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TI 84 Calculator: Part II

TI 84 Calculator: Part II. Macon State College Mary Dwyer Wolfe, Ph.D. Gaston Brouwer, Ph.D. July 2009 http://calculator.maconstate.edu. TI 84 Calculator . Solving equations using the Intersection of Graphs Method Scatterplots and Line Graphs Linear Regression Applications

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TI 84 Calculator: Part II

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  1. TI 84 Calculator: Part II Macon State College Mary Dwyer Wolfe, Ph.D. Gaston Brouwer, Ph.D. July 2009 http://calculator.maconstate.edu

  2. TI 84 Calculator • Solving equations using the Intersection of Graphs Method • Scatterplots and Line Graphs • Linear Regression • Applications • Quadratic Equations and the Vertex • Maximum/Minimum Applications

  3. Solving by the Intersections of Graphs Method The Intersections of Graphs method of solving equations is an alternate method of solving equations (not a replacement method). This method sometimes leads to approximate solutions whereas traditional symbolic methods can usually produce exact solutions. The Intersections of Graphs method is particularly useful when symbolic solutions are not possible.

  4. Solving by the Intersections of Graphs Method • Step 1: Enter the left side of the equation for Y1 and the right side of the equation for Y2 (Under Y=) • Step 2: Graph the equations in a window where the Intersection is visible. • Step 3: Compute the Intersection (2nd CALC 5) • Step 4: The x coordinate of the intersection point is the solution to the equation Solve: 2 – (3x + 5) = 8x + 17

  5. Solving by the Intersections of Graphs Method Solve: 2 – (3x + 5) = 8x + 17 Step 1: Step 2: Step 3: Step 3: continued … Step 4: The solution is approximately x = -1.818182 or x = -1.82 to the nearest hundredth

  6. Solving by the Intersections of Graphs Method Try this one: Solve 2x2 – 5x = 12 x = -1.5 or x = 4

  7. Solving by the Intersections of Graphs Method Try this one: Solve: ex+2= 52x for x

  8. Solving a Linear Inequality Graphically: Example 1 • Solve [2, 15, 1] by [2, 15, 1] S T E P 1 S T E P 2 First we find the intersection of the left and right side just as we do with equations. Note that the graphs intersect at the point (8.20, 7.59). Since this is an inequality, we must now determine if the correct sign is > or it flips to <. S T E P 3

  9. Solving a Linear Inequality Graphically: Example 1 -- continued • Solve Y1 Y2 S T E P 4 Note that the graphs intersect at the point (8.20, 7.59). So we center an x-value of 8. When x < 8.20, Y1 < Y2, but when x > 8.20, Y1 > Y2. Since our original equation was Y1 > Y2 , we know x > 8.20. Thus in interval notation the solution set is (8.20, ∞).

  10. Solving a Linear Inequality Graphically: Example 2 • Solve [10, 10, 1] by [10, 10, 1] S T E P 1 S T E P 2 Note that the graphs intersect at the point (1.36, 2.72). Since this is an inequality, we must now determine if the correct sign is > or it flips to <. S T E P 3

  11. Solving a Linear Inequality Graphically: Example 2 • Solve Y1 Y2 S T E P 4 Note that the graphs intersect at the point (1.36, 2.72). We find that x ≤ 1.36 when Y1 > Y2. Thus in interval notation the solution set is ( ∞, 1.36].

  12. Solving Compound Inequalities Example: Suppose the Fahrenheit temperature x miles above the ground level is given by T(x) = 88 – 32 x. Determine the altitudes where the air temp is from 300 to 400. • We must solve the inequality 30 < 88 – 32 x < 40 • Graph all three parts in the same window

  13. Solving Compound Inequalities • We must solve the inequality -- continued 30 < 88 – 32 x < 40 • Find the 2 intersection points Note: Use the down arrow to switch to the 2nd two equations. Symbolically, Between 1.5 and 1.8125 miles above ground level, the air temperature is between 30 and 40 degrees Fahrenheit.

  14. Scatterplots and Line Graphs Graph the set of data:

  15. Scatterplots and Line Graphs Scatterplot Line Graph

  16. Least Squares Regression (Line of Best Fit) Note that this data appears to be linear:

  17. Least Squares Regression (Line of Best Fit) f(x) = 2x - 3

  18. Least Squares Regression (Line of Best Fit) Using this regression equation, f(x) = 2x - 3, what y (or f(x)) is paired with x = 10? That is, find f(10). f(10) = 17

  19. Least Squares Regression (Line of Best Fit) Using this regression equation, f(x) = 2x - 3, what x is paired with y = f(x) = 10? That is, solve 10 = 2x - 3. x = 6.5 is paired with y = 10

  20. More Linear Regression Find the equation of the line that passes through the points (2, -3) and (-5, -4). We could use point-slope form and find the equation symbolically or … We could use linear regression.

  21. Nearly Linear Data – An Application From NCTM.org: Predict the maximum height for a bike that weighs 21.5 pounds if all other factors are held constant.

  22. Nearly Linear Data – An Application Predict the maximum height for a bike that weighs 21.5 pounds if all other factors are held constant. A height of 9.994 inches is expected for a weight of 21.5 pounds.

  23. Nearly Linear Data – An Application Your turn! You already have the model in your calculator! Predict the maximum weight for a bike that so that it can reach a height of 10.5 inches if all other factors are held constant.

  24. Nearly Linear Data – An Application Predict the maximum weight for a bike that so that it can reach a height of 10.5 inches if all other factors are held constant. A weight of about 18.3 pounds is expected for a height of 10.5 inches.

  25. Finding a Vertex (Max/Min Point) Find the vertex of y = 2x2 – 7x - 1 The vertex is approximately (1.75, -7.125)

  26. Find the Vertex Your Turn! Find the vertex of y = -3x2 + 5x - 4 The vertex is approximately (0.833, -1.912)

  27. Application – Find a Maximum A home owner has 200 feet of fencing to make a rectangular garden in his yard that is protected from the rabbits and deer. He decides to use the long side of the house as one side of the fenced area so a larger area can be obtained as less fencing is needed. That way he can walk out the back door into the garden. What of the dimensions of garden that maximize the area for planting? 200 – 2x x x The house! A(x) = x(200 – 2x)

  28. Application – Find a Maximum A(x) = x(200 – 2x) X = 50 200 – 2x = 200 – 2(50) = 100 The dimensions are 50 by 100 feet.

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