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PilotLPTM [1.2K]
4 years ago
13

Which of the tables below could represent a linear function?

Mathematics
1 answer:
drek231 [11]4 years ago
6 0

Answer:

1. Is not linear 2. Is not linear 3. Is linear 4. Is linear

Step-by-step explanation:

A function is said to be linear if it is in the form y = mx + c, where x and y represent its ordered pair (x,y) and m is the gradient and c its y- intercept. For function to be linear, for any three consecutive points (x₁,y₁), (x₂,y₂) and (x₃,y₃) the gradient, m will be constant. This follows that \frac{y_{2} -y_{1}  }{x_{2} - x_{1} } = \frac{y_{3} -y_{2}  }{x_{3} - x_{2} } for these points. We now insert the 3 points in each pair of data in the order x₁, x₂, x₃ and the corresponding y = f(x) as y₁, y₂, y₃ respectively.

For set 1

x₁ = 0, x₂ = 2, x₃ = 4, y₁= f(x₁) = -11, y₂ = f(x₂)= -13, y₃ = f(x₃) = -10.

So, \frac{[-13-(-11)]}{2-0} = -1 L.H.S\\ \frac{[-10-(-13)] }{4-2} = \frac{3}{2} R.H.S  

     Since -1 ≠ \frac{3}{2}

    The function is not linear.

For set 2

x₁ = 1, x₂ = 2, x₃ = 3, y₁= f(x₁) = 33, y₂ = f(x₂)= 11, y₃ = f(x₃) = 49.

So, \frac{[11-33)]}{2-1} = -22  L.H.S\\ \frac{[49-11)] }{3-2} = 38  R.H.S  

     Since -22 ≠ 38

    The function is not linear.

For set 3

x₁ = 0, x₂ = 10, x₃ = 20, y₁= f(x₁) = 8, y₂ = f(x₂)= 18, y₃ = f(x₃) = 28.

So, \frac{[18-8)]}{10-0} = 1    L.H.S\\ \frac{[28-18)] }{20-10} = 1     R.H.S  

     Since 1 (L.H.S) = 1 (R.H.S)

    The function is linear.

For set 4

x₁ = 0, x₂ = 5, x₃ = 10, y₁= f(x₁) = -20, y₂ = f(x₂)= -25, y₃ = f(x₃) = -30.

So, \frac{[-25-(-20)]}{5-0} = -1    L.H.S\\ \frac{[-30 -(-25)]}{10-5} = -1     R.H.S  

     Since -1 (L.H.S) = -1 (R.H.S)

    The function is linear.

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