W A T K I N S - J O H N S O N C O M P A N Y Semiconductor Equipment Group

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Chabot Mathematics
8.3 QuadraticFcn Graphs
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu
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Review

• Any QUESTIONS About
• 8.2 ? Quadratic Eqn Applications
• Any QUESTIONS About HomeWork
• 8.2 ? HW-40

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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 parabolas 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.

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Example ? Graph f(x) 2x2

• SolutionMake T-Table andConnect-Dots

x y (x, y)
0 1 1 2 2 0 2 2 8 8 (0, 0) (1, 2) (1, 2) (2, 8) (2, 8)

• x 0 is Axis of Symm
• (0,0) is Vertex

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Example ? Graph f(x) -3x2

• SolutionMake T-Table andConnect-Dots

x y (x, y)
0 1 1 2 2 0 3 3 12 12 (0, 0) (1, 3) (1, 3) (2, 12) (2, 12)

• Same Axis Vertex but open DOWNward

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Examples of ax2 Parabolas
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5
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3
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Graphing f(x) ax2

• The graph of f(x) ax2 is a parabola with
• x 0 as its axis of symmetry.
• The Origin as its vertex.
• For a gt 0, the parabola opens upward.
• For a lt 0,the parabola opens downward.
• If a is greater than 1, the parabola is
  narrower than y x2.
• If a is between 0 and 1, the parabola is wider
  than y x2.

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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.
• This allows us to observe similarities to the
  graphs drawn in previous slides.

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Example ? Graph f(x) 2(x-2)2

• SolutionMake T-Table andConnect-Dots

x y (x, y)
0 1 1 2 3 4 4 1 9 0 1 4 (0, 4) (1, 1) (1, 9) (2, 0) (3, 1) (4, 4)
vertex

• The Vertex SHIFTED 2-Units to the Right

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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.
• 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.

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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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Example ? Graph

• Make T-Table andConnect-Dots

x y (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

• The Vertex SHIFTED 3-Units Left and1-Unit Down

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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
  gt 0, the parabola opens up, and if a lt 0, the
  parabola opens down.

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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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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 gt 0, the parabola opens up.
• If a lt 0, the parabola opens down.
• Find the vertex. The vertex is (h, k).
• If a gt 0 (or a lt 0), the function f has a
  minimum (or a maximum) value k at x h

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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.
  If the solutions are real numbers, they are the
  xintercepts. If not, the parabola either lies
  above the xaxis (when a gt 0) or below the
  xaxis (when a lt 0).

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Graphing f(x) a(x h)2 k

• Find the y-intercept
• Find the y-intercept by replacing x with 0. Then
  f(0) ah2 k is the y-intercept.
• Sketch the graph
• Plot the points found in Steps 35 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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Example ? Graph

• SOLUTION

Step 1 a 2, h 3, and k 8 Step 2 a 2, a gt
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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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.
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Example ? Graph

• SOLUTION cont.
• Sketch GraphUsing the 4 points
• Vertex
• Two x-Intercepts
• One y-Intercept

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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.

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Example ? Graph

• SOLUTION

f (x) x2 2x 1
(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

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Example ? Graph

• SOLUTION

f (x) 2x2 6x 3
CompleteSquare
2(x2 3x) 3
2(x2 3x 9/4 9/4) 3
2(x2 3x 9/4) 3 18/4
2(x 3/2)2 3/2

• The vertex is at (3/2, 3/2)

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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)

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Graphing f(x) ax2 bx c

• The graph is a parabola. Identify a, b, and c
• Determine how the parabola opens
• If a gt 0, the parabola opens up.
• If a lt 0, the parabola opens down
• Find the vertex (h, k). Use the formula

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Graphing f(x) ax2 bx c

• Find the xinterceptsLet 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 xaxis when a gt 0
• below the xaxis when a lt 0

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Graphing f(x) ax2 bx c

• Find the yintercept. 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.

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Example ? Graph

• SOLUTION

Step 1 a 2, b 8, and c 5 Step 2 a 2, a
lt 0, the parabola opens down. Step 3 Find (h, k).
Maximum value of 3 at x 2
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Example ? Graph

• SOLUTION

Step 4 Let f (x) 0.
Step 5 Let x 0.
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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 yintercept
(0, 5) with respect to the axis x 2 is (4, 5).
Step 7 The parabola passing through the points
found in Steps 36 is sketched on the next slide.
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Example ? Graph

• SOLUTION cont.
• Sketch GraphUsing the pointsJust Determined

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Find Domain Range

• Find the domain and range for f(x)
• SOLUTION Examine the Graph to find that the
• Domain is (-8, 8)
• Range is (-8, 3

• Given the graph of f(x) -2x2 8x 5

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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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All Done for Today
GeometricCompleteTheSquare
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Chabot Mathematics
Appendix
Bruce Mayer, PE Licensed Electrical Mechanical
EngineerBMayer_at_ChabotCollege.edu

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Graph y x

• Make T-table

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