Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §8.3 Quadratic Fcn Graphs Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

8.2 Review § Any QUESTIONS About Any QUESTIONS About HomeWork MTH 55 Review § Any QUESTIONS About §8.2 → Quadratic Eqn Applications Any QUESTIONS About HomeWork §8.2 → HW-38

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

Making Complete Plots            Arrows in POSITIVE Direction Only Label x & y axes on POSITIVE ends Mark and label at least one unit on each axis Use a ruler for Axes & Straight-Lines Label significant points or quantities         

Graphs of Quadratic Eqns All quadratic functions have graphs similar to y = x2. 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) = x2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.

Example  Graph f(x) = 2x2 Solution: Make T-Table and Connect-Dots x y (-1,2 ) (2,8) (-2,8) -5 -4 -3 -2 -1 1 2 3 4 5 4 3 6 2 5 1 (1,2) (0, 0) -1 -2 7 8 x y (x, y) 1 –1 2 –2 8 (0, 0) (1, 2) (–1, 2) (2, 8) (–2, 8) x = 0 is Axis of Symm (0,0) is Vertex

Example  Graph f(x) = −3x2 y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 Solution: Make T-Table and Connect-Dots x y (x, y) 1 –1 2 –2 –3 –12 (0, 0) (1, –3) (–1, –3) (2, –12) (–2, –12) Same Axis & Vertex but opens DOWNward

Examples of ax2 Parabolas y 6 5 4 3 2 1 x

Graphing f(x) = ax2 The graph of f(x) = ax2 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 = x2. If |a| is between 0 and 1 e.g., ¼, the parabola is wider (broader) than y = x2.

The Graph of f(x) = a(x – h)2 We could next consider graphs of f(x) = ax2 + 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.

Example  Graph f(x) = (x−2)2 y -5 -4 -3 -2 -1 1 2 3 4 5 4 3 6 2 5 1 -1 -2 7 8 Solution: Make T-Table and Connect-Dots x y (x, y) 1 –1 2 3 4 9 (0, 4) (1, 1) (–1, 9) (2, 0) (3, 1) (4, 4) vertex The Vertex SHIFTED 2-Units to the Right

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 = ax2. Shift to Value of x that makes (x – h) =0 If h is positive, the graph of y = ax2 is shifted h units to the right. If h is negative, the graph of y = ax2 is shifted |h| units to the left. The vertex is (h, 0) and the axis of symmetry is x = h.

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

Example  Graph Make T-Table and Connect-Dots x y (x, y) –1 –2 –3 –4 -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 -4 -5 -6 -7 -8 Make T-Table and Connect-Dots x y (x, y) –1 –2 –3 –4 –5 -11/2 –3/2 (0, -11/2) (–1, –3) (–2, –3/2) (–3, –1) (–4, –3/2) (–5, –3) vertex The Vertex SHIFTED 3-Units Left and 1-Unit Down

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.

Graphing f(x) = a(x – h)2 + k The graph is a parabola. Identify a, h, and k Determine how the parabola opens. If a > 0 (positive), the parabola opens up. If a < 0 (negative), the parabola opens down. 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

Graphing f(x) = a(x – h)2 + k 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).

Graphing f(x) = a(x – h)2 + k Find the y-intercept Find the y-intercept by replacing x with 0. Then y = f(0) = ah2 + k is the y-intercept. 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

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.

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 = 2x2 shifted three units right and eight units down.

Example  Graph SOLUTION cont. Sketch Graph Using the 4 points Vertex Two x-Intercepts One y-Intercept

Translation of Graphs Graph y = f(x) = x2. Make T-Table & Connect-Dots Select integers for x, starting with −2 and ending with +2. The T-table:

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)

Axes Translation Now Move UP the graph of y = x2 by two units as shown (−2,6) (2,6) (−2,4) (2,4) (−1,3) (1,3) What is the Equation for the new Curve? (0,2) (−1,1) (1,1) (0,0)

Axes Translation Compare ordered pairs on the graph of with the corresponding ordered pairs on the new curve:

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 = x2+2 true, and every point off the new curve makes the equation y = x2+2 false. An equation for the new curve is thus y = x2+2

Axes Translation Similarly, if every point on the graph of y = x2 were is moved 2 units down, an equation of the new curve is y = x2−2

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.

Vertical Translation Given the Graph of y = f(x), and c > 0 The graph of y = f(x) + c is a vertical translation c-units UP The graph of y = f(x) − c is a vertical translation c-units DOWN

Horizontal Translation What if every point on the graph of y = x2 were moved 5 units to the right as shown below. (−2,4) (2,4) (3,4) (7,4) What is the eqn of the new curve?

Horizontal Translation Compare ordered pairs on the graph of with the corresponding ordered pairs on the new curve:

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

Horizontal Translation Given the Graph of y = f(x), and c > 0 The graph of y = f(x−c) is a horizontal translation c-units to the RIGHT The graph of y = f(x+c) is a horizontal translation c-units to the LEFT

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 = x2, but is translated 3-unit RIGHT and 2-units DOWN. In the y = (x−3)2−2, call 3 and −2 TRANSLATORS

Example  Plot by Translation The graphs of y=x2 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

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

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).

Example  Plot by Translation Locate the Origin of the Translated Axes Set using the translator values Move: 3-Right, 2-Down

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

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

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

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}

Graphing Using Translation Let y = f(x) Remove the x-value & y-value “translators” to form an x’y’ eqn. Find ordered pair solutions of the x’y’ eqn Use the translation rules to find the origin of the x’y’-plane. Draw dashed x’ and y’ axes through the translated origin. Plot the ordered pairs of the x’y’ equation on the x’y’-plane, and use the points to draw an appropriate graph.

Example  Graph Step-1 Step-2 Step-3 → T-table in x’y’

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:

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

Notice IDENTICAL Shapes Compare: Notice IDENTICAL Shapes

Completing the Square By completing the square, we can rewrite any polynomial ax2 + 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.

Example  Graph SOLUTION f (x) = x2 – 2x – 1 = (x2 – 2x) – 1 Adding ZERO f (x) = x2 – 2x – 1 x y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 = (x2 – 2x) – 1 = (x2 – 2x + 1 – 1) – 1 = (x2 – 2x + 1) – 1 – 1 = (x – 1)2 – 2 The vertex is at (1, −2) The Parabola Opens UP

Example  Graph SOLUTION f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3 y -5 -4 -3 -2 -1 1 2 3 4 5 -3 2 -2 3 -1 1 6 5 4 -4 -5 f (x) = –2x2 + 6x – 3 Complete Square = –2(x2 – 3x) – 3 = –2(x2 – 3x + 9/4 – 9/4) – 3 = –2(x2 – 3x + 9/4) – 3 + 18/4 Vertex by complete-sq; x & y intercepts also = –2(x – 3/2)2 + 3/2 The vertex is at (3/2, 3/2) The Parabola Opens DOWN

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) = ax2 + 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])

Graphing f(x) = ax2 + bx + c The graph is a parabola. Identify a, b, and c Determine how the parabola opens If a > 0, the parabola opens up. If a < 0, the parabola opens down Find the vertex (h, k). Use the formula

Graphing f(x) = ax2 + bx + c Find the x-intercepts Let y = f(x) = 0. Find x by solving the equation ax2 + 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

Graphing f(x) = ax2 + bx + c Find the y-intercept. Let x = 0. The result f(0) = c is the y-intercept. The parabola is symmetric with respect to its axis, x = −b/(2a) Use this symmetry to find additional points. Draw a parabola through the points found in Steps 3-6.

Example  Graph Maximum value of y = 3 at x = 2 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

Example  Graph SOLUTION Step 4 Let f (x) = 0. Step 5 Let x = 0.

Example  Graph SOLUTION cont. Sketch Graph Using the points Just Determined

Find Domain & Range Given the graph of f(x) = −2x2 +8x − 5 Find the domain and range for f(x) SOLUTION  Examine the Graph to find that the: Domain is (−∞, ∞) Range is (−∞, 3]

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.

Geometric Complete The Square All Done for Today Geometric Complete The Square

Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu –

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 Expand (x3)^2 to give y = 3x^2 + 18x + 31

Graph y = |x| Make T-table

Translate Up or Down Make Graphs for Notice: Of the form of VERTICAL Translations

Translate Left or Right Make Graphs for Notice: Of the form of HORIZONTAL Translations

Example ReCall Graph y = |x| Make T-table

Example Graph y = |x+2|+3 Step-1 Step-2 Step-3 → T-table in x’y’

Example Graph y = |x+2|+3 Step-4: the x’y’-plane origin is 2 units LEFT and 3 units UP from xy-plane Left 2 Up 3

Example Graph y = |x+2|+3 Step-5: Remember that the graph of y = |x| is V-Shaped: Left 2 Up 3

Over 2 Up 3

Over 2 Up 3

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 = 3x2+2 y = 3x2 y = 3x2−3

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 = 3x2 y = 3(x-2)2 y = 3(x+2)2

Graphing by Translation Let y = f(x) Remove the translators to form an x’y’ eqn Find ordered pair solutions of the x’y’ eqn Use the translation rules to find the origin of the x’y’-plane. Draw the dashed x’ and y’ axes through the translated origin. Plot the ordered pairs of the x’y’ equation on the x’y’-plane, and use the points to draw an appropriate graph.