Question

1. a partially completed input - output table is shown. complete the table so that it represents a function.

Explanation:

Step1: Recall Function Definition

A function is a relation where each input (x - value) has exactly one output (y - value). So, we need to assign unique inputs (or reuse existing inputs with the same output, but since we are completing the table, we can choose new inputs or use the rule if there is a pattern. Wait, the given table has some inputs: -3, -1, 3, 10 and some outputs. Wait, maybe we can assume a linear function? Let's check the given points. Wait, the input -1 has output 1. Let's see if we can find a pattern. Alternatively, maybe the function is \(y = x + 2\)? Wait, when \(x=-1\), \(y=-1 + 2=1\), which matches. Let's check \(x = - 3\), if \(y=-3 + 2=-1\)? But the output for -3 is given as 4? Wait, maybe my initial assumption is wrong. Wait, maybe the table is:

Input | Output
-3 | 4
? | 6
-1 | 1
? | 4
3 |?
? |?
10 |?

Wait, no, the image is a bit unclear. Wait, maybe the key is that each input must have a unique output, and each output can have multiple inputs. So, to complete the table, we need to assign inputs such that no input is repeated (or if repeated, same output). Let's assume we can use any inputs as long as each input has one output. Let's pick inputs that are not already used (-3, -1, 3, 10). Let's say the missing inputs: let's take 0, 2, 5, 7 (for example). And assign outputs. But maybe there is a pattern. Wait, the input -1 gives output 1. Let's see the difference: 1 - (-1)=2. Input -3: output 4. 4 - (-3)=7. No, that's not linear. Alternatively, maybe the function is \(y=x^{2}+0\)? No, -1 squared is 1, which matches. -3 squared is 9, but output is 4. No. Wait, maybe the problem is just to ensure that each input has one output. So, for the missing input rows, we can choose any input values that are not already in the input column (or if we reuse, same output). Let's list the existing inputs: -3, -1, 3, 10. So new inputs can be 0, 2, 4, 5, etc. And assign outputs. But maybe the intended function is linear. Wait, let's re - examine. Suppose the input -1 has output 1. Let's suppose the function is \(y = x+2\). Then:

For input -3: \(y=-3 + 2=-1\), but the output is given as 4. So that's not it. Maybe \(y=-x + 0\)? -1 gives 1, which matches. -3 gives 3, but output is 4. No. Alternatively, maybe the table is:

Input | Output
-3 | 4
0 | 6 (since 0 + 6? No, maybe 0 as input, output 6)
-1 | 1
2 | 4 (2 + 2 = 4)
3 | 5 (3 + 2 = 5)
5 | 7 (5 + 2 = 7)
10 | 12 (10 + 2 = 12)

But this is just a guess. Wait, the main idea is that each input has one output. So, to complete the table, we can choose inputs (x - values) that are not already used ( - 3, -1, 3, 10) and assign corresponding y - values (outputs) such that each x has one y. For example:

Input | Output
-3 | 4
0 | 6
-1 | 1
2 | 4
3 | 5
6 | 8
10 | 12

This works because each input ( - 3, 0, -1, 2, 3, 6, 10) has a unique output, and no input is repeated.

Step2: Implement the Function Rule

Since the key is the function definition (each input → one output), we can choose inputs and outputs as long as the function property holds. Let's pick a simple linear function for clarity, say \(y=x + 2\) (even though -3 gives 4 which is -3 + 7, maybe another rule). Wait, maybe the given output for -3 is 4, -1 is 1. Let's find the slope between (-3,4) and (-1,1): \(m=\frac{1 - 4}{-1-(-3)}=\frac{-3}{2}=-1.5\). Then the equation is \(y - 4=-1.5(x + 3)\). \(y=-1.5x-4.5 + 4=-1.5x-0.5\). Let's check x=-1: \(y=-1.5(-1)-0.5 = 1.5 - 0.5 = 1\), which matches. Good! So the function is \(y=-1.5x-0.5\) or \(y=-\frac{3}{2}x-\frac{1}{2}\).

Now, let's complete the table:

For the missing input (let's say the one with output 6): solve \(6=-\frac{3}{2}x-\frac{1}{2}\). Multiply both sides by 2: \(12=-3x - 1\) → \(3x=-13\) → \(x=-\frac{13}{3}\approx - 4.33\). But maybe we can use integer inputs. Alternatively, use the function to find outputs for given inputs:

Input 3: \(y=-\frac{3}{2}(3)-\frac{1}{2}=-\frac{9}{2}-\frac{1}{2}=-5\)

Input 10: \(y=-\frac{3}{2}(10)-\frac{1}{2}=-15-\frac{1}{2}=-15.5\)

For the input with output 4: solve \(4=-\frac{3}{2}x-\frac{1}{2}\). Multiply by 2: \(8=-3x - 1\) → \(3x=-9\) → \(x=-3\), but -3 is already used with output 4, so that's okay (same input, same output).

So, to complete the table, we can use the function \(y =-\frac{3}{2}x-\frac{1}{2}\) to find the missing values.

Step3: Fill in the Table

Using the function \(y=-\frac{3}{2}x-\frac{1}{2}\):

- For input \(x = 0\): \(y=-\frac{1}{2}\)
- For input \(x = 2\): \(y=-\frac{3}{2}(2)-\frac{1}{2}=-3-\frac{1}{2}=-\frac{7}{2}\)
- But maybe the problem just requires that each input has a unique output, so we can choose any valid inputs and outputs as long as the function property is satisfied. The key is that no input is associated with more than one output.

So, to complete the table, we can assign inputs that are not already in the input column ( - 3, -1, 3, 10) and calculate their outputs using a consistent rule (like the linear function we found) or any rule, as long as each input has one output.

Answer:

To complete the table to represent a function, we ensure each input has exactly one output. Using the linear function \(y =-\frac{3}{2}x-\frac{1}{2}\) (derived from the points \((-3,4)\) and \((-1,1)\)):

• For input \(x = 3\), output \(y=-\frac{3}{2}(3)-\frac{1}{2}=-5\)
• For input \(x = 10\), output \(y=-\frac{3}{2}(10)-\frac{1}{2}=-15.5\)

For the missing input with output \(6\), solve \(6=-\frac{3}{2}x-\frac{1}{2}\) to get \(x =-\frac{13}{3}\)

(Note: The table can be completed with any valid inputs/outputs as long as the function definition is satisfied. The above is one possible completion using a linear function fit to the given points.)