Question

svlc algebra 1a - standard (15260)
representing relationships
x | f(x)
-2 | -8
-1 | -3
0 | -2
1 | 4
2 | 1
3 | 3
what ordered pair is closest to a local minimum of the function, f(x)?
(1, 4) (-1, -3)
(2, 1) (0, -2)

Explanation:

Step1: Understand local minimum

A local minimum is a point where the function value is lower than its neighboring points. We analyze the given \( x \) and \( f(x) \) values.

Step2: Analyze each ordered pair

- For \( (1, 4) \): \( f(1) = 4 \). Neighboring points: \( f(0) = -2 \), \( f(2) = 1 \). 4 is higher than both, so not a minimum.
- For \( (-1, -3) \): \( f(-1) = -3 \). Neighboring points: \( f(-2) = -8 \), \( f(0) = -2 \). -3 is higher than \( f(-2) \) (-8) and lower than \( f(0) \) (-2). But let's check others.
- For \( (2, 1) \): \( f(2) = 1 \). Neighboring points: \( f(1) = 4 \), \( f(3) = 3 \). 1 is lower than both? No, 1 is lower than 4 and 3? Wait, 1 is less than 4 and 3? Wait, 1 is less than 4 (yes) and less than 3 (no, 1 < 3? Wait 1 is less than 3? Yes, 1 < 3. Wait, no, 1 is less than 3? Wait 1 is 1, 3 is 3. So 1 < 3. But let's check the other point.
- For \( (0, -2) \): \( f(0) = -2 \). Neighboring points: \( f(-1) = -3 \), \( f(1) = 4 \). -2 is higher than \( f(-1) \) (-3) and lower than \( f(1) \) (4). Now compare \( f(-1) = -3 \), \( f(0) = -2 \), \( f(2) = 1 \). Wait, the local minimum should be around where the function changes from decreasing to increasing. Let's list the \( f(x) \) values in order of \( x \): \( x=-2, f=-8 \); \( x=-1, f=-3 \); \( x=0, f=-2 \); \( x=1, f=4 \); \( x=2, f=1 \); \( x=3, f=3 \). Wait, from \( x=-2 \) to \( x=-1 \): \( f \) increases from -8 to -3. From \( x=-1 \) to \( x=0 \): \( f \) increases from -3 to -2. From \( x=0 \) to \( x=1 \): \( f \) increases to 4. Then decreases to 1 at \( x=2 \), then increases to 3 at \( x=3 \). Wait, but the local minimum would be around the lowest point in a small interval. Wait, the values: \( f(-2) = -8 \) (very low), but the options don't have \( (-2, -8) \). The options are \( (1,4) \), \( (-1,-3) \), \( (2,1) \), \( (0,-2) \). Wait, maybe I misread. Wait the question is "closest to a local minimum". Wait, maybe the function's behavior: let's see the trend. From \( x=-2 \) to \( x=0 \), \( f(x) \) is increasing (from -8 to -3 to -2). Then from \( x=0 \) to \( x=1 \), it increases to 4. Then decreases to 1 at \( x=2 \), then increases to 3 at \( x=3 \). Wait, but the local minimum in the given options: the lowest \( f(x) \) among the options is \( (-1, -3) \) with \( f=-3 \), then \( (0, -2) \) with \( f=-2 \), then \( (2,1) \), then \( (1,4) \). But wait, maybe the local minimum is around where the function changes direction. Wait, maybe I made a mistake. Wait, let's check the values again. The \( f(x) \) values:

\( x=-2 \): -8

\( x=-1 \): -3

\( x=0 \): -2

\( x=1 \): 4

\( x=2 \): 1

\( x=3 \): 3

So from \( x=-2 \) to \( x=0 \), the function is increasing (since -8 < -3 < -2). Then from \( x=0 \) to \( x=1 \), it increases to 4. Then decreases to 1 at \( x=2 \), then increases to 3 at \( x=3 \). So the local minimum in the interval around the given points: the lowest value among the options is \( (-1, -3) \) with \( f=-3 \), but wait \( x=-2 \) has \( f=-8 \) which is lower, but it's not an option. Wait the options are \( (1,4) \), \( (-1,-3) \), \( (2,1) \), \( (0,-2) \). Wait, maybe the question is about local minimum in the visible points. Wait, maybe I misinterpret. Wait, a local minimum is a point where the function value is less than or equal to its immediate neighbors. Let's check each option's neighbors:

- \( (-1, -3) \): neighbors \( x=-2 \) (f=-8) and \( x=0 \) (f=-2). So \( f(-1) = -3 \) is greater than \( f(-2) = -8 \), so it's not a local minimum (since left neighbor is lower).
- \( (0, -2) \): neighbors \( x=-1 \) (f=-3) and \( x=1 \) (f=4). So \( f(0) = -2 \) is greater than \( f(-1) = -3 \) (left neighbor) and less than \( f(1) = 4 \) (right neighbor). So it's not a local minimum (left neighbor is lower).
- \( (2, 1) \): neighbors \( x=1 \) (f=4) and \( x=3 \) (f=3). So \( f(2) = 1 \) is less than \( f(1) = 4 \) and less than \( f(3) = 3 \)? Wait 1 < 3? Yes, 1 < 3. So 1 is less than both neighbors? Wait \( f(1)=4 \), \( f(2)=1 \), \( f(3)=3 \). So 1 is less than 4 (yes) and less than 3 (yes? 1 < 3, yes). Wait, but is that a local minimum? Wait, but let's check \( (-1, -3) \) again. Wait \( x=-1 \): left neighbor \( x=-2 \) (f=-8), right neighbor \( x=0 \) (f=-2). So \( f(-1) = -3 \) is between -8 and -2. So it's a local maximum? No, wait, if the function is increasing from \( x=-2 \) to \( x=0 \) (since -8 < -3 < -2), then \( x=-1 \) is part of an increasing interval. Then from \( x=0 \) to \( x=1 \), it increases to 4, then decreases to 1 at \( x=2 \), then increases to 3 at \( x=3 \). So the function decreases from \( x=1 \) to \( x=2 \) (4 to 1) and increases from \( x=2 \) to \( x=3 \) (1 to 3). So at \( x=2 \), the function changes from decreasing to increasing, so \( (2, 1) \) is a local minimum? Wait, but \( f(2)=1 \), neighbors \( f(1)=4 \) and \( f(3)=3 \). So 1 is less than both? Wait 1 < 3? Yes, 1 is less than 3. So \( (2, 1) \) would be a local minimum? But wait earlier \( x=-2 \) has lower value. Wait, maybe the question is about the local minimum in the given options, considering the closest. Wait, maybe I made a mistake. Wait let's re-express the \( f(x) \) values:

\( x \): -2, -1, 0, 1, 2, 3

\( f(x) \): -8, -3, -2, 4, 1, 3

So the function values: starts at -8 (x=-2), increases to -3 (x=-1), increases to -2 (x=0), increases to 4 (x=1), decreases to 1 (x=2), increases to 3 (x=3). So the local minimum would be where the function changes from decreasing to increasing. The function decreases from x=1 to x=2 (4 to 1) and increases from x=2 to x=3 (1 to 3). So x=2 is a local minimum (since it's lower than both neighbors). But wait, the options: (2,1) is an option. But wait, also, x=-2 is a minimum, but not an option. Wait the options are (1,4), (-1,-3), (2,1), (0,-2). Wait, maybe I messed up. Wait, let's check the y-values: -8 (x=-2), -3 (x=-1), -2 (x=0), 4 (x=1), 1 (x=2), 3 (x=3). The lowest y-value among the options is -3 (x=-1), then -2 (x=0), then 1 (x=2), then 4 (x=1). But a local minimum is a point where it's lower than its immediate neighbors. Let's check each option's immediate neighbors (the x-values adjacent to the x in the ordered pair):

- For (-1, -3): immediate neighbors are x=-2 (f=-8) and x=0 (f=-2). So f(-1) = -3 is greater than f(-2) = -8 (left neighbor) and less than f(0) = -2 (right neighbor). So it's not a local minimum (left neighbor is lower).
- For (0, -2): immediate neighbors are x=-1 (f=-3) and x=1 (f=4). So f(0) = -2 is greater than f(-1) = -3 (left neighbor) and less than f(1) = 4 (right neighbor). Not a local minimum (left neighbor is lower).
- For (2, 1): immediate neighbors are x=1 (f=4) and x=3 (f=3). So f(2) = 1 is less than f(1) = 4 (left neighbor) and less than f(3) = 3 (right neighbor)? Wait 1 < 3? Yes, 1 is less than 3. So 1 is less than both neighbors. So this would be a local minimum (since it's lower than both immediate neighbors).
- For (1, 4): immediate neighbors are x=0 (f=-2) and x=2 (f=1). So f(1) = 4 is greater than both, so not a minimum.

Wait, but earlier, the function at x=2: left neighbor x=1 (f=4), right neighbor x=3 (f=3). So 1 is less than 4 and less than 3? Wait 1 < 3, yes. So (2,1) is a local minimum? But wait, the value at x=-2 is -8, which is lower, but it's not an option. The question is "closest to a local minimum". Wait, maybe the local minimum is around x=2, so (2,1) is the closest among the options. Wait, but let's check again. Wait, maybe I made a mistake in the neighbor for (2,1). x=2's neighbors are x=1 and x=3. f(1)=4, f(3)=3. So f(2)=1 is less than both 4 and 3? Yes, 1 < 4 and 1 < 3. So it's a local minimum. Alternatively, maybe the local minimum is around x=0? No, because f(0)=-2 is higher than f(-1)=-3. Wait, maybe the question is considering the local minimum in the given points, and (2,1) is the one that is a local minimum (since it's lower than its neighbors). Or maybe I messed up. Wait, let's check the answer options again. Wait, maybe the correct answer is (-1, -3)? No, because f(-2) is lower. Wait, the problem says "closest to a local minimum". So the local minimum is at x=-2 (f=-8), but it's not an option. The closest x to -2 among the options is x=-1 (distance 1 from x=-2) and x=0 (distance 2 from x=-2). So x=-1 is closer to x=-2 (distance 1) than x=0 (distance 2). So the ordered pair closest to the local minimum (at x=-2) would be (-1, -3), since x=-1 is closer to x=-2 than x=0, x=1, or x=2. Wait, that makes sense. Because the local minimum is at x=-2 (f=-8), and the closest x among the options is x=-1 (distance 1), so the ordered pair (-1, -3) is closest to the local minimum at (-2, -8).

Step3: Conclusion

Among the options, (-1, -3) is the closest to the local minimum (which is near x=-2, f=-8) as x=-1 is the closest x-value to x=-2 among the given options.

Answer:

\boxed{(-1, -3)}