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

2. 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

3. 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

4. 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         

5. 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.

6. Example  Graph f(x) = 2x2 Solution: Make T-Table and Connect-Dots x y
(-1,2 )
(2,8)
(-2,8) 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

7. Example  Graph f(x) = −3x2y -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

8. Examples of ax2 Parabolasy 6 5 4 3 2 1 x

9. 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.

10. 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.

11. Example  Graph f(x) = (x−2)2y 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

12. 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.

13. 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

14. Example  Graph Make T-Table and Connect-Dots x y (x, y) –1 –2 –3 –4 -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

15. 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.

16. 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

17. 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).

18. 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

19. 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.

20. 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.

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

22. 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:

23. 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)

24. 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)

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

26. 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

27. 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

28. 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.

29. 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

30. 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?

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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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).

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

39. 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

40. 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

41. 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

42. 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}

43. 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.

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

45. 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:

46. 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

47. Notice IDENTICAL Shapes
Compare:
Notice IDENTICAL Shapes

48. 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.

49. Example  Graph SOLUTION f (x) = x2 – 2x – 1 = (x2 – 2x) – 1
Adding ZERO
f (x) = x2 – 2x – 1 x y -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

50. Example  Graph SOLUTION f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3y -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) – /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

51. 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])

52. 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

53. 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

54. 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.

55. 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

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

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

58. 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]

59. 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.

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

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

62. 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

63. Graph y = |x|
Make T-table

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

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

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

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

68. 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

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

70. Over 2
Up 3

71. Over 2
Up 3

72. 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

73. 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

74. 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.