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BMayer@ChabotCollege.edu 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 BMayer@ChabotCollege.edu Chabot Mathematics §8.3 Quadratic Fcn Applications

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BMayer@ChabotCollege.edu 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-39 8.3 MTH 55

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

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BMayer@ChabotCollege.edu 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

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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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

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BMayer@ChabotCollege.edu 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.

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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.

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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.

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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’

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BMayer@ChabotCollege.edu 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.

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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)

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BMayer@ChabotCollege.edu 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 + 140 a = –1, b = 14, and c = 140.

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BMayer@ChabotCollege.edu 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.

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BMayer@ChabotCollege.edu 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 2 + 60t 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

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BMayer@ChabotCollege.edu 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 2 + 60t 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 1.875 sec h = −16(1.875) 2 + 60(1.875) = 56.25

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BMayer@ChabotCollege.edu 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 2 + 60t the Vertex is at (1.875, 56.25) Max Height Time at Which Max Height Occurs Interpreting Vertex Information find: The maximum height is 56.25 feet, The max height occurs 1.875 seconds after rocket launch

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BMayer@ChabotCollege.edu 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 2 + 60t 0 = − 4t(4t − 15) −4t = 0 or 4t − 15 = 0 t = 0 or t = 3.75

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BMayer@ChabotCollege.edu 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 2 + 60t Max Hgt ≈ 56ft Crash-Down @ ≈ 3.8 sec

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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)

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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:

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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

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BMayer@ChabotCollege.edu 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 BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

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

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

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3.3 Analyzing Graphs of Quadratic Functions

3.3 Analyzing Graphs of Quadratic Functions

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