1. Finding the equation of a straight line is easy Just use y = mx + c Finding the equation of a quadratic curve from a set of data is more difficult. We have to convert it into a straight line – this is called Linearising Linearising a Quadratic Curve 1

2. Graph of distance against time -2 0 2 4 6 8 10 12 14 00.20.40.60.811.2 time Distance Linearising a Quadratic Curve 1 TimeDistance 00 0.10.04 0.20.23 0.30.69 0.41.45 0.52.57 0.64 0.75.6 0.87.62 0.910.12 112.77

3. Using your knowledge of completing the square the equation is y = a(x + b) 2 + c a = vertical stretch b = horizontal translation c = vertical translation In our case b = 0 No Horizontal translation So y = ax 2 + c -2 0 2 4 6 8 10 12 14 0 0.20.40.60.811.2 time Graph of distance against time Distance

4. To find a and c make a new table of values of x 2 where x 2 = X So y = aX + c Time xDistance yTime 2 X = x 2 Distance y 0000 0.10.040.010.04 0.20.230.040.23 0.30.690.090.69 0.41.450.161.45 0.52.570.252.57 0.640.364 0.75.60.495.6 0.87.620.647.62 0.910.10.8110.1 112.81

5. Now a straight line is obtained The gradient is m =12.812 The y intercept is c = –0.3852 The equation is y = 12.812X – 0.3852 Replace X by x 2 y = 12.812x 2 – 0.3852 So distance = 12.812  time 2 – 0.3852 Graph of distance against time 2 y = 12.812 X - 0.3852 -2 0 2 4 6 8 10 12 14 0 0.511.5 X distance y

6. Finding the equation of a quadratic curve 2 No. of days x No. of cases y 00 122 234 354 467 569 686 788 895 9102 10105 11103 12100 1389 Graph of No. of cold cases against no. of days -40 -20 0 20 40 60 80 100 120 0510152025 days No. of cases Plot the points and workout the horizontal translation of the maximum

7. The horizontal translation is x = 10 as the maximum occurs at x = 10 It has also been flipped, stretched vertically and translated vertically Using your knowledge of completing the square The equation is y = a(x + b) 2 + c a = vertical stretch b = horizontal translation c = vertical translation Graph of No. of cold cases against no. of days -40 -20 0 20 40 60 80 100 120 0510152025 days No. of cases

8. In our case b = –10 as it has been translated 10 to the right y = a(x - 10) 2 + c To find a and c make a new table of values of (x–10) 2 So y = aX + cwhere X = (x–10) 2

9. No. of days x No. of cases y X (x–10) 2 No. of cases y 001000 1228122 2346434 3544954 4673667 5692569 6861686 7889 8954 91021 101050 111031 121004 13899 Now plot a new graph of X = (x–10) 2 and y = No. of cases Find the equation of the line of best fit Do this by hand, Excel and the calculator to provide the required checks

10. Graph of No. of cold cases against (x-10) 2 y = -0.8447x + 96.577 -40 -20 0 20 40 60 80 100 120 050100150200 (x-10) 2 no. cold cases Now a straight line is obtained The gradient is m = –0.8447 The y intercept is c = 96.577

11. The equation is y = –0.8447X + 96.577 Replace X by (x – 10) 2 y = –0.8447(x – 10) 2 + 96.577 Multiply out the brackets y = –0.8447(x – 10)(x – 10) + 96.577 y = –0.8447(x 2 – 20x + 100) + 96.577 y = –0.8447x 2 + 16.894x – 84.77 + 96.577 y = –0.8447x 2 + 16.894x + 11.807

12. No. of cases = –0.8447  No. of days 2 + 16.894  No. of days + 11.807 This is very similar to the quadratic curve of best fit in the original graph This equation can be used to work out the no. of cases if you are given the no. of days Eg If days =10 then No. cases = –0.8447  10 2 + 16.894  10 + 11.807 = 96.27 Actual value = 105 % error =