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Willis analyzed the following table to determine if the function it represents is linear or non-linear. First he found the differences in the y-values as 7 – 1 = 6, 17 – 7 = 10, and 31 – 17 = 14. Then he concluded that since the differences of 6, 10, and 14 are increasing by 4 each time, the function has a constant rate of change and is linear. What was Willis’s mistake?

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nobillionaireNobley
His mistake is that he used the differences of 6, 10 and 14 instead of the differences of 7, 17 and 31.

i.e. for a linear function, the y values should have a constant rate of change and not the differences.

Answer:

He got the values of y for different values of x as:

1    7    17    31

Now Willis found the difference in y-values and analyze those y-values are increasing by 4, and said that the function has a constant rate and hence the function is linear.

But his approach is completely wrong as for determining the function is linear we see that rate of change is determined by the slope of a graph and is given as:

[tex]\dfrac{y_{i+1}-y_{i}}{x_{i+1}- x_{i}}[/tex]

where ([tex]x_i,y_i[/tex]) are different interpolating points and its corresponding value.

So Willis must have calculated [tex]\dfrac{y_{i+1}-y_{i}}{x_{i+1}- x_{i} }[/tex] and check that it is equal for different pair of points  in order to say that the function is linear.




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