Question

(b) which functions have a range of y < 0? check all that apply.
□function 1 □function 2 □function 3 □none of the functions
(c) which functions have at least one y-intercept? check all that apply.
□function 1 □function 2 □function 3 □none of the functions
(d) for which functions does the graph fall to the left? that is, which functions have the end behavior that as x decreases without bound, y decreases without bound? check all that apply.
□function 1 □function 2 □function 3 □none of the functions

Explanation:

Step1: Analyze Function 1

Function 1 is a downward - opening parabola (from the graph). The vertex is above the x - axis, and the parabola opens down. The range of a downward - opening parabola is \(y\leq k\) where \(k\) is the y - coordinate of the vertex. Since the vertex is above the x - axis, the range will include values greater than 0 (the vertex y - value is positive) and some values less than or equal to that positive value. But it will have \(y\) values both positive and negative? Wait, no, a downward - opening parabola with vertex above the x - axis: if the vertex is at \(y = a,a>0\), then the parabola will cross the x - axis (since it opens down), so the range is \(y\leq a\). But does it have \(y < 0\)? Let's see, if it crosses the x - axis, then between the two x - intercepts, the parabola is below the x - axis (since it opens down). Wait, no, if the vertex is above the x - axis and it opens down, the parabola will have two x - intercepts, and between them, the parabola is above the x - axis? Wait, no, the vertex is the maximum point. So if the vertex is above the x - axis, the parabola opens down, so the graph is above the x - axis at the vertex and goes down to cross the x - axis on both sides. Wait, no, the standard form of a parabola is \(y=a(x - h)^2 + k\). If \(a<0\) (opens down) and \(k>0\) (vertex above x - axis), then when \(y = 0\), \(a(x - h)^2+k=0\), \((x - h)^2=-\frac{k}{a}\). Since \(a < 0\) and \(k>0\), \(-\frac{k}{a}>0\), so there are two real roots. The graph is above the x - axis at \(x = h\) (vertex) and decreases to cross the x - axis at two points. So between the two x - intercepts, the graph is above the x - axis? Wait, no, when \(x\) is between the two roots, \((x - h)^2<-\frac{k}{a}\), so \(y=a(x - h)^2 + k\). Since \(a<0\) and \((x - h)^2<-\frac{k}{a}\), then \(a(x - h)^2>-k\), so \(y=a(x - h)^2 + k>-k + k = 0\)? Wait, that can't be. Wait, let's take an example: \(y=-(x - 3)^2+4\). The vertex is at \((3,4)\), opens down. When \(y = 0\), \(-(x - 3)^2+4=0\), \((x - 3)^2 = 4\), \(x-3=\pm2\), \(x = 1\) or \(x = 5\). So the graph is at \(x = 3,y = 4\) (maximum), and at \(x = 1\) and \(x = 5,y = 0\). For \(x=2\), \(y=-(2 - 3)^2+4=-1 + 4=3>0\). For \(x=0\), \(y=-(0 - 3)^2+4=-9 + 4=-5<0\). Oh! So my previous reasoning was wrong. So a downward - opening parabola with vertex above the x - axis will have \(y\) values less than 0 outside the interval between the two x - intercepts. So for \(x<1\) or \(x>5\) in the example, \(y<0\), and between \(1\) and \(5\), \(y>0\). So Function 1 (the parabola) has \(y < 0\) for some \(x\) values.

Step2: Analyze Function 2

Function 2 is a linear function (from the graph, it's a straight line with a positive slope, since as \(x\) increases, \(y\) increases). A linear function has the form \(y=mx + b\). Since it has a positive slope (\(m>0\)) and we can see that when \(x\) is large enough, \(y\) is large and positive, and as \(x\) decreases, \(y\) will decrease. But a linear function with a positive slope will have \(y\) values that go from \(-\infty\) to \(+\infty\) (if \(m eq0\)). Wait, no, if it's a linear function with a positive slope, the range is all real numbers. But does it have \(y < 0\)? Let's see the graph: the line starts at some point (maybe \(x = 3\)) and goes up. If the line has a positive slope, then as \(x\) decreases below \(3\), \(y\) will decrease. So there will be some \(x\) values where \(y<0\). Wait, but maybe the line is \(y=m(x - 3)\) with \(m>0\). So when \(x<3\), \(y<0\), and when \(x>3\), \(y>0\). So Function 2 has \(y < 0\) for \(x<3\).

Step3: Analyze Function 3

Function 3 is a rational function (from the graph, it has a vertical asymptote, maybe \(x = 3\), and as \(x\) approaches \(3\) from the right, \(y\) goes to \(-\infty\), and as \(x\) approaches \(3\) from the left, \(y\) goes to \(+\infty\)? Wait, no, the graph shows that as \(x\) approaches the vertical asymptote (let's say \(x = 3\)) from the right, \(y\) goes down (negative direction), and as \(x\) approaches from the left, \(y\) is positive (since the left side of the asymptote has a horizontal asymptote at \(y = 0\) from above? Wait, the graph on the right: the function has a vertical asymptote, and a horizontal asymptote at \(y = 0\) (the x - axis) from above (as \(x\) approaches \(\pm\infty\), \(y\) approaches \(0\) from above). And near the vertical asymptote (say \(x = 3\)), as \(x\) approaches \(3\) from the right, \(y\) goes to \(-\infty\) (so \(y < 0\) there), and as \(x\) approaches \(3\) from the left, \(y\) goes to \(+\infty\) (so \(y>0\) there). So Function 3 has \(y < 0\) for \(x\) near the vertical asymptote on the right.

Wait, but the question is "Which functions have a range of \(y < 0\)?". Wait, range is the set of all \(y\) values. So we need to find functions where all \(y\) values are less than \(0\).

Let's re - analyze:

Function 1: The parabola (Function 1) has \(y\) values both less than \(0\) (outside the interval between x - intercepts) and greater than \(0\) (between x - intercepts). So its range is not \(y < 0\) (it has \(y>0\) values).

Function 2: The linear function with positive slope. Its range is all real numbers (since as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\)). So it has \(y>0\) and \(y < 0\) and \(y = 0\). So its range is not \(y < 0\).

Function 3: The rational function. Let's see the graph: it has a horizontal asymptote at \(y = 0\) (from above, as \(x\to\pm\infty\), \(y\) approaches \(0\) from above, so \(y>0\) for large \(|x|\)). And near the vertical asymptote, on the right side, \(y\) goes to \(-\infty\) (so \(y < 0\)), but on the left side of the vertical asymptote, \(y\) goes to \(+\infty\) (so \(y>0\)). So Function 3 has \(y>0\) (left of asymptote, and for large \(|x|\)) and \(y < 0\) (right of asymptote). So its range is not \(y < 0\) (it has \(y>0\) values).

So none of the functions have a range of \(y < 0\).

For part (c):

Step1: Analyze y - intercept

A y - intercept is a point where \(x = 0\). So we need to check if \(x = 0\) is in the domain of the function and find \(y\) when \(x = 0\).

Function 1: The parabola. Let's see the domain: it's a parabola, domain is all real numbers. So \(x = 0\) is in the domain. So it has a y - intercept.

Function 2: The linear function. Domain is all real numbers (since it's a line). So \(x = 0\) is in the domain, so it has a y - intercept.

Function 3: The rational function. We need to check if \(x = 0\) is in the domain. The vertical asymptote is at \(x = 3\) (from the graph), so \(x = 0\) is in the domain (since \(0 eq3\)). So we can find \(y\) when \(x = 0\). So it has a y - intercept.

Wait, but maybe the graphs are different. Wait, the first graph (Function 1) is a parabola, the second (Function 2) is a line, the third (Function 3) is a rational function.

Wait, for a y - intercept, we look at \(x = 0\).

Function 1: The parabola: when \(x = 0\), is there a point? The parabola is defined for all \(x\), so yes, it has a y - intercept.

Function 2: The line: defined for all \(x\), so when \(x = 0\), there is a \(y\) - value, so it has a y - intercept.

Function 3: The rational function: \(x = 0\) is not the vertical asymptote (vertical asymptote is at \(x = 3\)), so \(x = 0\) is in the domain, so it has a y - intercept.

Wait, but maybe the graphs are such that:

Wait, the first graph (Function 1): the parabola, when \(x = 0\), is \(x = 0\) in the visible domain? The x - axis has marks from 1 to 6, but the domain of a parabola is all real numbers, so \(x = 0\) is in the domain.

Function 2: The line, \(x = 0\) is in the domain (since it's a line, domain is \(\mathbb{R}\)).

Function 3: The rational function, \(x = 0\) is in the domain (vertical asymptote at \(x = 3\)), so \(x = 0\) is allowed.

So all three functions have at least one y - intercept? Wait, no, maybe the line (Function 2) starts at \(x = 3\)? Wait, the graph of Function 2: the x - axis has a blue marker at \(x = 5\), and the line starts at \(x = 3\)? If the domain of Function 2 is \(x\geq3\), then \(x = 0\) is not in the domain, so it has no y - intercept.

Ah, that's a key point! The blue markers on the x - axis: for Function 1, the blue marker is at \(x = 6\), but the parabola is drawn from \(x = 1\) to \(x = 6\)? No, the graphs have a horizontal axis with marks 1,2,3,4,5,6, and a blue marker (maybe indicating the domain).

Function 1: The parabola is drawn between \(x = 1\) and \(x = 6\)? No, the domain of a parabola is all real numbers, but maybe the graph is only showing a part. Wait, the blue marker is at \(x = 6\) for Function 1, Function 2 at \(x = 5\), Function 3 at \(x = 1\).

Wait, maybe the domain of Function 2 is \(x\geq3\) (since the line starts at \(x = 3\)). So if the domain of Function 2 is \(x\geq3\), then \(x = 0\) is not in the domain, so it has no y - intercept.

Function 1: If the domain is all real numbers (parabola), then \(x = 0\) is in the domain, so y - intercept exists.

Function 3: If the domain is \(x eq3\), and \(x = 0\) is in the domain (since \(0 eq3\)), then y - intercept exists.

But this is getting complicated. Maybe the intended answer for (b) is "None of the functions", for (c):

Function 1: parabola, has a y - intercept (since it's a function defined for all real numbers, so \(x = 0\) is in domain).

Function 2: if it's a line with domain \(x\geq3\), then \(x = 0\) is not in domain, so no y - intercept.

Function 3: rational function, domain \(x eq3\), so \(x = 0\) is in domain, so y - intercept.

But maybe the graphs are:

Function 1: parabola, domain all real numbers, so y - intercept.

Function 2: line, domain \(x\geq3\), so no y - intercept (since \(x = 0<3\) is not in domain).

Function 3: rational function, domain \(x eq3\), so \(x = 0\) is in domain, so y - intercept.

But this is unclear. Alternatively, maybe all three functions have a y - intercept. But let's assume the standard case.

For part (d):

End behavior: as \(x\to-\infty\) (x decreases without bound), \(y\to-\infty\) (y decreases without bound).

Function 1: parabola, downward - opening. The end behavior of a parabola \(y=ax^2+bx + c\) with \(a<0\) is as \(x\to\pm\infty\), \(y\to-\infty\). Wait, no: for \(y = ax^2+bx + c\), if \(a<0\), as \(x\to\pm\infty\), \(ax^2\) dominates, so \(y\to-\infty\) (since \(a<0\) and \(x^2\) is positive and large). So for Function 1 (downward - opening parabola), as \(x\to-\infty\), \(y\to-\infty\) (since \(a<0\), \(x^2\) is large positive, \(ax^2\) is large negative). So end behavior: as \(x\to-\infty\), \(y\to-\infty\) (falls to the left).

Function 2: linear function with positive slope (\(y=mx + b,m>0\)). As \(x\to-\infty\), \(y=mx + b\to-\infty\) (since \(m>0\), multiplying by \(-\infty\) gives \(-\infty\)). Wait, no: if \(m>0\), as \(x\to-\infty\), \(mx\to-\infty\), so \(y\to-\infty\). So as \(x\) decreases without bound, \(y\) decreases without bound? Wait, no: if \(m>0\), \(y=mx + b\), when \(x\) increases, \(y\) increases; when \(x\) decreases, \(y\) decreases. So as \(x\to-\infty\), \(y\to-\infty\) (falls to the left).

Function 3: rational function. Let's assume it's of the form \(y=\frac{k}{x - 3}\). As \(x\to-\infty\), \(y=\frac{k}{x - 3}\to0\) (since \(x-3\to-\infty\), so \(\frac{k}{x - 3}\to0\) if \(k\) is a constant). So as \(x\to-\infty\), \(y\) approaches \(0\) (from above, since \(k>0\) maybe), so it doesn't fall to the left (it approaches \(0\)).

But this is getting too complex. Since the problem is about function graphs and their properties, the discipline is Mathematics, subfield is Calculus (or Algebra, specifically Functions).

But let's focus on part (b) first.

For part (b):

We need functions where all \(y\) - values are less than \(0\) (range \(y < 0\)).

- Function 1: Parabola (downward - opening). It has a maximum point (vertex) above the x - axis (since it's a downward - opening parabola with a peak). So it will have \(y\) - values greater than \(0\) (at the vertex and between x - intercepts) and less than \(0\) (outside the x - intercepts). So its range includes \(y>0\), so not \(y < 0\).

Answer:

Step1: Analyze Function 1

Function 1 is a downward - opening parabola (from the graph). The vertex is above the x - axis, and the parabola opens down. The range of a downward - opening parabola is \(y\leq k\) where \(k\) is the y - coordinate of the vertex. Since the vertex is above the x - axis, the range will include values greater than 0 (the vertex y - value is positive) and some values less than or equal to that positive value. But it will have \(y\) values both positive and negative? Wait, no, a downward - opening parabola with vertex above the x - axis: if the vertex is at \(y = a,a>0\), then the parabola will cross the x - axis (since it opens down), so the range is \(y\leq a\). But does it have \(y < 0\)? Let's see, if it crosses the x - axis, then between the two x - intercepts, the parabola is below the x - axis (since it opens down). Wait, no, if the vertex is above the x - axis and it opens down, the parabola will have two x - intercepts, and between them, the parabola is above the x - axis? Wait, no, the vertex is the maximum point. So if the vertex is above the x - axis, the parabola opens down, so the graph is above the x - axis at the vertex and goes down to cross the x - axis on both sides. Wait, no, the standard form of a parabola is \(y=a(x - h)^2 + k\). If \(a<0\) (opens down) and \(k>0\) (vertex above x - axis), then when \(y = 0\), \(a(x - h)^2+k=0\), \((x - h)^2=-\frac{k}{a}\). Since \(a < 0\) and \(k>0\), \(-\frac{k}{a}>0\), so there are two real roots. The graph is above the x - axis at \(x = h\) (vertex) and decreases to cross the x - axis at two points. So between the two x - intercepts, the graph is above the x - axis? Wait, no, when \(x\) is between the two roots, \((x - h)^2<-\frac{k}{a}\), so \(y=a(x - h)^2 + k\). Since \(a<0\) and \((x - h)^2<-\frac{k}{a}\), then \(a(x - h)^2>-k\), so \(y=a(x - h)^2 + k>-k + k = 0\)? Wait, that can't be. Wait, let's take an example: \(y=-(x - 3)^2+4\). The vertex is at \((3,4)\), opens down. When \(y = 0\), \(-(x - 3)^2+4=0\), \((x - 3)^2 = 4\), \(x-3=\pm2\), \(x = 1\) or \(x = 5\). So the graph is at \(x = 3,y = 4\) (maximum), and at \(x = 1\) and \(x = 5,y = 0\). For \(x=2\), \(y=-(2 - 3)^2+4=-1 + 4=3>0\). For \(x=0\), \(y=-(0 - 3)^2+4=-9 + 4=-5<0\). Oh! So my previous reasoning was wrong. So a downward - opening parabola with vertex above the x - axis will have \(y\) values less than 0 outside the interval between the two x - intercepts. So for \(x<1\) or \(x>5\) in the example, \(y<0\), and between \(1\) and \(5\), \(y>0\). So Function 1 (the parabola) has \(y < 0\) for some \(x\) values.

Step2: Analyze Function 2

Function 2 is a linear function (from the graph, it's a straight line with a positive slope, since as \(x\) increases, \(y\) increases). A linear function has the form \(y=mx + b\). Since it has a positive slope (\(m>0\)) and we can see that when \(x\) is large enough, \(y\) is large and positive, and as \(x\) decreases, \(y\) will decrease. But a linear function with a positive slope will have \(y\) values that go from \(-\infty\) to \(+\infty\) (if \(m
eq0\)). Wait, no, if it's a linear function with a positive slope, the range is all real numbers. But does it have \(y < 0\)? Let's see the graph: the line starts at some point (maybe \(x = 3\)) and goes up. If the line has a positive slope, then as \(x\) decreases below \(3\), \(y\) will decrease. So there will be some \(x\) values where \(y<0\). Wait, but maybe the line is \(y=m(x - 3)\) with \(m>0\). So when \(x<3\), \(y<0\), and when \(x>3\), \(y>0\). So Function 2 has \(y < 0\) for \(x<3\).

Step3: Analyze Function 3

Function 3 is a rational function (from the graph, it has a vertical asymptote, maybe \(x = 3\), and as \(x\) approaches \(3\) from the right, \(y\) goes to \(-\infty\), and as \(x\) approaches \(3\) from the left, \(y\) goes to \(+\infty\)? Wait, no, the graph shows that as \(x\) approaches the vertical asymptote (let's say \(x = 3\)) from the right, \(y\) goes down (negative direction), and as \(x\) approaches from the left, \(y\) is positive (since the left side of the asymptote has a horizontal asymptote at \(y = 0\) from above? Wait, the graph on the right: the function has a vertical asymptote, and a horizontal asymptote at \(y = 0\) (the x - axis) from above (as \(x\) approaches \(\pm\infty\), \(y\) approaches \(0\) from above). And near the vertical asymptote (say \(x = 3\)), as \(x\) approaches \(3\) from the right, \(y\) goes to \(-\infty\) (so \(y < 0\) there), and as \(x\) approaches \(3\) from the left, \(y\) goes to \(+\infty\) (so \(y>0\) there). So Function 3 has \(y < 0\) for \(x\) near the vertical asymptote on the right.

Wait, but the question is "Which functions have a range of \(y < 0\)?". Wait, range is the set of all \(y\) values. So we need to find functions where all \(y\) values are less than \(0\).

Let's re - analyze:

Function 1: The parabola (Function 1) has \(y\) values both less than \(0\) (outside the interval between x - intercepts) and greater than \(0\) (between x - intercepts). So its range is not \(y < 0\) (it has \(y>0\) values).

Function 2: The linear function with positive slope. Its range is all real numbers (since as \(x\to-\infty\), \(y\to-\infty\) and as \(x\to+\infty\), \(y\to+\infty\)). So it has \(y>0\) and \(y < 0\) and \(y = 0\). So its range is not \(y < 0\).

Function 3: The rational function. Let's see the graph: it has a horizontal asymptote at \(y = 0\) (from above, as \(x\to\pm\infty\), \(y\) approaches \(0\) from above, so \(y>0\) for large \(|x|\)). And near the vertical asymptote, on the right side, \(y\) goes to \(-\infty\) (so \(y < 0\)), but on the left side of the vertical asymptote, \(y\) goes to \(+\infty\) (so \(y>0\)). So Function 3 has \(y>0\) (left of asymptote, and for large \(|x|\)) and \(y < 0\) (right of asymptote). So its range is not \(y < 0\) (it has \(y>0\) values).

So none of the functions have a range of \(y < 0\).

For part (c):

Step1: Analyze y - intercept

A y - intercept is a point where \(x = 0\). So we need to check if \(x = 0\) is in the domain of the function and find \(y\) when \(x = 0\).

Function 1: The parabola. Let's see the domain: it's a parabola, domain is all real numbers. So \(x = 0\) is in the domain. So it has a y - intercept.

Function 2: The linear function. Domain is all real numbers (since it's a line). So \(x = 0\) is in the domain, so it has a y - intercept.

Function 3: The rational function. We need to check if \(x = 0\) is in the domain. The vertical asymptote is at \(x = 3\) (from the graph), so \(x = 0\) is in the domain (since \(0
eq3\)). So we can find \(y\) when \(x = 0\). So it has a y - intercept.

Wait, but maybe the graphs are different. Wait, the first graph (Function 1) is a parabola, the second (Function 2) is a line, the third (Function 3) is a rational function.

Wait, for a y - intercept, we look at \(x = 0\).

Function 1: The parabola: when \(x = 0\), is there a point? The parabola is defined for all \(x\), so yes, it has a y - intercept.

Function 2: The line: defined for all \(x\), so when \(x = 0\), there is a \(y\) - value, so it has a y - intercept.

Function 3: The rational function: \(x = 0\) is not the vertical asymptote (vertical asymptote is at \(x = 3\)), so \(x = 0\) is in the domain, so it has a y - intercept.

Wait, but maybe the graphs are such that:

Wait, the first graph (Function 1): the parabola, when \(x = 0\), is \(x = 0\) in the visible domain? The x - axis has marks from 1 to 6, but the domain of a parabola is all real numbers, so \(x = 0\) is in the domain.

Function 2: The line, \(x = 0\) is in the domain (since it's a line, domain is \(\mathbb{R}\)).

Function 3: The rational function, \(x = 0\) is in the domain (vertical asymptote at \(x = 3\)), so \(x = 0\) is allowed.

So all three functions have at least one y - intercept? Wait, no, maybe the line (Function 2) starts at \(x = 3\)? Wait, the graph of Function 2: the x - axis has a blue marker at \(x = 5\), and the line starts at \(x = 3\)? If the domain of Function 2 is \(x\geq3\), then \(x = 0\) is not in the domain, so it has no y - intercept.

Ah, that's a key point! The blue markers on the x - axis: for Function 1, the blue marker is at \(x = 6\), but the parabola is drawn from \(x = 1\) to \(x = 6\)? No, the graphs have a horizontal axis with marks 1,2,3,4,5,6, and a blue marker (maybe indicating the domain).

Function 1: The parabola is drawn between \(x = 1\) and \(x = 6\)? No, the domain of a parabola is all real numbers, but maybe the graph is only showing a part. Wait, the blue marker is at \(x = 6\) for Function 1, Function 2 at \(x = 5\), Function 3 at \(x = 1\).

Wait, maybe the domain of Function 2 is \(x\geq3\) (since the line starts at \(x = 3\)). So if the domain of Function 2 is \(x\geq3\), then \(x = 0\) is not in the domain, so it has no y - intercept.

Function 1: If the domain is all real numbers (parabola), then \(x = 0\) is in the domain, so y - intercept exists.

Function 3: If the domain is \(x
eq3\), and \(x = 0\) is in the domain (since \(0
eq3\)), then y - intercept exists.

But this is getting complicated. Maybe the intended answer for (b) is "None of the functions", for (c):

Function 1: parabola, has a y - intercept (since it's a function defined for all real numbers, so \(x = 0\) is in domain).

Function 2: if it's a line with domain \(x\geq3\), then \(x = 0\) is not in domain, so no y - intercept.

Function 3: rational function, domain \(x
eq3\), so \(x = 0\) is in domain, so y - intercept.

But maybe the graphs are:

Function 1: parabola, domain all real numbers, so y - intercept.

Function 2: line, domain \(x\geq3\), so no y - intercept (since \(x = 0<3\) is not in domain).

Function 3: rational function, domain \(x
eq3\), so \(x = 0\) is in domain, so y - intercept.

But this is unclear. Alternatively, maybe all three functions have a y - intercept. But let's assume the standard case.

For part (d):

End behavior: as \(x\to-\infty\) (x decreases without bound), \(y\to-\infty\) (y decreases without bound).

Function 1: parabola, downward - opening. The end behavior of a parabola \(y=ax^2+bx + c\) with \(a<0\) is as \(x\to\pm\infty\), \(y\to-\infty\). Wait, no: for \(y = ax^2+bx + c\), if \(a<0\), as \(x\to\pm\infty\), \(ax^2\) dominates, so \(y\to-\infty\) (since \(a<0\) and \(x^2\) is positive and large). So for Function 1 (downward - opening parabola), as \(x\to-\infty\), \(y\to-\infty\) (since \(a<0\), \(x^2\) is large positive, \(ax^2\) is large negative). So end behavior: as \(x\to-\infty\), \(y\to-\infty\) (falls to the left).

Function 2: linear function with positive slope (\(y=mx + b,m>0\)). As \(x\to-\infty\), \(y=mx + b\to-\infty\) (since \(m>0\), multiplying by \(-\infty\) gives \(-\infty\)). Wait, no: if \(m>0\), as \(x\to-\infty\), \(mx\to-\infty\), so \(y\to-\infty\). So as \(x\) decreases without bound, \(y\) decreases without bound? Wait, no: if \(m>0\), \(y=mx + b\), when \(x\) increases, \(y\) increases; when \(x\) decreases, \(y\) decreases. So as \(x\to-\infty\), \(y\to-\infty\) (falls to the left).

Function 3: rational function. Let's assume it's of the form \(y=\frac{k}{x - 3}\). As \(x\to-\infty\), \(y=\frac{k}{x - 3}\to0\) (since \(x-3\to-\infty\), so \(\frac{k}{x - 3}\to0\) if \(k\) is a constant). So as \(x\to-\infty\), \(y\) approaches \(0\) (from above, since \(k>0\) maybe), so it doesn't fall to the left (it approaches \(0\)).

But this is getting too complex. Since the problem is about function graphs and their properties, the discipline is Mathematics, subfield is Calculus (or Algebra, specifically Functions).

But let's focus on part (b) first.

For part (b):

We need functions where all \(y\) - values are less than \(0\) (range \(y < 0\)).

• Function 1: Parabola (downward - opening). It has a maximum point (vertex) above the x - axis (since it's a downward - opening parabola with a peak). So it will have \(y\) - values greater than \(0\) (at the vertex and between x - intercepts) and less than \(0\) (outside the x - intercepts). So its range includes \(y>0\), so not \(y < 0\).