[email protected] MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot...
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Transcript of [email protected] MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot...
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
§8.3 Quadratic§8.3 QuadraticFcn GraphsFcn Graphs
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt2
Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §8.2 → Quadratic Eqn Applications
Any QUESTIONS About HomeWork• §8.2 → HW-38
8.2 MTH 55
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt3
Bruce Mayer, PE Chabot College Mathematics
Graphs of Quadratic EqnsGraphs 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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt4
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph ff((xx) = 2) = 2xx22
Solution:Make 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
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
78
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt5
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph ff((xx) = ) = −−33xx22
Solution:Make T-Table andConnect-Dots
Same Axis & Vertex but opens DOWNward
x y (x, y)
0
1
–1
2
–2
0
–3
–3
–12
–12
(0, 0)
(1, –3)
(–1, –3)
(2, –12)
(–2, –12)
x
y
-5 -4 -3 -2 -1 1 2 3 4 5
-3
2
-2
3
-1
1
6
54
-4
-5
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt6
Bruce Mayer, PE Chabot College Mathematics
Examples of Examples of axax22 Parabolas Parabolas2( )f x x
4
3
6
2
5
1
2( ) 4f x x
21( )
4f x x
2( )f x x
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt7
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = axax22
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt8
Bruce Mayer, PE Chabot College Mathematics
The Graph of The Graph of ff((xx) = ) = aa((xx – – hh))22
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt9
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph ff((xx) = () = (xx−2−2))22
Solution:Make T-Table andConnect-Dots
The Vertex SHIFTED 2-Units to the Right
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)
x
y
-5 -4 -3 -2 -1 1 2 3 4 5
4
3
6
2
5
1
-1
-2
78
vertex
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt10
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = aa((xx−−hh))22
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt11
Bruce Mayer, PE Chabot College Mathematics
Graph of Graph of ff((xx) = ) = aa((xx – – hh))22 + + kk
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
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt12
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph
The Vertex SHIFTED 3-Units Left and1-Unit Down
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)
21( ) ( 3) 1.
2f x x
x
y
-5 -4 -3 -2 -1 1 2 3 4 5
-3
2
-2
3
-1
1
-4
-5-6
-7
-8vertex
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt13
Bruce Mayer, PE Chabot College Mathematics
Quadratic Fcn in Standard FormQuadratic Fcn in Standard Form
The Quadratic Function Written in STANDARD Form:
f x a x h 2 k , a 0
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt14
Bruce Mayer, PE Chabot College Mathematics
Example Example Find Quadratic Fcn 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
y a x h 2 k
y a x 3 2 4
y a x 3 2 4
7 a –4 3 2 4
a 3
Hence,
y 3 x 3 2 4
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt15
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = aa((xx – – hh))22 + + kk
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
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt16
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = aa((xx – – hh))22 + + kk
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).
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt17
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = aa((xx – – hh))22 + + kk
5. Find the y-intercept• Find the y-intercept by replacing x with 0.
Then y = f(0) = ah2 + 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
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt18
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph f x 2 x 3 2 8.
SOLUTION
Step 1 a = 2, h = 3, and k = –8Step 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.
0 2 x 3 2 8
8 2 x 3 2
4 x 3 2
x 3 2
x 5 or x 1
x-intercepts: 1 and 5
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt19
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph f x 2 x 3 2 8.
SOLUTION cont.Step 5 Replace x with 0.
f 0 2 0 3 2 8
2 9 8 10
y-intercept is 10 .
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt20
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph f x 2 x 3 2 8.
SOLUTION cont.• Sketch Graph
Using the 4 points– Vertex
– Two x-Intercepts
– One y-Intercept
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt21
Bruce Mayer, PE Chabot College Mathematics
Completing the SquareCompleting 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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt22
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph 2( ) 2 1.f x x x
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
x
y
-5 -4 -3 -2 -1 1 2 3 4 5
-3
2
-2
3
-1
1
6
54
-4
-5
Adding ZERO
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt23
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph SOLUTION
The vertex is at (3/2, 3/2)
2( ) 2 6 3.f x x x
f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3
= –2(x2 – 3x + 9/4) – 3 + 18/4
= –2(x2 – 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
1
6
54
-4
-5
CompleteSquare
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt24
Bruce Mayer, PE Chabot College Mathematics
The Vertex of a ParabolaThe 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:
24, or , .
2 2 2 4
b b b ac bf
a a a a
• 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])
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt25
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = axax22 + bx + c + 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
h, k b
2a, f
b
2a
.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt26
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = axax22 + bx + c + bx + c
4. Find the x-interceptsLet 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
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt27
Bruce Mayer, PE Chabot College Mathematics
Graphing Graphing ff((xx) = ) = axax22 + bx + c + 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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt28
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph SOLUTION
f x 2x2 8x 5.
Step 1 a = –2, b = 8, and c = –5Step 2 a = –2, a < 0, the parabola opens down.Step 3 Find (h, k).
h b
2a
8
2 2 2
k f 2 2 2 2 8 2 5 3
h, k 2, 3 Maximum value of y = 3 at x = 2
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt29
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph SOLUTION
f x 2x2 8x 5.
Step 4 Let f (x) = 0.
f 0 2 0 2 8 0 5
y-intercept is 5 .Step 5 Let x = 0.
2x2 8x 5 0
x 8 8 2 4 2 5
2 2 4 6
2
x-intercepts are 4 6
2 and
4 6
2.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt30
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph SOLUTION
f x 2x2 8x 5.
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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt31
Bruce Mayer, PE Chabot College Mathematics
Example Example Graph Graph f x 2x2 8x 5.
SOLUTION cont.• Sketch Graph
Using the pointsJust Determined
f x 2x2 8x 5
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt32
Bruce Mayer, PE Chabot College Mathematics
Find Domain & RangeFind 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]
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt33
Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard 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.
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt34
Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
GeometricComplete
TheSquare
22
2
52510
10
xxx
xx
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt35
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
AppendiAppendixx
–
srsrsr 22
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt36
Bruce Mayer, PE Chabot College Mathematics
Graph Graph yy = | = |xx||
Make T-tablex y = |x |
-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6
x
y
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt37
Bruce Mayer, PE Chabot College Mathematics
-3
-2
-1
0
1
2
3
4
5
-3 -2 -1 0 1 2 3 4 5
M55_§JBerland_Graphs_0806.xls -5
-4
-3
-2
-1
0
1
2
3
4
5
-10 -8 -6 -4 -2 0 2 4 6 8 10
M55_§JBerland_Graphs_0806.xls
x
y