2. Finding the Equation of a Quadratic Using QUADDATA recorded by rolling a ball up and down a ramp. The distance is measured from an arbitrary starting point. Distance from the starting point toward the motion detector is positive. DmDm Time sec © 2002 Jerel L. Welker, Lincoln High School Permission is granted to copy and distribute with copyright notation attached.

3. Finding a Quadratic Equation in Vertex Form y=a(x-h) 2 + k Step 1: Find the Vertex Use the TRACE function to find the vertex of the parabola. The vertex occurs at (h, k). The parabola has a vertex at (1.65 sec, 0.463 m). The equation is y = a(x - 1.65 sec ) 2 + 0.463 m

4. Finding a Quadratic Equation in Vertex Form y=a(x-h) 2 + k Step 2: Find the value of ‘a’ Substitute for (x, y) in the equation y = a(x - 1.65 sec ) 2 + 0.463 m. Solve algebraically for ‘a’. Use the TRACE function to find another point on the parabola. -0.583 m = a(2.73 sec - 1.65 sec ) 2 + 0.463 m -0.583 m - 0.463 m = a(2.73 sec - 1.65 sec ) 2 (-0.583 m - 0.463 m ) (2.73 sec - 1.65 sec ) 2 = a a = -0.897 m/sec 2

5. Finding a Quadratic Equation in Vertex Form y=a(x-h) 2 + k Step 3: Graph the Equation Substituting for ‘a’ in the equation y = a(x - 1.65 sec ) 2 + 0.463 m, one gets y = -0.897 m/sec 2 (x - 1.65 sec ) 2 + 0.463 m. Use the function grapher to graph: y = -0.897 m/sec 2 (x - 1.65 sec ) 2 + 0.463 m. If you want to try and improve the fit, use different values of a, h, and k to see how each changes the graph.

6. Finding a Quadratic Equation in Vertex Form y=a(x-h) 2 + k Step 4: Write the mathematical model. The mathematical model is: D m = -0.90 m/sec 2 (T sec - 1.65 sec ) 2 + 0.46 m. Step 5: Explain the mathematical model. The ball reached its point furthest from the starting point (origin) or closest to the motion detector 1.65 sec after data collection started. This point was 0.46 meters from the starting point. One-half of the acceleration was -0.90 m/sec 2. Up the ramp was positive, so the acceleration was down the ramp.

7. Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s) Before we begin Factored Form: Turn off the Vertex equation in the function grapher by placing the cursor on the = sign. Press ENTER to remove the highlight on the = sign. The function is stored in memory but is not graphed unless turned on.

8. Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s) Step 1: Find the x-intercepts, roots, or zeroes of the polynomial using the TRACE function. The values of ‘r’ and ‘s’ are the x-intercepts or where the graph crosses the x-axis. Find the closest point, average, or estimate the roots. Note that when r = 0.9 y is very close to 0. r = 0.9 sec The value of ‘s’ is between 2.34 and 2.37 but closer to 2.37. Estimate s = 2.36 sec

9. Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s) Step 2: Find the ‘a’ value. Substitute values of ‘r’ and ‘s’ into the equation. Choose a third point for ‘x’ and ‘y’ and solve for ‘a’. The vertex is a good choice for the third point. y = a (x - 0.9 sec )(x - 2.36 sec ) Substituting the vertex (1.65, 0.463): 0.463 m = a(1.65 sec - 0.9 sec )(1.65 sec - 2.36 sec ) Solving for a: a = 0.463 m ((1.65 sec - 0.9 sec )(1.65 sec - 2.36 sec )) a = -0.869 m/sec 2

10. Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s) Step 3: Graph the equation. y = -0.869 m/sec 2 (x - 0.9 sec )(x - 2.36 sec ) Place the equation in Y2 as Y1 has been turned off. If you want to try and improve the fit, use different values of ‘a’, ‘r’, and ‘s’ to see how each changes the graph.

11. Finding a Quadratic Equation in Factored Form y=a(x-r)(x-s) Step 4: Write a Mathematical Model. D m = -0.87 m/sec 2 (T sec - 0.9 sec )(T sec - 2.36 sec ) Step 5: Explain the Mathematical Model. The ball passes the point marked as the origin 0.9 seconds after data collection started. The ball was moving in a positive direction (up the ramp). 2.36 seconds after data collection started, the ball passed the origin on its way down the ramp. One-half of the acceleration is - 0.87 m/sec 2. Acceleration is in a negative direction (down the ramp).

12. Finding a Quadratic Equation in Quadratic Form y=ax 2 + bx + c Before we begin finding Quadratic Form! Turn off the Vertex equation in the function grapher by placing the cursor on the = sign. Press ENTER to remove the highlight on the = sign. The function is stored in memory but is not graphed unless turned on.

13. Finding a Quadratic Equation in Quadratic Form y=ax 2 + bx + c Quadratic form allows for practice expanding the factored and vertex forms of a parabola. Step 1: Expand (multiply out) the equation in factored form. D m = -0.87 m/sec 2 (T sec - 0.9 sec )(T sec - 2.36 sec ) D m = -0.87 (T 2 - 2.36T - 0.9T + 2.124) D m = -0.87 (T 2 - 3.26T + 2.124) D m = -0.87 m/sec 2 T 2 + 2.84 m/sec T - 1.85 m

14. Finding a Quadratic Equation in Quadratic Form y=ax 2 + bx + c Step 2: Graph the model. D m = -0.87 m/sec 2 T 2 + 2.84 m/sec T - 1.85 m Step 3: Explain the model. One-half of the acceleration is -0.87 m/sec 2 (down the ramp). If the ball had been in continuous motion, it would have had a velocity of 2.84 m/sec when data collection began (t=0). The ball would have been 1.85 meters away from the starting point in a negative direction when data collection started.

15. Finding a Quadratic Equation in Quadratic Form y=ax 2 + bx + c Finding a second quadratic form from vertex form of the parabola. Step 1: Expand the vertex form of the parabola. D m = -0.90 m/sec 2 (T sec - 1.65 sec ) 2 + 0.46 m. D m = -0.90 (T 2 - 2*1.65*T + 1.65 2 ) + 0.46 D m = -0.90 (T 2 - 3.30T + 2.7225) + 0.46 D m = -0.90T 2 + 2.97T - 2.45025 + 0.46 D m = -0.90 m/sec 2 T 2 + 2.97 m/sec T - 1.99 m

16. Finding a Quadratic Equation in Quadratic Form y=ax 2 + bx + c Step 2: Graph the expanded vertex form of the parabola. D m = -0.90 m/sec 2 T 2 + 2.97 m/sec T - 1.99 m Step 3: Explain the mathematical model. One-half of the accleration is -.90 m/sec 2 (down the ramp). If the ball had been in continuous motion, the initial velocity would have been 2.97 m/sec when data collection began (t=0). The position of the ball at t=0 would have been -1.99 m (below the origin on the ramp).

17. Using the Quadratic Regression y=ax 2 + bx + c Remember to turn off the equation in Y4!! Students who skip immediately to the quadratic regression will find that it does not fit well as a result of the erroneous data collected before the ball started moving and after the ball was stopped. The erroneous regression is shown graphed below.

18. Using the Quadratic Regression y=ax 2 + bx + c Step 1: Choose an appropriate part of the graph for a regression. From the PRGM menu, select the CHOOSE program. Press ENTER to run the program from the HOME screen. A descriptive message will appear asking you to use the cursor keys to select the left and right sides of the data. The program will keep the data between the two selected points.

19. Using the Quadratic Regression y=ax 2 + bx + c Step 1: Choose an appropriate part (con’t). Use the arrow keys to move the cursor along the data to a point on the left side of the data to be saved. All points to the left of this point will not be used in the regression. Press ENTER to continue. Continue moving the cursor to a point on the right side of the data to be saved. All data between these two points will be used in the regression. Points to the right of this point will not be used. Press ENTER to continue.

20. Using the Quadratic Regression y=ax 2 + bx + c Step 1: Choose an appropriate part (con’t). The original data set will remain in L1 and L2. The selected points will be copied to L3 and L4 as shown in the message box. Press ENTER to see the selected data set graphed on the screen in a new window.

21. Using the Quadratic Regression y=ax 2 + bx + c Step 2: Find a Quadratic Regression. Choose the QuadReg regression from the STAT - CALC menu The data for this regression is in L3 and L4. Enter these lists with comma’s. From the VARS - Y-Vars Function menu, choose Y5 from the list. The regression will be stored in Y5. Press ENTER to find the regression.

22. Using the Quadratic Regression y=ax 2 + bx + c Step 2: Find a Quadratic Regression (con’t). The regression in quadratic form will be displayed. If DiagnosticOn is selected, the correlation coefficient will also be displayed. Step 3: Graph the Quadratic Regression. Press GRAPH to display the graph with the selected data in L3 and L4.

23. Using the Quadratic Regression y=ax 2 + bx + c Step 4: Graph the regression with the original data set. Choose STAT PLOT (2nd Y=). Turn off Stat Plot 2 and turn Stat Plot 1 back on. Note that Stat Plot one graphs lists L1 and L2. Stat Plot 2 was turned on by the CHOOSE program and graphs L3 and L4. Set the WINDOW to show the original data set or press ZoomStat (Zoom-9) to view the original window.

24. Using the Quadratic Regression y=ax 2 + bx + c Step 4: Graph the regression (con’t). Press GRAPH to display the graph in the selected window. The equation graphed is shown in Y5 of the function grapher.

25. Using the Quadratic Regression y=ax 2 + bx + c Step 5: Write the mathematical model. D m = -0.87 m/sec 2 T 2 + 2.83 m/sec T - 1.84 m Step 6: Explain the mathematical model. One-half of the accleration is -.87 m/sec 2 (down the ramp). If the ball had been in continuous motion, the initial velocity would have been 2.83 m/sec when data collection began (t=0). The position of the ball at t=0 would have been -1.84 m (below the origin on the ramp). The End