe the function is a polynomial? No, it has a vertical asymptote, so it's a rational function.

Wait, maybe the original problem had a different graph. Since the user provided a graph with a vertical asymptote at \( x=-1 \), horizontal asymptotes: as \( x\to\infty \), \( y\to 0 \) (the red dashed line), and as \( x\to-\infty \), \( y\to 2 \) (the horizontal line that the left - most part of the blue curve is on). But the pre - filled boxes were \( -\infty \), maybe that's a mistake. But since the user is asking for help, let's proceed with the visual.

Wait, maybe the function is \( y=\frac{2(x + 1)-2}{x + 1}=\frac{2x+2 - 2}{x + 1}=\frac{2x}{x + 1}=2-\frac{2}{x + 1}\). Let's check: vertical asymptote at \( x=-1 \), horizontal asymptote \( y = 2 \) as \( x\to-\infty \) (because \( \frac{2}{x + 1}\to 0 \) as \( x\to-\infty \), so \( y\to 2 \)) and \( y\to 2 \) as \( x\to\infty \)? No, \( \frac{2}{x + 1}\to 0 \) as \( x\to\infty \), so \( y\to 2 \). But the red dashed line is \( y = 0 \), so maybe my function is wrong.

Alternatively, maybe the function is \( y=\frac{-2}{x + 1}\). Then vertical asymptote at \( x=-1 \), horizontal asymptote \( y = 0 \). As \( x\to-\infty \), \( \frac{-2}{x + 1}\to 0 \) (because \( x+1\to-\infty \), so \( \frac{-2}{x + 1}\to 0 \)). As \( x\to\infty \), \( \frac{-2}{x + 1}\to 0 \). But the blue curve: on the left ( \( x < - 1 \) ), if \( y=\frac{-2}{x + 1} \), when \( x\to-\infty \), \( y\to 0 \), and when \( x\) approaches \( - 1 \) from the left ( \( x\to - 1^{-} \) ), \( y\to\infty \), and from the right ( \( x\to - 1^{+} \) ), \( y\to-\infty \). But the blue curve in the graph: at \( x = 0 \), it has a point at \( (0,-1) \). Wait, maybe the function is \( y=\frac{x - 0}{x + 1}=\frac{x}{x + 1}=1-\frac{1}{x + 1}\). Then vertical asymptote at \( x=-1 \), horizontal asymptote \( y = 1 \). As \( x\to-\infty \), \( \frac{1}{x + 1}\to 0 \), so \( y\to 1 \). As \( x\to\infty \), \( y\to 1 \). But the blue curve at \( x = 0 \) is at \( y=-1 \)? No, that doesn't match.

Wait, maybe the graph is of a function with end behavior as \( x\to-\infty \), \( y\to -\infty \) and \( x\to\infty \), \( y\to -\infty \)? No, the right - hand side of the graph ( \( x\to\infty \) ) has the blue curve approaching \( y = 0 \) (the red dashed line).

I think there's a misinterpretation here. Let's go back to the problem. The user's graph: the blue curve, on the left ( \( x\to-\infty \) ), the \( y \) - value is constant (horizontal line) at \( y = 2 \), and on the right ( \( x\to\infty \) ), the \( y \) - value approaches \( 0 \). But the pre - filled boxes were \( -\infty \), maybe that's an error. However, since the problem is about describing end behavior, and assuming that maybe the intended answer is that as \( x\to-\infty \), \( y\to 2 \) and as \( x\to\infty \), \( y\to 0 \), but the pre - filled options were \( -\infty \), which is confusing.

Wait, maybe the function is a polynomial? No, it has a vertical asymptote, so it's a rational function. Let's try to answer based on the visual:

Step1: End Behavior as \( x\to-\infty \)

Look at the part of the graph where \( x \) is very small ( \( x\to-\infty \) ), left of the vertical asymptote \( x=-1 \). The blue curve here is a horizontal line at \( y = 2 \) (from the grid, each square is 1 unit). So as \( x\to-\infty \), \( y\to 2 \). But the pre - filled box was \( -\infty \), maybe that's wrong.

Step2: End Behavior as \( x\to\infty \)

Look at the part of the graph where \( x \) is very large ( \( x\to\infty \) ), right of the vertical asymptote \( x=-1 \). The blue curve here approaches the red dashed line \( y = 0 \). So as \( x\to\infty \), \( y\to 0 \).

But the user's pre - filled boxes were \( -\infty \) for both, which is incorrect. Maybe there was a mistake in the problem setup. Alternatively, maybe the graph is of a different function.

Wait, maybe the function is \( y=\frac{-2}{x + 1}\). Then:

- As \( x\to-\infty \): \( \frac{-2}{x + 1}\to 0 \) (because \( x+1\to-\infty \), so the fraction approaches 0).
- As \( x\to\infty \): \( \frac{-2}{x + 1}\to 0 \).

But the blue curve at \( x = 0 \) is at \( y=-1 \) (since \( \frac{-2}{0 + 1}=-2 \), no, that doesn't match. Wait, at \( x = 0 \), the point is \( (0,-1) \), so if \( y=\frac{-1\times x}{x + 1}=\frac{-x}{x + 1} \). Then at \( x = 0 \), \( y = 0 \), no. At \( x = 0 \), the point is \( (0,-1) \), so \( y=\frac{-1}{x + 1} \). At \( x = 0 \), \( y=-1 \), correct. Then:

- As \( x\to-\infty \): \( \frac{-1}{x + 1}\to 0 \) (because \( x+1\to-\infty \), so the fraction approaches 0).
- As \( x\to\infty \): \( \frac{-1}{x + 1}\to 0 \).

But the left - hand side of the graph ( \( x\to-\infty \) ) for \( y=\frac{-1}{x + 1} \) is a horizontal line approaching \( y = 0 \), and the right - hand side also approaches \( y = 0 \). But the blue curve on the left ( \( x < - 1 \) ) for \( y=\frac{-1}{x + 1} \) would be positive (since \( x+1<0 \) when \( x < - 1 \), so \( \frac{-1}{x + 1}>0 \)), but the blue curve on the left is at \( y = 2 \), which doesn't match.

I think there's a miscommunication here. Given that the pre - filled boxes were \( -\infty \), maybe the intended function is a polynomial with leading coefficient negative and even degree? No, it has a vertical asymptote.

Alternatively, maybe the problem is about a different graph. Given the confusion, and assuming that the intended answer (maybe a mistake in the problem) is that as \( x\to-\infty \), \( y\to -\infty \) and as \( x\to\infty \), \( y\to -\infty \), but that doesn't match the visual.

Wait, the blue curve: on the left ( \( x < - 1 \) ), it's a horizontal line at \( y = 2 \), on the right ( \( x > - 1 \) ), it has a "wave" and approaches \( y = 0 \) as \( x\to\infty \). So for the end behavior:

- As \( x\to-\infty \), \( y\to 2 \)
- As \( x\to\infty \), \( y\to 0 \)

But since the pre - filled boxes were \( -\infty \), maybe the problem had a different graph. Given the time I've spent, I'll assume that maybe the intended answer is that as \( x\to-\infty \), \( y\to 2 \) and as \( x\to\infty \), \( y\to 0 \), but I'm not sure.

Answer:

As \( x\to-\infty \), \( y\to 2 \); As \( x\to\infty \), \( y\to 0 \) (but this is based on the visual, the pre - filled \( -\infty \) may be incorrect)