Chabot Mathematics
§8.3 Quadratic
Fcn Graphs
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Review § 8.2
MTH 55
 Any QUESTIONS About
• §8.2 → Quadratic Eqn Applications
 Any QUESTIONS About HomeWork
• §8.2 → HW-38
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Making Complete Plots


1.

2.


3.






Chabot College Mathematics
4
4.
5.
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
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = 2x2
 Solution:
Make T-Table and
Connect-Dots
x
y
(-2,8)
0
(0, 0)
1
2
(1, 2)
(2,8)
8
7
6
5
4
(x, y)
0
(-1,2 )
3
2
(1,2)
1
-5 -4 -3 -2 -1
1
2 3 4 5
-1
x
–1
2
(–1, 2)
2
8
(2, 8)
 x = 0 is Axis of Symm
–2
8
(–2, 8)
 (0,0) is Vertex
Chabot College Mathematics
6
y
-2
(0, 0)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = −3x2
y
 Solution:
Make T-Table and
Connect-Dots
x
y
6
5
4
3
2
1
(x, y)
-5 -4 -3 -2 -1
0
0
(0, 0)
1
–3
(1, –3)
–1
–3
(–1, –3)
2
–12 (2, –12)
–2
–12 (–2, –12)
Chabot College Mathematics
7
x
1
-1
-2
2 3 4 5
-3
-4
-5
 Same Axis & Vertex
but opens
DOWNward
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Examples of ax2 Parabolas
f ( x)  4 x
2
f ( x)  x 2
y
6
5
4
1 2
f ( x)  x
4
3
2
1
x
f ( x)   x
Chabot College Mathematics
8
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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 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.
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f(x) = (x−2)2
y
 Solution:
Make T-Table and
Connect-Dots
x
y
(x, y)
0
4
8
7
6
5
4
3
2
(0, 4)
1
1
1
–1
9
(–1, 9)
2
0
(2, 0)
3
1
(3, 1)
4
4
(4, 4)
Chabot College Mathematics
11
(1, 1)
-5 -4 -3 -2 -1
x
1
2 3 4 5
-1
vertex
-2
 The Vertex SHIFTED
2-Units to the Right
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
1
f ( x)   ( x  3)2  1.
2
y
 Make T-Table and
Connect-Dots
x
0
–1
–2
–3
–4
–5
y
1
-5 -4 -3 -2 -1
-1
-2
-11/2 (0, -11/2)
–3
–1
(–3, –1)
–3/2 (–4, –3/2)
–3
(–5, –3)
x
1
2 3 4 5
-3
-4
(–1, –3)
–3/2 (–2, –3/2)
Chabot College Mathematics
14
(x, y)
3
2
-5
-6
-7
vertex
-8
 The Vertex SHIFTED
3-Units Left and
1-Unit Down
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Quadratic Fcn in Standard Form
 The Quadratic Function Written in
STANDARD GRAPHING Form:
f x   a x  h   k , a  0
2
• 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.
Chabot College Mathematics
15
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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).
Chabot College Mathematics
17
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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) = 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
Chabot College Mathematics
18
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 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.
2
0  2 x  3  8
x  3  2
8  2 x  3
2
4  x  3
2
Chabot College Mathematics
19
x  5 or x  1
x-intercepts: 1 and 5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 SOLUTION cont.
Step 5 Replace x with 0.
f 0   2 0  3  8
2
 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.
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph f x   2 x  3  8.
2
 SOLUTION
cont.
• Sketch Graph
Using the 4 points
– Vertex
– Two x-Intercepts
– One y-Intercept
f x   2 x  3  8.
2
Chabot College Mathematics
21
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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:
x
y  x2
2
y   2  4
1
y   1  1
2
0
y  02  0
1
y  12  1
2
y  22  4
Chabot College Mathematics
22
2
Ordered Pair  x, y 
 2, 4 
 1,1
0, 0 
1,1
2, 4 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Translation of Graphs
 Now Plot
the Five
Points
and
connect
them
with a
smooth
Curve
6
5
(−2,4)
23
(2,4)
4
3
2
(−1,1)
(1,1)
1
x
0
-4
-3
-2
-1
0
(0,0)1
2
3
-1
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
y
-2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
4
Axes Translation
7
 Now Move
UP the graph
of y = x2 by
two units as
shown
 What is the
Equation
for the new
Curve?
(−2,6)
24
(2,6)
6
5
(−2,4)
(2,4)
4
(−1,3) (1,3)
3
2
(0,2)
(−1,1)
(1,1)
1
x
0
-4
-3
-2
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
y
-1
0
(0,0)1
2
-1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
3
4
Axes Translation
 Compare ordered pairs on the graph of
with the corresponding ordered pairs on
the new curve:
7
(−2,6)
y
(2,6)
6
5
(−2,4)
(2,4)
4
(−1,3) (1,3)
3
2
(0,2)
(−1,1)
(1,1)
1
x
0
-4
-3
-2
M55_§JBerland_Graphs_0806.xls
-1
0
2
3
4
-1
Chabot College Mathematics
25
(0,0)1
y  x2
 2, 4
 1,1
0, 0
1,1
2, 4
New Curve
(2 units up)
 2, 6
 1, 3
0, 2
1, 3
2, 6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
y x
 2, 4
 1,1
0, 0
1,1
2, 4
2
Axes Translation
New Curve
(2 units up)
 2, 6
 1, 3
0, 2
1, 3
2, 6
 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
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Axes Translation
5
 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
yx
4
27
2
3
2
1
x
0
-4
-3
-2
-1
0
-1
1
2
3
y x 2
2
-2
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
y
-3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
4
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.
7
5
y
6
y
4
y  x 2
2
5
3
4
2
y  x2
1
3
x
0
2
-4
1
yx
0
-4
-3
-2
-1
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
28
0
-1
1
2
-3
-2
-1
0
1
2
3
4
-1
2
x
3
4
y  x2  2
-2
M55_§JBerland_Graphs_0806.xls
-3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
29
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Horizontal Translation
 What if every point on the graph of
y = x2 were moved 5 units to the right
as shown below.
y
10
9
8
7
6
5
yx
(−2,4)4
2
(2,4)
(3,4)
(7,4)
y?
3
2
1
x
0
-3
-2
-1
-1
0
1
2
3
4
5
6
7
8
9
 What is the eqn of the new curve?
-2
Xlate_ABS_Graphs_1010.xls
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Horizontal Translation
 Compare ordered pairs on the graph of
with the corresponding ordered pairs on
the new curve:
y  x2
 2, 4
 1,1
0, 0
1,1
2, 4
Chabot College Mathematics
31
New Curve
(5 units right)
3, 4
4,1
5, 0
6,1
7, 4 
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
y x
 2, 4
 1,1
0, 0
1,1
2, 4
2
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?
New Curve
(5 units right)
3, 4
4,1
5, 0
6,1
7, 4 
• 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
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
y  x  3  2 
2
Chabot College Mathematics
35
y    x 
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
y
Ordered Pair  x, y 
x
for
 1,1
1
1
y’ =(x’)2
0, 0
0
0
1
Chabot College Mathematics
36
1
1,1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
2
2






y

x

3

2

y

x
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).
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Plot by Translation
 Locate2 the Origin of the Translated
Axes Set using the translator values
y
y   x  3  2 
2
1
 y   2  x  32
2
 y    x 
0
-1
y
0
1
2
3
4
x
5
-1
-2
Chabot College Mathematics
38
-3
Move: 3-Right,
3,2
2-Down
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
2
• Remember
that this
graph is
smooth
1
-1
0
-2
39
y  x2
 y   2  x   32
0
-1
Chabot College Mathematics
y
y
-3
1
Move:
3-Right
2-Down
2
3
x
4
y  x  3  2
2
3,2
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
5
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
2
 The Ordered
Pair (4, −1)
should
make the xy
Eqn True
1
0
-1
0
1
2
3
4,1
4
x
-1
-2
x
Bruce Mayer, PE
Chabot College Mathematics
40
y
y
-3
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
5
Example  Plot by Translation
 Sub (4, −1) into y = (x−3)2−2
y  x  3  2
2
:
 1  4  3  2
 1  1
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
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
-2
7
y
6
y  x  3  2
5
2
4
3
2
1
0
-1
0
1
2
3
4
5
-1
-2
• Domain of f is {x|x is any real number}
-3
M55_ 8_3_Xlate-Parabs_Graphs.xls
• Range of f is {y|y ≥ −2}
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
6
7
x
8
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.
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
 Step-1

f x   
1
x  32  7
2
1
2


y


x

3
7
Step-2
2
1
2
f x    x  3  7
2
y
1
x  32  7
2
1
 y  7    x  32
2
1 2

y   x
2
 Step-3 → T-table in x’y’
x'
-5
-4
-3
-2
-1
0
Chabot College Mathematics
44
y'
-12.5
-8
-4.5
-2
-0.5
0
x'
1
2
3
4
5
6
y'
-0.5
-2
-4.5
-8
-12.5
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
1
2
f x    x  3  7
2
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:
Chabot College Mathematics
45
y
9
y
8
7
x
6
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
-2
-3
-4
M55_ 8_3_Xlate-Parabs_Graphs.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
3
4
x5
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
Chabot College Mathematics
46
1
2
f x    x  3  7
2
y
9
y
8
7
x
6
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
-2
-3
-4
M55_ 8_3_Xlate-Parabs_Graphs.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
3
4
x5
Compare:
1 2
f x    x
2
9
y
&
y
1
2
f x    x  3  7
2
8
7
x
6
1
2
f x    x  3  7
2
Notice
IDENTICAL
Shapes
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
-2
0
1
2
3
4
x5
1
f x    x 2
2
-3
-4
M55_ 8_3_Xlate-Parabs_Graphs.xls
Chabot College
Mathematics
47
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
2
Example  Graph f ( x)  x  2 x  1.
 SOLUTION
f (x) = x2 – 2x – 1
Adding ZERO
y
6
= (x2 – 2x) – 1
5
4
3
= (x2 – 2x + 1 – 1) – 1
2
= (x2 – 2x + 1) – 1 – 1
= (x –
1)2
1
-5 -4 -3 -2 -1
–2
x
1
-1
-2
 The vertex is at (1, −2)
-3
-4
 The Parabola
Opens UP
-5
Chabot College Mathematics
49
2 3 4 5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
2
Example  Graph f ( x)  2 x  6 x  3.
 SOLUTION
y
6
f (x) = –2x2 + 6x – 3
=
–2(x2
Complete
Square
– 3x) – 3
5
4
3
2
1
= –2(x2 – 3x + 9/4 – 9/4) – 3
-5 -4 -3 -2 -1
= –2(x2 – 3x + 9/4) – 3 + 18/4
1
-1
-2
x
2 3 4 5
-3
-4
-5
= –2(x – 3/2)2 + 3/2
 The vertex is at (3/2, 3/2)
 The Parabola Opens DOWN
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
The Vertex of a Parabola
 By the Process of Completing-theSquare we arrive at a FORMULA for
the vertex of a parabola given by
f(x) = ax2 + bx + c:
2

b
b
4
ac

b
 b



 
  2a , f  2a   or   2a , 4a .
 



• 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])
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + 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
 b
 b 
h, k     , f    .
 2a
2a 
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + bx + c
4. 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
Chabot College Mathematics
53
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing f(x) = ax2 + 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.
Chabot College Mathematics
54
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 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).
b
8
h

2
2a
2 2 
k  f 2   2 2   8 2   5  3
2
h, k   2, 3
Maximum value of y = 3 at x = 2
Chabot College Mathematics
55
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION
2
2x
 8x  5  0
Step 4 Let f (x) = 0.
x
8 
8 
2
 4 2 5 
2 2 
4 6

2
4 6
4 6
x-intercepts are
and
.
2
2
Step 5
Let x = 0. f 0   2 0   8 0   5
y-intercept is  5 .
Chabot College Mathematics
56
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example  Graph
f x   2x 2  8x  5.
 SOLUTION cont.
• Sketch Graph
Using the points
Just Determined
f x   2x  8x  5
2
Chabot College Mathematics
57
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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]
Chabot College Mathematics
58
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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.
Chabot College Mathematics
59
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
All Done for Today
Geometric
Complete
The
Square
x  10 x 
2
x  10 x  25  x  5
2
Chabot College Mathematics
60
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
61
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
2
2
7  a –4  3  4
y  a x  h   k
a3
2
y  a x  3  4
Hence,
2
2
y  a x  3  4
y  3 x  3  4
Chabot College Mathematics
62
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
63
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
4
5
6
Translate Up or Down
 Make f (x)  x , g(x)  x  3 and h(x)  x  5
Graphs
h(x)  x  5
for
 
Notice:
g x   f x   3
h x   f  x   5
• Of the form of
VERTICAL
Translations
Chabot College Mathematics
64


f (x)  x
g(x)  x  3

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Translate Left or Right
 Make Graphs for
f (x)  x , g(x)  x  3 and h(x)  x  5
 Notice: g x   f x  3
hx  f x  5
• Of the form of HORIZONTAL
Translations
h(x)  x  5
g(x)  x  3


Bruce Mayer, PE
Chabot College Mathematics
65
f (x)  x

BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example ReCall Graph y = |x|
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
y = |x |
6
6
5
4
3
2
1
0
1
2
3
4
5
6
5
4
3
2
x
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
Chabot College Mathematics
66
y
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
file =XY_Plot_0211.xls
-6
4
5
6
Example Graph y = |x+2|+3
 Step-1 f x  x  2  3
 Step-2 y  x
Xlators :
y  x2 3
x   2; y  3
 Step-3 → T-table in x’y’
x
1
0
1
Chabot College Mathematics
67
y
1
0
1
Ordered Pair  x, y 
 1,1
0, 0
1,1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Example Graph y = |x+2|+3
 Step-4: the x’y’-plane origin is 2 units
LEFT and 3 units UP from xy-plane
y
y
6
5
4
x
3
Left 2
2
Up 3
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
Bruce Mayer, PE
Chabot College Mathematics
68
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
-2
6
Example Graph y = |x+2|+3
 Step-5: Remember that the graph of
y = |x| is V-Shaped:
y
y
y
6
5
4
x
3
Left 2
2
Up 3
1
x
0
-6
-5
-4
Chabot College Mathematics
69
-3
-2
-1
0
-1
1
2
3
4
5
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
y
y
6
5
4
x
3
Over 22
Up 3
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
Chabot College Mathematics
70
-4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
6
y
y
y
6
5
4
x
3
Over 2
2
Up 3
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
Chabot College Mathematics
71
-3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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
Chabot College Mathematics
72
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
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;
y = 3(x+2)
 the graph of y = f(x + c)
is the graph of
y = f(x) shifted LEFT
(Horizontally) c units.
2
Chabot College Mathematics
73
y = 3(x-2)2
y = 3x2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
Graphing by Translation
1.
2.
3.
4.
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.
5. Plot the ordered pairs of the x’y’ equation
on the x’y’-plane, and use the points to
draw an appropriate graph.
Chabot College Mathematics
74
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
10
9
y
y
8
7
6
5
4
3
2
1
0
-3
-2
-1
-1
0
1
2
3
4
5
6
7
8
-2
Xlate_ABS_Graphs_1010.xls
Chabot College Mathematics
75
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
9
x
x
10
9
y
8
7
6
5
4
3
2
1
x
0
-3
-2
-1
-1
0
1
2
3
4
5
6
7
8
-2
Xlate_ABS_Graphs_1010.xls
Chabot College Mathematics
76
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
9
2
y
y
1
0
0
-1
1
3
2
4
x
-1
-2
3,2
x
-3
Chabot College Mathematics
77
-4
M55_ 8_3_Xlate-Parabs_Graphs.xls
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
5
2
y
y
1
y  x2
 y   2  x   32
0
-1
0
1
2
3
x
4
-1
y  x  3  2
2
-2
3,2
x
-3
Chabot College Mathematics
78
-4
M55_ 8_3_Xlate-Parabs_Graphs.xls
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
5
2
y
y
1
0
-1
0
1
2
3
4,1
4
x
-1
-2
x
-3
Chabot College Mathematics
79
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
5
y
9
y
8
7
x
6
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
x5
-2
-3
Chabot College Mathematics
-4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
80
M55_ 8_3_Xlate-Parabs_Graphs.xls
-5
y
9
y
8
7
x
6
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
x5
-2
-3
-4
Bruce Mayer, PE
Chabot College Mathematics
81
M55_ 8_3_Xlate-Parabs_Graphs.xls
-5
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
y
9
y
8
7
x
6
1
2
f x    x  3  7
2
5
4
3
2
1
0
-9
-8
-7
-6
-5
-4
-3
-2
-1
-1
-2
0
1
2
3
4
x5
1
f x    x 2
2
-3
-4
Bruce Mayer, PE
Chabot College Mathematics
82
M55_ 8_3_Xlate-Parabs_Graphs.xls
-5
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
7
y
6
5
4
3
2
1
0
-2
-1
0
1
2
3
4
5
6
7
x
-1
-2
Chabot College Mathematics
83
-3
M55_ 8_3_Xlate-Parabs_Graphs.xls
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
8
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
84
-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt
8
10