1. Definition of a Polynomial Function Let n be a nonnegative integer and let a n, a n-1,…, a 2, a 1, a 0, be real numbers with a n  0. The function defined by f (x)  a n x n  a n-1 x n-1  …  a 2 x 2  a 1 x  a 0 is called a polynomial function of x of degree n. The number a n, the coefficient of the variable to the highest power, is called the leading coefficient. Date: Topic: Linear and Quadratic Functions and Modeling (2.1)

2. Smooth, Continuous Graphs Two important features of the graphs of polynomial functions are that they are smooth and continuous. By smooth, we mean that the graph contains only rounded curves with no sharp corners.By continuous, we mean that the graph has no breaks and can be drawn without lifting your pencil from the rectangular coordinate system. These ideas are illustrated in the figure. Smooth rounded corner x x y y

3. Determine if the following functions are polynomial functions? Text Example f (x)  x4  4x3  4x2f (x)  x4  4x3  4x2 Yes, exponent positive integer. No, the exponent will be negative. There will be asymptotes, so the graph will not be continuous. No, exponent is not an integer. Fractional exponents are roots which are not continuous graphs for all real numbers.

4. Linear Functions and Their Graphs Constant Rate of Change Let (x 1, f(x 1 )) and (x 2, f(x 2 )) be distinct points on the graph of a function f. The average rate of change of f from x 1 to x 2 is this is also the slope formula of a function

5. Linear Correlation and Modeling The narrower the oval the stronger the linear correlation.

6. Modeling and Predicting

7. Enter and plot the data (scatter plot)

8. Find the regression model that fits the problem situation (equation of the line of best fit). Select two coordinates to create the equation for this data.

9. Find the regression model that fits the problem situation (equation of the line of best fit). Use the point-slope formula and (2.6,33710) and (3.6,18260)

10. Superimpose the graph of the regression model on the scatter plot, and observe the fit.

11. Use the regression model to make the predictions called for in the problem. Use the model to predict weekly cereal sales if the price is dropped to $2.00 or raised to $4.00 per box. If the price of the cereal is dropped to $2, the sales will rise to about 42,980 boxes. If the price of the cereal is raised to $4, the sales will fall to about 12,080 boxes.

12. Assignment: 2-1 pages 175-9 #1-6, 53, 65, 66 all; 7-11 odd

13. The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x 2 is positive, the parabola opens upward; negative, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point. Graphs of Quadratic Functions DAY 2

14. The Standard Vertex Form of a Quadratic Function f (x)  a(x  h) 2  k, a  0 f is a parabola whose vertex is the point (h, k). The parabola is symmetric to the line x  h (axis of symmetry). If a  0, the parabola opens upward; if a  0, the parabola opens downward.

15. Graphing Parabolas With Equations in Standard Form To graph f (x)  a(x  h) 2  k: 1.Determine whether the parabola opens upward or downward. If a  0, it opens upward. If a  0, it opens downward. Graph the quadratic function: f (x)   2(x  3) 2  8 a is -2. Thus, the parabola opens downward. 2.Determine the vertex and axis of symmetry of the parabola. The vertex is (h, k) and the axis of symmetry is x = h. Because h  3 and k  8, the parabola has its vertex at (3, 8) and the axis of symmetry is x = 3 3.Find any x-intercepts by replacing f(x) with 0. Solve the resulting quadratic equation for x.   2(x  3) 2  8

16. 0   2(x - 3) 2 + 8 Find the x-intercepts. Replace f (x) with 0 in f (x)   2(x  3) 2  4 = (x - 3) 2 + 3 + 3 The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0). - 8 - 8 Graph the quadratic function f (x)   2(x  3) 2  8.  8 =  2(x  3) 2  2  2 x  5 or x  1 ± 2 = x - 3 4.Find the y-intercept by replacing x with 0. f (0) = -2(0 - 3) 2 + 8 f (0) = -10 The y-intercept is –10. The parabola passes through (0,  10).

17. It opens downward. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is x  3. 5.Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

18. Graph the quadratic function (use completing the square to find the vertex). opens up vertex:( -1, -4) x-intercept: y-intercept:

19. The Vertex of a Parabola Whose Equation Is f (x)  ax 2  bx  c Consider the parabola defined by the quadratic function f (x)  ax 2  bx  c. The parabola's vertex is at

20. Find the vertex, write the equation in vertex form, find the x-intercepts of the quadratic function: f (x)   x 2  6x  Find the vertex. The x-coordinate of the vertex is x = -b/(2a) x = -6/2(-1) = 3 The x-coordinate of the vertex is 3. Substitute 3 for x in the equation of the function to find the y-coordinate: y = f(3) = -(3) 2 +6(3) - 2 = 7 The parabola has its vertex at (3,7). Write the equation in vertex form. Input the vertex (3,7) into the vertex form equation: f (x) = a(x - 3) 2 + 7 Use a from the original equation. f (x) =  (x - 3) 2 + 7

21. Find the vertex, write the equation in vertex form, find the x-intercepts of the quadratic function: f (x)   x 2  6x  Find the x-intercepts. Replace f (x) with 0 in f (x)   x 2  6x  2. 0 =  x 2  6x  2 a  1,b  6,c  2 x   b  b 2  4ac 2a   6  6 2  4(  1)(  2) 2(  1)   6  36  8  2 28   6   2   6  27  2  3  7 x x = 5.65 and x = 0.35 The x-intercepts are (5.65, 0) and (0.35, 0).

22. Minimum and Maximum: Quadratic Functions Consider f(x) = ax 2 + bx +c. The maximum or minimum occurs at the vertex. If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)) If a < 0, then f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a))

23. Find the maximum or minimum value of the quadratic function. Step 1 Determine if it is a maximum or minimum by how the parabola opens. a is -15358.93 a  0; this negative value tells us that the parabola opens down so this function has a maximum. Step 2 The maximum or minimum value occurs at the vertex. The x-coordinate of the vertex is x = -b/(2a) x = -73622.5/2(-15358.93) = 2.40 Substitute 2.40 for x in the equation of the function to find the y-coordinate or maximum value: y = f(3) = -15358.93(2.40) 2 +73622.5(2.40) = 88277.56 The maximum revenue is $88,277.56. Revenue can be found by multiplying the price per box, x, by the number of boxes sold, y. So the total revenue is given by R(x)  xy =  x 2  73,622.50x

24. The Position Formula for a Free-Falling Object Near Earth’s Surface An object that is falling or vertically projected into the air has its height above the ground given by s(t) = -1/2gt 2 + v 0 t + s 0 where s is the height, g = 32ft/sec 2 = 9.8m/sec 2 is the acceleration due to gravity, v 0 is the initial velocity (original velocity) of the object, t is the time that the object is in motion in seconds, and s 0 is the initial height (original height) of the object. The object has its vertical velocity given by v(t) = -gt + v 0

25. The winner of a promotion for the Houston Astros downtown ballpark threw a ball with an initial vertical velocity of 92ft/sec and it landed on the infield grass. The ball was thrown from the upper deck of seats, 83 ft above field level. Find the maximum height of the baseball, its time in the air, and its vertical velocity when it hits the ground. maximum height of the baseball The maximum height of the baseball is about 215 ft.

26. The winner of a promotion for the Houston Astros downtown ballpark threw a ball with an initial vertical velocity of 92ft/sec and it landed on the infield grass. The ball was thrown from the upper deck of seats, 83 ft above field level. Find the maximum height of the baseball, its time in the air, and its vertical velocity when it hits the ground. its time in the air Need to find the time when the ball hits the ground. its vertical velocity when it hits the ground The ball spends about 6.54 seconds in the air. The balls downward rate is 117 ft/sec when it hits the ground. ball height on the ground is 0

27. Modeling Vertical free-fall motion Ball thrown upward above CBR and it landed directly on the face of the device DAY 3

28. Make a scatter plot of the data. Using quadratic regression on the calculator (STAT, CALC) to find the model for the height of the ball: Predict the maximum height of the ball. Adjust graphing window to [0, 1.2] by [-0.5, 2.0] Using the graphing calculator, trace the line. Then use the CALC button to choose CALCULATE maximum. (.402, 1.800) The maximum height the ball achieved was about 1.80 meters above the face of the CBR. Predict the vertical velocity when it hits the face of the CBR. Using the graphing calculator, trace the line. Then use the CALC button to choose CALCULATE zero. (1.022, 0) use to calculate vertical velocity g/2

29. You have 120 feet of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area? x y 2x + 2y = 120 Fencing: Area:A(x)=xy -2x 2y = 120 - 2x 2 y = 60 - x =x(60 - x) A(x)=-x 2 + 60x Maximum Area: vertex The dimensions are 30 feet by 30 feet. The maximum area is 900 square feet. A(x)=(30)(30)=900

30. vertex Write the equation for the quadratic function whose graph contains the vertex (-2, -3) and point (1, 2). Calculate a using the point (1, 2):

31. Use completing the square to describe the graph of the function. Support your answers graphically. vertex:( 2, 4) Describe the graph of the function. Translate y = x 2 horizontally, right 2 units Vertically stretch by 3 Translate vertically, up 4 units

32. Assignment: 2-2 pages 176-8 #15-30 x3; 13,23,45,48,55,62 all 2-3 pages 176-9 #33-42 x3; 43,54,64,65,67,74,75 all

33. Linear and Quadratic Functions and Modeling