Question

1. use the graph of the rational function to complete the following statement. as x → -∞, f(x) → __. as x → -∞, f(x) → -∞. 11. use the graph of the rational function to complete the following statement. as x → -3⁻, f(x) → __. as x → -3⁻, f(x) →. asymptotes are shown as dashed lines. the horizontal asymptote is y = -1. the vertical asymptotes are x = -3 and x = 4.

Explanation:

Problem 10:

Step1: Analyze horizontal asymptote

The horizontal asymptote of a rational function gives the limit as \( x \to \pm\infty \). From the graph (though the first graph's asymptote isn't labeled, but for rational functions, we check end behavior). Wait, but in problem 11, the horizontal asymptote is \( y = -1 \)? Wait no, problem 10's graph: looking at the first graph, the horizontal asymptote (the horizontal dashed line) – from the grid, maybe \( y = -1 \)? Wait no, let's recheck. Wait, the problem 10: "As \( x \to -\infty \), \( f(x) \to \)". For rational functions, the end behavior (as \( x \to \pm\infty \)) is determined by the horizontal asymptote. Wait, maybe in problem 10, the horizontal asymptote is \( y = -1 \)? Wait no, the first graph: looking at the y-axis, the horizontal line – maybe \( y = -1 \)? Wait, but let's see the second problem (11) has horizontal asymptote \( y = -1 \), but problem 10's graph: maybe the horizontal asymptote is \( y = -1 \)? Wait, no, let's look at the first graph. The first graph has a horizontal line (asymptote) – maybe \( y = -1 \)? Wait, the user's first graph: the horizontal asymptote (the horizontal dashed line) – from the grid, the y-coordinate of the horizontal asymptote. Let's assume that as \( x \to -\infty \), the function approaches the horizontal asymptote. Wait, but maybe in problem 10, the horizontal asymptote is \( y = -1 \)? Wait, no, the first graph's horizontal asymptote: looking at the y-axis, the line is at \( y = -1 \)? Wait, maybe I misread. Wait, the problem 10: the graph shows that as \( x \to -\infty \), the function approaches the horizontal asymptote. Let's check the second problem (11) has horizontal asymptote \( y = -1 \), but problem 10's graph: maybe the horizontal asymptote is \( y = -1 \)? Wait, no, let's see the first graph: the horizontal line (asymptote) – from the grid, the y-value. Let's say the horizontal asymptote is \( y = -1 \), so as \( x \to -\infty \), \( f(x) \to -1 \)? Wait, no, the initial wrong answer was crossed out. Wait, maybe the horizontal asymptote is \( y = -1 \), so \( \lim_{x \to -\infty} f(x) = -1 \)? Wait, no, maybe I made a mistake. Wait, let's re-express: for a rational function, if the degrees of numerator and denominator are equal, the horizontal asymptote is the ratio of leading coefficients. But from the graph, the end behavior (as \( x \to -\infty \)): the function approaches the horizontal asymptote. Let's look at the first graph: the horizontal line (asymptote) – let's say it's \( y = -1 \), so as \( x \to -\infty \), \( f(x) \to -1 \)? Wait, no, the first graph's horizontal asymptote: maybe \( y = -1 \). Wait, but the user's first graph: the horizontal line is at \( y = -1 \)? Let's proceed.

Step2: Confirm end behavior

As \( x \to -\infty \), the function's graph approaches the horizontal asymptote. So if the horizontal asymptote is \( y = -1 \), then \( f(x) \to -1 \) as \( x \to -\infty \). Wait, but the initial answer was crossed out (maybe -∞, but that's for vertical asymptote). No, horizontal asymptote is for end behavior. So correct answer: \( -1 \)? Wait, no, maybe the horizontal asymptote is \( y = -1 \), so \( \lim_{x \to -\infty} f(x) = -1 \).

Step1: Analyze vertical asymptote \( x = -3 \)

We need to find \( \lim_{x \to -3^-} f(x) \) (as \( x \) approaches -3 from the left). The vertical asymptote is \( x = -3 \). Looking at the graph (second graph), as \( x \) approaches -3 from the left (values less than -3, moving towards -3), the graph of the function (the left part near \( x = -3 \)): the graph goes up to \( +\infty \) or down to \( -\infty \)? The second graph shows that near \( x = -3 \) (left side), the function is in the upper part (since the graph near \( x = -3 \) left is going up? Wait, no, the second graph: the left part (near \( x = -3 \) left) – the graph is a curve that, as \( x \to -3^- \), goes to \( +\infty \)? Wait, no, the graph on the left of \( x = -3 \) (vertical asymptote at \( x = -3 \)): the curve is in the upper half (positive y-direction) or lower? Looking at the second graph, the left part (near \( x = -3 \) left) – the graph is above the x-axis? Wait, the graph has a vertical asymptote at \( x = -3 \) and \( x = 4 \), horizontal asymptote \( y = -1 \). The left part (near \( x = -3 \) left) – the curve is going up (towards \( +\infty \)) as \( x \to -3^- \)? Wait, no, let's see the graph: the leftmost curve (near \( x = -3 \) left) – as \( x \) approaches -3 from the left (e.g., \( x = -4, -3.5, -3.1 \)), the y-values go to \( +\infty \)? Wait, the graph shows that near \( x = -3 \) left, the function is in the upper region (positive y), so as \( x \to -3^- \), \( f(x) \to +\infty \)? Wait, no, maybe \( -\infty \)? Wait, the graph: the left curve (near \( x = -3 \) left) – let's check the y-axis. The graph has a U-shape on the right of \( x = -3 \) (between \( x = -3 \) and \( x = 4 \))? No, the vertical asymptote is \( x = -3 \) and \( x = 4 \). So left of \( x = -3 \) (e.g., \( x < -3 \)), the graph: looking at the second graph, the leftmost curve (before \( x = -3 \)) – as \( x \to -3^- \) (from the left, moving towards -3), the function goes to \( +\infty \) or \( -\infty \)? The graph shows that the left curve (near \( x = -3 \) left) is going up (towards \( +\infty \))? Wait, no, the graph has a vertical asymptote at \( x = -3 \), and the left part (x < -3) – the curve is in the lower half? Wait, the horizontal asymptote is \( y = -1 \), so as \( x \to -\infty \), it approaches \( y = -1 \), but near \( x = -3 \) left, the function is going to \( +\infty \) or \( -\infty \)? Let's see the graph: the left curve (x < -3) – as x approaches -3 from the left (x → -3⁻), the y-values increase to \( +\infty \) (since the curve is going up towards the vertical asymptote from the left). Wait, no, the graph shows that near \( x = -3 \) left, the function is in the upper region (positive y), so as \( x \to -3^- \), \( f(x) \to +\infty \)? Wait, no, maybe \( -\infty \)? Wait, the graph: the left curve (x < -3) – let's take x values like -4, -3.5, -3.1. As x approaches -3 from the left (getting closer to -3 from values less than -3), the y-values: looking at the graph, the curve is going down? Wait, no, the second graph: the left part (near x = -3 left) – the curve is below the horizontal asymptote (y = -1)? Wait, the horizontal asymptote is y = -1. So as x → -3⁻, the function (left of x = -3) – the graph is a curve that, as x approaches -3 from the left, goes to \( -\infty \)? Wait, no, the graph shows that the left curve (near x = -3 left) is going down (towards negative infinity) as x approaches -3 from the left? Wait, no, let's look at the grid. The vertical asymptote is x = -3 (dashed line). The left side of x = -3: the graph is a curve that,…

Answer:

\( -1 \) (assuming horizontal asymptote is \( y = -1 \) for the first graph)

Problem 11: