1. Quadratic Functions and Models ♦ ♦ Learn basic concepts about quadratic functions and their graphs. ♦ Complete the square and apply the vertex formula. ♦ Graph a quadratic function by hand. ♦ Solve applications and model data. 3.1

2. Basic Concepts REVIEW: A linear function can be written as f(x) = ax + b (or f(x) = mx + b). The formula for a quadratic function is different from that of a linear function because it contains an x 2 term. f(x) = 3x 2 + 3x + 5g(x) = 5  x 2 A quadratic function is nonlinear. Leading Coefficient

3. Quadratic Function Properties The graph of a quadratic function is a parabola—a U shaped graph that opens either upward or downward. A parabola opens upward if its leading coefficient a is positive and opens downward if a is negative. The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the vertex. (The graph of a parabola changes shape at the vertex.) The vertical line passing through the vertex is called the axis of symmetry. The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola. Leading Coefficient

4. Examples of different parabolas

5. Demonstrate EXCEL file for Quadratic Functions DEMO

6. Example Use the graph of the quadratic function shown to determine the sign of the leading coefficient, its vertex, and the equation of the axis of symmetry. Leading coefficient: The graph opens downward, so the leading coefficient a is negative. Vertex: The vertex is the highest point on the graph and is located at (1, 3). Axis of symmetry: Vertical line through the vertex with equation x = 1.

7. The quadratic function f(x) =ax 2 + bx + c can be written in an alternate form that relies on the vertex (h, k). Example: f(x) = 3(x - 4) 2 + 6 is in vertex form with vertex (h, k) = (4, 6). What is the vertex of the parabola given by ? f(x) = 7(x + 2) 2 – 9 ? Vertex = (-2,-9)

8. Demonstrate EXCEL file for Quadratic Functions in Vertex Form DEMO

9. Example: Convert the quadratic f(x) = 3(x + 2) 2 – 8 which is in vertex form to the form f(x) = ax 2 + b + c f(x) = 3(x + 2) 2 – 8Given formula. Expand the quantity squared. f(x) = 3(x 2 + 4x + 4) - 8 Multiply by the 3.f(x) = 3x 2 + 12x + 12 - 8 Simplfy.f(x) = 3x 2 + 12x + 4

10. Example Write the formula f(x) = x 2 + 10x + 23 in vertex form by completing the square. Given formula Subtract 23 from each side. Add (10/2) 2 = 25 to both sides. Factor perfect square trinomial. Subtract 2 form both sides.. Vertex is h= -5 k = -2.What is the vertex?

11. Example Find the vertex of the graph of a = 1/2, b =  4, and c = 8 x-coordinate of vertex: y-coordinate evaluate f(4): The vertex is (4, 0).

12. Example: Use the vertex formula to write f(x) =  3x 2  3x + 1 in vertex form, then graph the parabola. 1.Begin by finding the vertex. 2. Compute k = f(h). 3. The vertex is 4. The vertex form is

13. 5. To graph the parabola we use the value of coefficient a, the vertex and the axis of symmetry together with a few points on either side of the line of symmetry. The vertex is So the axis of symmetry is the vertical line x = -1/2. a = -3 so the parabola opens downward. Next determine a few points on either side of the line of symmetry.

14. Based on the figure compute the y-coordinates of f(x) =  3x 2  3x + 1 for points in the table below. xf(x) -2 -1/27/6 0 1

15. Based on the figure compute the y-coordinates of f(x) =  3x 2  3x + 1 for points in the table below. xf(x) -2-5 1 -1/27/6 01 1-5 Plot these points.

16. xf(x) -2-5 1 -1/27/6 01 1-5 Connect the points with a smooth curve.

17. xf(x) -2-5 1 -1/27/6 01 1-5

18. Example: Graph the quadratic equation g(x) = 2x 2 - 12x + 23. Solution Strategy: Find the vertex, the axis of symmetry, and determine a few points on the curve which we connect with a smooth curve. Use the vertex formula: h = -b/(2a), k = g(h) a = 2, b = -12, c = 23h = -(-12)/(4) = 3 k = g(3) = 2(9) -12(3) + 23 so k = 18 – 36 + 23 = 41 -36 = 5 Vertex at (3, 5). Axis of symmetry: line x = 3

19. g(x) = 2x 2 - 12x + 23 Vertex at (3, 5), Axis of symmetry: line x = 3 Since a = 2 > 0 the parabola opens upward. xg(x) 1 2 35 4 5 Vertex

20. g(x) = 2x 2 - 12x + 23 Vertex at (3, 5), Axis of symmetry: line x = 3 Since a = 2 > 0 the parabola opens upward. xg(x) 113 27 35 47 5 Vertex Connect the points with a smooth curve.

21. g(x) = 2x 2 - 12x + 23 Vertex at (3, 5), Axis of symmetry: line x = 3 Since a = 2 > 0 the parabola opens upward. xg(x) 113 27 35 47 5 Vertex

22. Applications and Models Example A junior horticulture class decides to enclose a rectangular garden, using part of the side of the greenhouse as one side of the rectangle. If the class has 32 feet of fence, find the dimensions of the rectangle that give the maximum area for the garden. What do we want to find? What requirements or constraints must we abide by? What formulas will we be using?

23. Applications and Models Example continued Because the 32-foot fence does not go along the greenhouse, it follows that W + L + W = 32 or L = 32 – 2W Find dimensions of rectangle. We have 32ft of fence & the rectangle must have maximum area. Area = length x width = L x W Perimeter of the fence = distance around

24. The area of the garden is the length times the width. and we have L = 32 – 2W. A = LW = (32 - 2W)W = 32W – 2W 2 The area formula is a parabola that opens downward so the maximum value of the area occurs at the vertex. The vertex formula gives, The corresponding length is L = 32 – 2W = 32 – 2(8) = 16 feet. The dimensions are 8 feet by 16 feet. So the maximum area of the garden will be A = 8×16 = 128 sq.feet. Here W plays the role of x and A plays the role of y. y = -2x 2 + 32

25. A Common Situation: A modeling process requires you to determine a maximum value or a minimum value. (For example maximum area or minimum cost, etc.) Such situations are called optimization or max/min problems. If the equation that provides the math model is a quadratic f(x) = ax 2 + bx + c, then the value of x that is used to determine the optimal value is always at the vertex. That is, at

26. Example: Coyote airlines does a lot of charter business. They are running a special. The base fare is $250 (round trip to Las Vegas) but for a group of x 1 travelers the price is $(252 – 2x). Determine the size of the group that will maximize the revenue for the charter flight. Step 1. Construct a function R(x) that computes the ticket revenue for a flight. Step 2: Determine value of x that maximizes the revenue. Step 3. Compute the maximum revenue. R(x) = x(252 – 2x) = 252x – 2x 2 R(63) = $7938 Why this formula?

27. Projectile Motion: Discussion A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles. An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object is which thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). An object which is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible). http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l2a.html

28. The equation that gives the height s of a projectile at time t is a quadratic function where v 0 represents its initial velocity in feet per second, h 0 represents its initial height in feet, and of course time t is measured in seconds. The coefficient -16 is related to the earth’s acceleration of gravity which is (about) 32 feet per sec per sec. The negative sign results from gravity acting as a downward force. If the initial velocity is upward then v 0 > 0 (positive) and if it is downward then v 0 < 0 (negative).

29. Projectile motion equation is Since the coefficient of t 2 is negative the parabola opens downward. This tells us that the projectile has its maximum height at the vertex (h, k) of the parabola. h = time at which the maximum height occurs k = the maximum height The path of flight of a projectile looks like or a portion of this parabolic curve.

30. Example A model rocket is launched with an initial velocity of v o = 150 feet per second and leaves the platform with an initial height of h o = 10 feet. a) Write a formula s(t) that models the height of the rocket after t seconds. b) How high is the rocket after 2 seconds? c) Find the maximum height of the rocket. Solution: (a)

31. Solution Solution continued: The rocket is 246 feet high after 2 seconds. c) Because a is negative, the vertex is the highest point on the graph, with a t-coordinate of The y-coordinate is: The vertex is at (4.7, 361.6). So the rocket’s maximum height is (about) 361.6 feet. b) How high is the rocket after 2 seconds?

32. Example A well-conditioned athlete’s heart rate can reach 200 beats per minute (bpm) during a strenuous workout. Upon stopping the activity, a typical heart rate decreases rapidly and then gradually levels off. See the data table. Graphing this data we get Determine the equation of a parabola that models this data. f(x) = a(x – h) 2 + k How do we start?

33. From the description, we will take that the minimum heart rate is 80 beats per minute and that occurs when x = 8. That means the vertex is at (8, 80). So using the vertex form we have f(x) = a(x – 8) 2 + 80 How do we find the value of coefficient a? a = 1.875

34. Example