Question

which table represents a linear function?

Explanation:

Step1: Recall linear - function property

A linear function has a constant rate of change (slope). The slope formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).

Step2: Check first table

For the first table with points \((0,1)\), \((1,3)\), \((2, - 3)\), \((3,-1)\):
The slope between \((0,1)\) and \((1,3)\) is \(m_1=\frac{3 - 1}{1 - 0}=2\).
The slope between \((1,3)\) and \((2,-3)\) is \(m_2=\frac{-3 - 3}{2 - 1}=-6\). Since \(m_1
eq m_2\), it is not a linear function.

Step3: Check second table

For the second table with points \((0,0)\), \((1,1)\), \((2,4)\), \((3,9)\):
The slope between \((0,0)\) and \((1,1)\) is \(m_1=\frac{1 - 0}{1 - 0}=1\).
The slope between \((1,1)\) and \((2,4)\) is \(m_2=\frac{4 - 1}{2 - 1}=3\). Since \(m_1
eq m_2\), it is not a linear function.

Step4: Check third table

For the third table with points \((0,3)\), \((1,1)\), \((2,-1)\), \((3,-3)\):
The slope between \((0,3)\) and \((1,1)\) is \(m_1=\frac{1 - 3}{1 - 0}=-2\).
The slope between \((1,1)\) and \((2,-1)\) is \(m_2=\frac{-1 - 1}{2 - 1}=-2\).
The slope between \((2,-1)\) and \((3,-3)\) is \(m_3=\frac{-3+1}{3 - 2}=-2\). Since the slope is constant (\(m = - 2\)), it is a linear function.

Step5: Check fourth table

For the fourth table with points \((0,0)\), \((1,-1)\), \((2,4)\), \((3,-9)\):
The slope between \((0,0)\) and \((1,-1)\) is \(m_1=\frac{-1 - 0}{1 - 0}=-1\).
The slope between \((1,-1)\) and \((2,4)\) is \(m_2=\frac{4 + 1}{2 - 1}=5\). Since \(m_1
eq m_2\), it is not a linear function.

Answer:

The third table (with \(x = 0,1,2,3\) and \(y = 3,1,-1,-3\)) represents a linear function.