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NON-LINEAR RELATIONSHIPS. In previous sections we learned about linear correlations (relations that could be modelled by lines).

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Presentation on theme: "NON-LINEAR RELATIONSHIPS. In previous sections we learned about linear correlations (relations that could be modelled by lines)."— Presentation transcript:

1 NON-LINEAR RELATIONSHIPS

2 In previous sections we learned about linear correlations (relations that could be modelled by lines).

3 It’s fairly common to have definite patterns that are not lines.

4 If you try to model non-linear data with a line, you get a bad model that will have a value for “r” close to 0. The line is not close to most of the points.

5 Often, though, shapes other than lines will fit well with the data.

6 One of the most common non- linear relationships is quadratic, where the points come close to the shape of a parabola.

7 Instead of finding a line of best fit, we will find the equation for a parabola of best fit.

8 Let’s first find out all the basics about quadratic functions …

9 The general form for a quadratic function is f(x) = ax 2 + bx + c

10 The graph of every quadratic function is a parabola. Exactly what the parabola looks like depends on a, b, and c.

11 Most important fact about parabolas and their equations:

12 The bigger “a” is, the steeper (thinner) the parabola is.

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14 Because parabolas are symmetric, if one point is on the parabola, another point the same distance from the axis of symmetry is also on it.

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16 You could find the roots by a variety or methods, among them the quadratic formula:

17 It’s fairly easy to find a parabola of best fit on a graphing calculator. Everything works much the same as with linear regressions.

18 Example: Luxury seats at a sports stadium or arena generate a large amount of the venue’s profit.

19 Stadiums need to be careful in setting the prices for these seats. If they are too expensive, fewer people will buy the tickets, but if they are too cheap, they won’t make as much money.

20 Suppose a stadium has determined that they will make these profits for different ticket prices:

21 Find a function that models this data.

22 Since the profits go up and then come back down, this a quadratic function (parabola) is a good model.

23 First put the data in L1 and L2 of your calculator. STAT – EDIT

24 Go to STAT – CALC – QuadReg to find the equation.

25 Notice “a” is negative and very small.  It’s a fat parabola that opens downward.

26

27 So the parabola of best fit would be … f(x) = -.0000374x 2 +.0255x -.5271

28 You could use this to estimate what the profits would be at other price points. For instance, how much would they make if the tickets sold for $500 each?

29 f(x) = -.0000374x 2 +.0255x -.5271

30 How much would they make if the tickets sold for $500 each? f(x) = -.0000374x 2 +.0255x -.5271 The answer is about $2.87 million

31 Your calculator stores the exact values of a, b, and c. This can be useful if you need a more accurate estimate or if you want to know specific information about the parabola. For instance...

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33 In the VARS menu, go to Statistics. Press ENTER, and then choose EQ. a, b, and c are among the options.

34 The vertex of the parabola would be (341, 3.825) This is close to the top of the data.

35 You can even use the VARS feature to graph this parabola on your calculator (together with the data points, if you want).

36 Before you hit GRAPH, you may want a window that makes sense with this data.

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38 Once the function is graphed, it is easy to use it to find values. To do this hit 2 nd – TRACE (CALC), and choose VALUE.

39 Press ENTER. Choose the value of x you want to evaluate (like the $500 ticket price). Press ENTER again.

40 Note that this answer ($2.88 million) is slightly different than what we calculated with the formula ($2.7 million) because we rounded.


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