Question

functions and lines
identifying linear functions given ordered pairs
for each function, state whether it is linear.
function 1
{(0, 1), (1, -2), (2, 1), (3, 4)}
○ linear
○ not linear
function 2
{(-5, 0), (-1, 1), (3, 8), (7, 27)}
○ linear
○ not linear
function 3

x	y
-1	-1
2	1
5	5
8	8

○ linear
○ not linear
function 4

x	y
1	-5
4	-4
7	-3
10	-2

○ linear
○ not linear

Explanation:

Function 1:

Step1: Calculate slopes between points

Slope between \((0,1)\) and \((1, -2)\): \(m_1=\frac{-2 - 1}{1 - 0}=\frac{-3}{1}=-3\)
Slope between \((1, -2)\) and \((2, 1)\): \(m_2=\frac{1 - (-2)}{2 - 1}=\frac{3}{1}=3\)
Since \(m_1 eq m_2\), Function 1 is not linear.

Function 2:

Step1: Calculate slopes between points

Slope between \((-5,0)\) and \((-1,1)\): \(m_1=\frac{1 - 0}{-1 - (-5)}=\frac{1}{4}=0.25\)
Slope between \((-1,1)\) and \((3,8)\): \(m_2=\frac{8 - 1}{3 - (-1)}=\frac{7}{4}=1.75\)
Since \(m_1 eq m_2\), Function 2 is not linear.

Function 3:

Step1: Calculate slopes between points

Slope between \((-1,-1)\) and \((2,1)\): \(m_1=\frac{1 - (-1)}{2 - (-1)}=\frac{2}{3}\)
Slope between \((2,1)\) and \((5,5)\): \(m_2=\frac{5 - 1}{5 - 2}=\frac{4}{3}\)
Since \(m_1 eq m_2\), wait, correction: Wait, \((-1,-1)\) to \((2,1)\): \(\frac{1 - (-1)}{2 - (-1)}=\frac{2}{3}\); \((2,1)\) to \((5,5)\): \(\frac{5 - 1}{5 - 2}=\frac{4}{3}\); \((5,5)\) to \((8,8)\): \(\frac{8 - 5}{8 - 5}=1\). Wait, no, wait \((-1,-1)\), \((2,1)\): \(x\) difference \(3\), \(y\) difference \(2\); \((2,1)\), \((5,5)\): \(x\) difference \(3\), \(y\) difference \(4\); \((5,5)\), \((8,8)\): \(x\) difference \(3\), \(y\) difference \(3\). Wait, no, I made a mistake. Wait, \((-1,-1)\), \((2,1)\): \(x\) from \(-1\) to \(2\) is \(+3\), \(y\) from \(-1\) to \(1\) is \(+2\); \((2,1)\) to \((5,5)\): \(x +3\), \(y +4\); \((5,5)\) to \((8,8)\): \(x +3\), \(y +3\). So slopes are \(\frac{2}{3}\), \(\frac{4}{3}\), \(1\) – not equal. Wait, but wait the table: \(x=-1,y=-1\); \(x=2,y=1\); \(x=5,y=5\); \(x=8,y=8\). Wait, maybe I miscalculated. Wait, \((-1,-1)\) to \((2,1)\): \(y\) change \(2\), \(x\) change \(3\) → slope \(2/3\). \((2,1)\) to \((5,5)\): \(y\) change \(4\), \(x\) change \(3\) → slope \(4/3\). \((5,5)\) to \((8,8)\): \(y\) change \(3\), \(x\) change \(3\) → slope \(1\). So not linear? Wait, no, wait the points: \((-1,-1)\), \((2,1)\), \((5,5)\), \((8,8)\). Let's check if \(y = x\) when \(x=-1\), \(y=-1\) (yes); \(x=2\), \(y=1\) (no, \(2 eq1\)). Wait, so my mistake. Wait the table: \(x=-1,y=-1\); \(x=2,y=1\); \(x=5,y=5\); \(x=8,y=8\). So \((-1,-1)\): \(y=x\); \((2,1)\): \(y eq x\); \((5,5)\): \(y=x\); \((8,8)\): \(y=x\). So it's not linear. Wait, but maybe I messed up. Wait, let's recalculate slopes correctly. Between \((-1,-1)\) and \((2,1)\): \(m=\frac{1 - (-1)}{2 - (-1)}=\frac{2}{3}\). Between \((2,1)\) and \((5,5)\): \(m=\frac{5 - 1}{5 - 2}=\frac{4}{3}\). Between \((5,5)\) and \((8,8)\): \(m=\frac{8 - 5}{8 - 5}=1\). Since slopes are not equal, Function 3 is not linear? Wait, but the table looks like maybe a typo? Wait, no, the user's table: Function 3 has \(x=-1,y=-1\); \(x=2,y=1\); \(x=5,y=5\); \(x=8,y=8\). So it's not linear. Wait, but maybe I made a mistake. Let's check Function 4.

Function 4:

Step1: Calculate slopes between points

Slope between \((1,-5)\) and \((4,-4)\): \(m_1=\frac{-4 - (-5)}{4 - 1}=\frac{1}{3}\)
Slope between \((4,-4)\) and \((7,-3)\): \(m_2=\frac{-3 - (-4)}{7 - 4}=\frac{1}{3}\)
Slope between \((7,-3)\) and \((10,-2)\): \(m_3=\frac{-2 - (-3)}{10 - 7}=\frac{1}{3}\)
All slopes are equal (\(\frac{1}{3}\)), so Function 4 is linear.

Wait, I think I messed up Function 1 earlier. Let's recheck Function 1: \(\{(0,1),(1,-2),(2,1),(3,4)\}\). Slope between \((0,1)\) and \((1,-2)\): \(m=\frac{-2 - 1}{1 - 0}=-3\). Slope between \((1,-2)\) and \((2,1)\): \(m=\frac{1 - (-2)}{2 - 1}=3\). Slope between \((2,1)\) and \((3,4)\): \(m=\frac{4 - 1}{3 - 2}=3\). Wait, first slope \(-3\), then \(3\), then \(3\). So not all slopes equal, so Function 1 is not linear.

Function 2: \(\{(-5,0),(-1,1),(3,8),(7,27)\}\). Slope between \((-5,0)\) and \((-1,1)\): \(m=\frac{1 - 0}{-1 - (-5)}=\frac{1}{4}\). Slope between \((-1,1)\) and \((3,8)\): \(m=\frac{8 - 1}{3 - (-1)}=\frac{7}{4}\). Not equal, so not linear.

Function 3: As above, slopes not equal, so not linear.

Function 4: Slopes between all consecutive points are \(\frac{1}{3}\), so linear.

So correcting:

Function 1: Not linear (slopes -3, 3, 3 – not all equal).
Function 2: Not linear (slopes 1/4, 7/4 – not equal).
Function 3: Wait, wait, let's re-express Function 3's points: \((-1,-1)\), \((2,1)\), \((5,5)\), \((8,8)\). Let's check the difference in \(x\) and \(y\):

From \(-1\) to \(2\): \(x\) +3, \(y\) +2.
From \(2\) to \(5\): \(x\) +3, \(y\) +4.
From \(5\) to \(8\): \(x\) +3, \(y\) +3.

Since the \(y\)-differences are not constant (2, 4, 3) for constant \(x\)-difference (3), it's not linear.

Function 4:

From \(1\) to \(4\): \(x\) +3, \(y\) +1 (from -5 to -4).
From \(4\) to \(7\): \(x\) +3, \(y\) +1 (from -4 to -3).
From \(7\) to \(10\): \(x\) +3, \(y\) +1 (from -3 to -2).

So constant \(x\)-difference (3) and constant \(y\)-difference (1), so slope is constant (\(1/3\)), so linear.

So summarizing:

Function 1: Not linear (slopes -3, 3, 3 – not all same).
Function 2: Not linear (slopes 1/4, 7/4 – not same).
Function 3: Not linear (slopes 2/3, 4/3, 1 – not same).
Function 4: Linear (slopes 1/3, 1/3, 1/3 – same).

Answer:

Function 1: Not linear
Function 2: Not linear
Function 3: Not linear
Function 4: Linear

(For each function, the correct option is:
Function 1: Not linear
Function 2: Not linear
Function 3: Not linear
Function 4: Linear)