MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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

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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 3 Bruce Mayer, PE Chabot College Mathematics GRAPH BY PLOTTING POINTS  Step1. Make a representative T-table of solutions of the equation.  Step 2. Plot the solutions (the “dots”) as ordered pairs in the Cartesian coordinate plane.  Step 3. Connect the solutions (dots) in Step 2 by a smooth curve

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 4 Bruce Mayer, PE Chabot College Mathematics Making Complete Plots 1.Arrows in POSITIVE Direction Only 2.Label x & y axes on POSITIVE ends 3.Mark and label at least one unit on each axis 4.Use a ruler for Axes & Straight-Lines 5.Label significant points or quantities           

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 5 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 6 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 7 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)

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 8 Bruce Mayer, PE Chabot College Mathematics Examples of ax 2 Parabolas 4 3 6 2 5 1 x y

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 9 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 10 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 more 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 11 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 12 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.  Shift to Value of x that makes (x – h) =0 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 13 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 14 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 15 Bruce Mayer, PE Chabot College Mathematics Quadratic Fcn in Standard Form  The Quadratic Function Written in STANDARD GRAPHING 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 16 Bruce Mayer, PE Chabot College Mathematics Graphing y = 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 17 Bruce Mayer, PE Chabot College Mathematics Graphing y = 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: AdditionPrin + MultPrin + SqRtPrin + AdditionPrin –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).

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 18 Bruce Mayer, PE Chabot College Mathematics Graphing y = 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 19 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 20 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Graph  SOLUTION cont. Sketch Graph Using the 4 points –Vertex –Two x-Intercepts –One y-Intercept

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 22 Bruce Mayer, PE Chabot College Mathematics Translation of Graphs  Graph y = f(x) = x 2. Make T-Table & Connect-Dots  Select integers for x, starting with −2 and ending with +2. The T-table:

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 23 Bruce Mayer, PE Chabot College Mathematics Translation of Graphs  Now Plot the Five Points and connect them with a smooth Curve (−2,4)(2,4) (−1,1)(1,1) (0,0)

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 24 Bruce Mayer, PE Chabot College Mathematics Axes Translation  Now Move UP the graph of y = x 2 by two units as shown (−2,4)(2,4) (−1,1)(1,1) (0,0) (−2,6) (2,6) (−1,3) (1,3) (0,2)  What is the Equation for the new Curve?

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 25 Bruce Mayer, PE Chabot College Mathematics Axes Translation  Compare ordered pairs on the graph of with the corresponding ordered pairs on the new curve:

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 26 Bruce Mayer, PE Chabot College Mathematics Axes Translation  Notice that the x-coordinates on the new curve are the same, but the y-coordinates are 2 units greater  So every point on the new curve makes the equation y = x 2 +2 true, and every point off the new curve makes the equation y = x 2 +2 false.  An equation for the new curve is thus y = x 2 +2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 27 Bruce Mayer, PE Chabot College Mathematics Axes Translation  Similarly, if every point on the graph of y = x 2 were is moved 2 units down, an equation of the new curve is y = x 2 −2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 28 Bruce Mayer, PE Chabot College Mathematics Axes Translation  When every point on a graph is moved up or down by a given number of units, the new graph is called a vertical translation of the original graph.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 29 Bruce Mayer, PE Chabot College Mathematics Vertical Translation  Given the Graph of y = f(x), and c > 0 1.The graph of y = f(x) + c is a vertical translation c-units UP 2.The graph of y = f(x) − c is a vertical translation c-units DOWN

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 30 Bruce Mayer, PE Chabot College Mathematics Horizontal Translation  What if every point on the graph of y = x 2 were moved 5 units to the right as shown below.  What is the eqn of the new curve? (−2,4)(2,4)(3,4)(7,4)

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 31 Bruce Mayer, PE Chabot College Mathematics Horizontal Translation  Compare ordered pairs on the graph of with the corresponding ordered pairs on the new curve:

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 32 Bruce Mayer, PE Chabot College Mathematics Horizontal Translation  Notice that the y-coordinates on the new curve are the same, but the x-coordinates are 5 units greater.  Does every point on the new curve make the equation y = (x+5) 2 true? No; for example if we input (5,0) we get 0 = (5+5) 2, which is false. But if we input (5,0) into the equation y = (x−5) 2, we get 0 = (5−5) 2, which is TRUE.  In fact, every point on the new curve makes the equation y = (x−5) 2 true, and every point off the new curve makes the equation y = (x−5) 2 false. Thus an equation for the new curve is y = (x−5) 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 33 Bruce Mayer, PE Chabot College Mathematics Horizontal Translation  Given the Graph of y = f(x), and c > 0 1.The graph of y = f(x−c) is a horizontal translation c-units to the RIGHT 2.The graph of y = f(x+c) is a horizontal translation c-units to the LEFT

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Use Translation to graph f(x) = (x−3) 2 −2  LET y = f(x) → y = (x−3) 2 −2  Notice that the graph of y = (x−3) 2 −2 has the same shape as y = x 2, but is translated 3-unit RIGHT and 2-units DOWN.  In the y = (x−3) 2 −2, call 3 and −2 TRANSLATORS

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 35 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  The graphs of y=x 2 and y=(x−3) 2 −2 are different; although they have the Same shape they have different locations  Now remove the translators by a substitution of x’ (“x-prime”) for x, and y’ (“y-prime”) for y  Remove translators for an (x’,y’) eqn

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 36 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Since the graph of y=(x−3) 2 −2 has the same shape as the graph of y’ =(x’) 2 we can use ordered pairs of y’ =(x’) 2 to determine the shape  T-table for y’ =(x’) 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 37 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Next use the TRANSLATION RULES to find the origin of the x’y’-plane. Draw the x’-axis and y’-axis through the translated origin The origin of the x’y’-plane is 3 units right and 2 units down from the origin of the xy-plane. Through the translated origin, we use dashed lines to draw a new horizontal axis (the x’-axis) and a new vertical axis (the y’-axis).

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 38 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Locate the Origin of the Translated Axes Set using the translator values Move: 3-Right, 2-Down

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 39 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Now Plot the ordered pairs of the x’y’ equation on the x’y’-plane, and use the points to draw an appropriate graph. Remember that this graph is smooth Move: 3-Right 2-Down

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 40 Bruce Mayer, PE Chabot College Mathematics Example Plot by Translation  Perform a partial-check to determine correctness of the last graph. Pick any point on the graph and find its (x,y) CoOrds; e.g., (4, −1) is on the graph  The Ordered Pair (4, −1) should make the xy Eqn True

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 41 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  Sub (4, −1) into y = (x−3) 2 −2 :  Thus (4, −1) does make y = (x−3) 2 −2 true. In fact, all the points on the translated graph make the original Eqn true, and all the points off the translated graph make the original Eqn false

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 42 Bruce Mayer, PE Chabot College Mathematics Example  Plot by Translation  What are the Domain & Range of y = (x−3) 2 −2?  To find the domain & range of the xy-eqn, examine the xy-graph (not the x’y’ graph).  The xy graph showns Domain of f is {x|x is any real number} Range of f is {y|y ≥ −2}

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 43 Bruce Mayer, PE Chabot College Mathematics Graphing Using Translation 1.Let y = f(x) 2.Remove the x-value & y-value “translators” to form an x’y’ eqn. 3.Find ordered pair solutions of the x’y’ eqn 4.Use the translation rules to find the origin of the x’y’-plane. Draw dashed x’ and y’ axes through the translated origin. 5.Plot the ordered pairs of the x’y’ equation on the x’y’-plane, and use the points to draw an appropriate graph.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 44 Bruce Mayer, PE Chabot College Mathematics Example  Graph  Step-2  Step-3 → T-table in x’y’  Step-1

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 45 Bruce Mayer, PE Chabot College Mathematics Example  Graph  Step-4: The origin of the x’y’ -plane is 3 units left (to the MINUS side) and 7 units up from the origin of the xy-plane:

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 46 Bruce Mayer, PE Chabot College Mathematics Example  Graph  Step-5: We know that the basic shape of this graph is Parabolic. Thus we can sketch the graph using Fewer Points on the translated axis using the T-Table

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 47 Bruce Mayer, PE Chabot College Mathematics Compare: Notice IDENTICAL Shapes

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 48 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 49 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 50 Bruce Mayer, PE Chabot College Mathematics Example  Graph  SOLUTION  The vertex is at (3/2, 3/2)  The Parabola Opens DOWN 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 51 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])

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 52 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 53 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 54 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 55 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 56 Bruce Mayer, PE Chabot College Mathematics Example  Graph  SOLUTION Step 4 Let f (x) = 0. Step 5 Let x = 0.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 57 Bruce Mayer, PE Chabot College Mathematics Example  Graph  SOLUTION cont. Sketch Graph Using the points Just Determined

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 58 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]

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 59 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.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 60 Bruce Mayer, PE Chabot College Mathematics All Done for Today Geometric Complete The Square

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 61 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 62 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

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 63 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 64 Bruce Mayer, PE Chabot College Mathematics Translate Up or Down  Make Graphs for  Notice: Of the form of VERTICAL Translations

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 65 Bruce Mayer, PE Chabot College Mathematics Translate Left or Right  Make Graphs for  Notice: Of the form of HORIZONTAL Translations

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 66 Bruce Mayer, PE Chabot College Mathematics Example  ReCall Graph y = |x|  Make T-table

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 67 Bruce Mayer, PE Chabot College Mathematics Example  Graph y = |x+2|+3  Step-1  Step-2  Step-3 → T-table in x’y’

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 68 Bruce Mayer, PE Chabot College Mathematics Example  Graph y = |x+2|+3  Step-4: the x’y’-plane origin is 2 units LEFT and 3 units UP from xy-plane Up 3 Left 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 69 Bruce Mayer, PE Chabot College Mathematics Example  Graph y = |x+2|+3  Step-5: Remember that the graph of y = |x| is V-Shaped: Up 3 Left 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 70 Bruce Mayer, PE Chabot College Mathematics Up 3 Over 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 71 Bruce Mayer, PE Chabot College Mathematics Up 3 Over 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 72 Bruce Mayer, PE Chabot College Mathematics Quadratic Fcn Translation  Given the graph of the equation y = f(x), and c > 0,  the graph of y = f(x) + c is the graph of y = f(x) shifted UP (vertically) c units;  the graph of y = f(x) – c is the graph of y = f(x) shifted DOWN (vertically) c units y = 3x 2 y = 3x 2 − 3 y = 3x 2 +2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 73 Bruce Mayer, PE Chabot College Mathematics Quadratic Fcn Translation  Given the graph of the equation y = f(x), and c > 0,  the graph of y = f(x– c) is the graph of y = f(x) shifted RIGHT (Horizontally) c units;  the graph of y = f(x + c) is the graph of y = f(x) shifted LEFT (Horizontally) c units. y = 3x 2 y = 3(x-2) 2 y = 3(x+2) 2

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 74 Bruce Mayer, PE Chabot College Mathematics Graphing by Translation 1.Let y = f(x) 2.Remove the translators to form an x’y’ eqn 3.Find ordered pair solutions of the x’y’ eqn 4.Use the translation rules to find the origin of the x’y’-plane. Draw the dashed x’ and y’ axes through the translated origin. 5.Plot the ordered pairs of the x’y’ equation on the x’y’-plane, and use the points to draw an appropriate graph.

BMayer@ChabotCollege.edu MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 75 Bruce Mayer, PE Chabot College Mathematics

MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.

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Presentation on theme: "MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical."— Presentation transcript:

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