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MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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Presentation on theme: "MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical."— Presentation transcript:

1 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §8.3 Quadratic Fcn Applications

2 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §8.3 → Quadratic Eqn Graphs  Any QUESTIONS About HomeWork §8.3 → HW MTH 55

3 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 3 Bruce Mayer, PE Chabot College Mathematics Parabolas with Vertical Shifts

4 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 4 Bruce Mayer, PE Chabot College Mathematics Parabolas with Vertical Shifts  The Graph of F(x) = x 2 + k has the same SHAPE as f(x) = x 2, with the shape shifted VERTICALLY: k units UP when k > 0 –i.e.; k is POSITIVE |k| units DOWN when k < 0 –i.e., k is NEGATIVE The Vertex is (0, k) k > 0 produces shift up k units k < 0 produces shift down k units

5 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 5 Bruce Mayer, PE Chabot College Mathematics Parabolas with Horizontal Shifts

6 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 6 Bruce Mayer, PE Chabot College Mathematics Parabolas with Horizontal Shifts  The Graph of F(x) = (x − h) 2 has the same SHAPE as f(x) = x 2, with the shape shifted HORIZONTALLY: h units RIGHT when h > 0 –POSTIVE h |h| units LEFT when h < 0 –NEGATIVE h The Vertex is (h,0) h > 0 produces shift right h units h < 0 produces shift left h units

7 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 7 Bruce Mayer, PE Chabot College Mathematics Caveat: Shifting Parabolas  Horizontal Shifts are EASY to move in the WRONG Direction  To determine the size & direction of a Horizontal shift, find the value of x that makes x−h = 0. Some Examples F(x) = (x − 5) 2 shifts RIGHT 5-units as x = +5 causes x − 5 to be zero F(x) = (x + 7) 2 shifts LEFT 7-units as x = −7 causes x + 7 to be zero

8 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 8 Bruce Mayer, PE Chabot College Mathematics Parabolas with ↨ & ↔ Shifts

9 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  F(x) = (x+3) 2 − 2  Cast F(x) = (x+3) 2 − 2 into the form F(x) = (x−h) 2 + k  Since h < 0, there is a shift to the left, and since k < 0, there is a shift down.

10 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 10 Bruce Mayer, PE Chabot College Mathematics Graphs of f(x) = a(x – h) 2 + k  The Graph of f(x) = a(x – h) 2 + k has the same shape as the graph of y = a(x – h) 2 If k is positive, the graph of y = a(x – h) 2 is shifted k units up. If k is negative, the graph of y = a(x – h) 2 is shifted |k| units down The vertex is (h, k), and the axis of symmetry is x = h. For a > 0, k is the MINimum function value For a < 0, k is the MAXimum function value

11 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Graph  SOLUTION: Make T-Table, ID Vertex and Maximum xy(x, y) 0 –1 –2 –3 –4 –5 -11/2 –3 –3/2 –1 –3/2 –3 (0, -11/2) (–1, –3) (–2, –3/2) (–3, –1) (–4, –3/2) (–5, –3) vertex maximum (−1) LoS Concave DOWN

12 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 12 Bruce Mayer, PE Chabot College Mathematics Maximum & Minimum Probs  We have seen that for any quadratic function f, the value of f(x) at the vertex is either a maximum or a minimum.  Thus problems in which a quantity must be maximized or minimized can be solved by finding the coordinates of the vertex, assuming the problem can be modeled with a quadratic function.

13 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Maximum  A farmer has 200 ft of fence with which to form a rectangular pen on his farm. If an existing fence forms one side of the rectangle, what dimensions will MAXIMIZE the size of the area? 1.Familiarize - make a drawing and label it, letting w = Rectangle Width l = Rectangle Length Existing fence ww l

14 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Maximum  Recall that for Rectangles Area = lw Perimeter = 2w + 2l  Since the existing fence forms one length of the rectangle, the fence will comprise three sides. Thus

15 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  Maximum 2.Translate. Now have two equations: One guarantees that all 200 ft of fence will be used; the other expresses area in terms of length and width. Existing fence ww l

16 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Maximum 3.CarryOut - Now need to express A as a function of l or w but not both. To do so, we solve for l in the first equation to obtain l = 200 – 2w. Substituting for l in the second equation, yields a quadratic function.

17 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Maximum  Factoring and completing the square to Obtain  Then by the Max-at-Vertex Criteria:  Find l max from the Perimeter Constraint

18 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Maximum 4.Check - The check is left for us to do Later. 5.State - The dimensions for the largest rectangular area for the pen that can be enclosed is 50 ft by 100 ft. Existing fence 50’ 100’ 50’

19 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Sniping Siblings  A widower with 10 children marries a widow who also has children. After their marriage, they have their own children. If the total number of children is 24, and we assume that the children of the same parents do not fight.  Find the maximum possible number of fights among the children. In this example, a fight between Sean and Misty, no matter how many times they fight, is considered as ONE fight.

20 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Sniping Siblings  SOLUTION: Suppose the widow had x number of children before marriage. Then the couple has 24 − 10 − x = 14 − x additional children after their marriage.  Since the children of the same parents do not fight, there are no fights among the 10 children the widower brought into the marriage, or among the x children the widow brought into the marriage, or among the 14 − x children of the parents

21 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Sniping Siblings  Then The possible number of fights among the children of 1.the widower (10 children) and the widow (x children) is 10x. 2.the widower (10 children) and the couple (14 – x children) is 10(14 – x), and 3.the widow (x children) and the couple (14 – x children) is x(14 – x)

22 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Sniping Siblings  TRANSLATE: The possible number y of all fights is given by  In the Quadratic Function: y = f (x) = –x 2 +14x a = –1, b = 14, and c = 140.

23 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Sniping Siblings  The vertex (h, k) is given by  Since, a = −1 < 0, the function f opens DOWN and has MAXIMUM value k. Hence, the maximum possible number of fights among the children is 189.

24 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  A Model Rocket is launched straight up with an initial velocity of 60 feet per sec.  The equation h ≈ −16t t describes the height, h, of the rocket, t seconds after launch, FIND: the maximum height that the rocket reaches. the amount of time that the rocket is in the air; i.e., find the total flite-time

25 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  SOLUTION - Since the graph of h ≈ −16t t is a parabola that opens down, the maximum height occurs at its vertex:  Use the quadratic equation to find the height at the vertex time of sec h = −16(1.875) (1.875) = 56.25

26 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  SOLUTION: for h ≈ −16t t the Vertex is at (1.875, 56.25) Max Height Time at Which Max Height Occurs  Interpreting Vertex Information find: The maximum height is feet, The max height occurs seconds after rocket launch

27 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  The rocket Flite-Time is from launch until it returns to the ground.  At launch and upon returning to the ground, the rocket’s height is 0, so we need to find t when h = 0 0 = −16t t 0 = − 4t(4t − 15) −4t = 0 or 4t − 15 = 0 t = 0 or t = 3.75

28 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  The height is 0 when t = 0 and when t = 3.75 seconds, so the rocket is in the air for 3.75 seconds.  Check by Graphing h = −16t t Max Hgt ≈ 56ft ≈ 3.8 sec

29 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  Graph the quadratic function f(x) = x 2 + 2x + 2 and solve each quadratic inequality. a. x 2 + 2x + 2 > 0 b. x 2 + 2x + 2 < 0  SOLUTION – Analyze Eqn Parameters Step 1: a = 1, b = 2, and c = 2 Step 2: a = 1 > 0, the parabola opens UP Step 3: Find (h, k) → Next Slide

30 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  Find h by Formula  Find k = f(h)  With the Vertex of (−1,1) The Max for f(x) = 1, which occurs at x = −1  Next find x-intercepts → f(x) = 0

31 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 31 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  Setting f(x) = 0:  The solutions are not real, the graph does not intersect the x-axis

32 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 32 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  Find y-intercept by setting x = 0 in f(x)  From the (−1,1) Vertex recognize the Line of Symmetry (LoS) at x = −1 Moving 1-unit to the Left & Right of the LoS produces points (−2, 2), and (0, 2) which are symmetric with respect to the axis of symmetry

33 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  To Draw Graph use Opens UP Vertex (−1,1) NO x-intercepts y-intercept = 2 Pts Symmetric about the LoS –(−2, 2) –(0, 2)

34 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  Graph InEquality  Use Graph to Assess InEqualities  The entire graph lies above the x-axis. Thus, y is always > 0. a)x 2 +2x + 2 > 0 is always true, the solution is (−∞,∞) b)x 2 +2x + 2 < 0 is never true, the solution is Ø

35 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 35 Bruce Mayer, PE Chabot College Mathematics Fitting Quadratic Functions  Whenever a certain quadratic function fits a situation, that function can be determined if THREE inputs and their outputs are known.  Each of the given ordered pairs is called a DATA POINT

36 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 36 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Fitting  Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data.  SOLUTION – Need to Find a function of the form f(x) = ax 2 + bx + c given that f(0) = 10.4, f(3) = 16.8, and f(6) = 12.6  Thus Need a, b, & c Such That: a(0) 2 + b(0) + c = 10.4 a(3) 2 + b(3) + c = 16.8 a(6) 2 + b(6) + c = 12.6

37 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 37 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Fitting  This amounts to Solving a SYSTEM of Three Eqns for Unknowns a, b, & c c = 10.4 (1) 9a + 3b + c = 16.8 (2) 36a + 6b + c = 12.6 (3)  Substituting c = 10.4 into eqns (2) & (3) and solving the resulting system yields

38 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 38 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Fitting  Thus the data set (0, 10.4), (3, 16.8), and (6, 12.6) produces a Quadratic Fit:

39 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 39 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §8.3 Exercise Set 58, 68  A Quadratic (and Linear) Fit for Fish Relationship between centrum radius and precaudal length for eastern North Pacific salmon sharks (Lamna ditropis), showing significant fits given by linear and quadratic equations (sexes combined, n=182). PCL = precaudal length, CR = centrum radius

40 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 40 Bruce Mayer, PE Chabot College Mathematics All Done for Today Quadratic Production Function

41 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 41 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

42 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 42 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

43 MTH55_Lec-52_Fa08_sec_8-3b_Quadratic_Fcn_Apps.ppt 43 Bruce Mayer, PE Chabot College Mathematics


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