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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §8.3 Quadratic Fcn Graphs

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review § Any QUESTIONS About §8.2 → Quadratic Eqn Applications Any QUESTIONS About HomeWork §8.2 → HW-38 8.2 MTH 55

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 3 Bruce Mayer, PE Chabot College Mathematics Graphs of Quadratic Eqns All quadratic functions have graphs similar to y = x 2. Such curves are called parabolas. They are U-shaped and symmetric with respect to a vertical line known as the parabola’s line of symmetry or axis of symmetry. For the graph of f(x) = x 2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example Graph f(x) = 2x 2 Solution: Make T-Table and Connect-Dots xy(x, y) 0 1 –1 2 –2 0228802288 (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) x = 0 is Axis of Symm (0,0) is Vertex

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example Graph f(x) = −3x 2 Solution: Make T-Table and Connect-Dots Same Axis & Vertex but opens DOWNward xy(x, y) 0 1 –1 2 –2 0 –3 –12 (0, 0) (1, –3) (–1, –3) (2, –12) (–2, –12)

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 6 Bruce Mayer, PE Chabot College Mathematics Examples of ax 2 Parabolas 4 3 6 2 5 1

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 7 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = ax 2 The graph of f(x) = ax 2 is a parabola with x = 0 as its axis of symmetry. The Origin, (0,0) as its vertex. For Positive a the parabola opens upward For Negative a the parabola opens downward If |a| is greater than 1; e.g., 4, the parabola is narrower (tighter) than y = x 2. If |a| is between 0 and 1 e.g., ¼, the parabola is wider (broader) than y = x 2.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 8 Bruce Mayer, PE Chabot College Mathematics The Graph of f(x) = a(x – h) 2 We could next consider graphs of f(x) = ax 2 + bx + c, where b and c are not both 0. It turns out to be convenient to first graph f(x) = a(x – h) 2, where h is some constant; i.e., h is a NUMBER This allows us to observe similarities to the graphs drawn in previous slides.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example Graph f(x) = (x−2) 2 Solution: Make T-Table and Connect-Dots The Vertex SHIFTED 2-Units to the Right xy(x, y) 0 1 –1 2 3 4 419014419014 (0, 4) (1, 1) (–1, 9) (2, 0) (3, 1) (4, 4) vertex

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 10 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = a(x−h) 2 The graph of y = f(x) = a(x – h) 2 has the same shape as the graph of y = ax 2. If h is positive, the graph of y = ax 2 is shifted h units to the right. If h is negative, the graph of y = ax 2 is shifted |h| units to the left. The vertex is (h, 0) and the axis of symmetry is x = h.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 11 Bruce Mayer, PE Chabot College Mathematics Graph of f(x) = a(x – h) 2 + k Given a graph of f(x) = a(x – h) 2, what happens if we add a constant k? Suppose we add k = 3. This increases f(x) by 3, so the curve moves up If k is negative, the curve moves down. The axis of symmetry for the parabola remains x = h, but the vertex will be at (h, k), or equivalently (h, f(h)) f(h) = a([h] – h) 2 + k = 0 + k → f(h) = k

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example Graph The Vertex SHIFTED 3-Units Left and 1-Unit Down Make T-Table and Connect-Dots 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

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 13 Bruce Mayer, PE Chabot College Mathematics Quadratic Fcn in Standard Form The Quadratic Function Written in STANDARD Form: The graph of f is a parabola with vertex (h, k). The parabola is symmetric with respect to the line x = h, called the axis of the parabola. If a > 0, the parabola opens up, and if a < 0, the parabola opens down.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example Find Quadratic Fcn Find the standard form of the quadratic function whose graph has vertex (−3, 4) and passes through the point ( −4, 7). SOLUTION: Let y = f(x) be the quadratic function. Then

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 15 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = a(x – h) 2 + k 1.The graph is a parabola. Identify a, h, and k 2.Determine how the parabola opens. If a > 0 (positive), the parabola opens up. If a < 0 (negative), the parabola opens down. 3.Find the vertex. The vertex is (h, k). If a > 0 (or a < 0), the function f has a minimum (or a maximum) value k at x = h

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 16 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = a(x – h) 2 + k 4.Find the x-intercepts. Find the x-intercepts (if any) by setting f(x) = 0 and solving the equation a(x – h) 2 + k = 0 for x. –Solve by one of: [Factoring + ZeroProducts], Quadratic Formula, CompleteSquare –If the solutions are real numbers, they are the x-intercepts. –If the solutions are NOT Real Numbers, the parabola either lies above the x–axis (when a > 0) or below the x–axis (when a < 0).

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 17 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = a(x – h) 2 + k 5.Find the y-intercept Find the y-intercept by replacing x with 0. Then y = f(0) = ah 2 + k is the y-intercept. 6.Sketch the graph Plot the points found in Steps 3-5 and join them by a parabola. –If desired, show the axis of symmetry, x = h, for the parabola by drawing a dashed vertical line

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION Step 1 a = 2, h = 3, and k = –8 Step 2 a = 2, a > 0, the parabola opens up. Step 3 (h, k) = (3, –8); the function f has a minimum value –8 at x = 3. Step 4 Set f (x) = 0 and solve for x.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION cont. Step 5 Replace x with 0. Step 6 axis: x = 3, opens up, vertex: (3, –8), passes through (1, 0), (5, 0) and (0, 10), the graph is y = 2x 2 shifted three units right and eight units down.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION cont. Sketch Graph Using the 4 points –Vertex –Two x-Intercepts –One y-Intercept

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 21 Bruce Mayer, PE Chabot College Mathematics Completing the Square By completing the square, we can rewrite any polynomial ax 2 + bx + c in the form a(x – h) 2 + k. Once that has been done, the procedures just discussed enable us to graph any quadratic function.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION f (x) = x 2 – 2x – 1 = (x 2 – 2x) – 1 = (x 2 – 2x + 1) – 1 – 1 = (x 2 – 2x + 1 – 1) – 1 = (x – 1) 2 – 2 The vertex is at (1, −2) The Parabola Opens UP x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Adding ZERO

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION The vertex is at (3/2, 3/2) f (x) = –2x 2 + 6x – 3 = –2(x 2 – 3x) – 3 = –2(x 2 – 3x + 9/4) – 3 + 18/4 = –2(x 2 – 3x + 9/4 – 9/4) – 3 = –2(x – 3/2) 2 + 3/2 x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 1 6 5 4 -4 -5 Complete Square

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 24 Bruce Mayer, PE Chabot College Mathematics The Vertex of a Parabola By the Process of Completing-the- Square we arrive at a FORMULA for the vertex of a parabola given by f(x) = ax 2 + bx + c: The x-coordinate of the vertex is −b/(2a). The axis of symmetry is x = −b/(2a). The second coordinate of the vertex is most commonly found by computing f(−b/[2a])

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 25 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = ax 2 + bx + c 1.The graph is a parabola. Identify a, b, and c 2.Determine how the parabola opens If a > 0, the parabola opens up. If a < 0, the parabola opens down 3.Find the vertex (h, k). Use the formula

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 26 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = ax 2 + bx + c 4.Find the x-intercepts Let y = f(x) = 0. Find x by solving the equation ax 2 + bx + c = 0. If the solutions are real numbers, they are the x-intercepts. If not, the parabola either lies –above the x–axis when a > 0 –below the x–axis when a < 0

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 27 Bruce Mayer, PE Chabot College Mathematics Graphing f(x) = ax 2 + bx + c 5.Find the y-intercept. Let x = 0. The result f(0) = c is the y-intercept. 6.The parabola is symmetric with respect to its axis, x = −b/(2a) Use this symmetry to find additional points. 7.Draw a parabola through the points found in Steps 3-6.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION Step 1 a = –2, b = 8, and c = –5 Step 2 a = –2, a < 0, the parabola opens down. Step 3 Find (h, k). Maximum value of y = 3 at x = 2

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION Step 4 Let f (x) = 0. Step 5 Let x = 0.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION Step 6 Axis of symmetry is x = 2. Let x = 1, then the point (1, 1) is on the graph, the symmetric image of (1, 1) with respect to the axis x = 2 is (3, 1). The symmetric image of the y–intercept (0, –5) with respect to the axis x = 2 is (4, –5). Step 7 The parabola passing through the points found in Steps 3–6 is sketched on the next slide.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 31 Bruce Mayer, PE Chabot College Mathematics Example Graph SOLUTION cont. Sketch Graph Using the points Just Determined

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 32 Bruce Mayer, PE Chabot College Mathematics Find Domain & Range Given the graph of f(x) = −2x 2 +8x − 5 Find the domain and range for f(x) SOLUTION Examine the Graph to find that the: Domain is (−∞, ∞) Range is (−∞, 3]

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 33 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §8.3 Exercise Set 4, 16, 22, 30 The Directrix of a Parabola A line perpendicular to the axis of symmetry used in the definition of a parabola. A parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix.

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 34 Bruce Mayer, PE Chabot College Mathematics All Done for Today Geometric Complete The Square

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 35 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-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 36 Bruce Mayer, PE Chabot College Mathematics Graph y = |x| Make T-table

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BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 37 Bruce Mayer, PE Chabot College Mathematics

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