Graphing and Calculator Skills

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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
B. GRAPHING AND MANIPULATION OF FUNCTIONS
1.
Parent Graphs
Ex 1:
Basic Types of Functions
(1.)
A polynomial function is any function of the form
f (x)  ax n  bx n1  cx n2  dx n3 L  C .
The domain of a polynomial function is always the set of real numbers. The power n
defines the degree of the polynomial and all powers must be positive integers. The degree
of the polynomial affects the range.
(2.)
A root function is any function of the form
y  n x  x1/n .
The value n defines the root, the nth root of x in the case above. The domain and range
are dependent on the root of the function.
Ex 2:
(3.)
A rational function is any function that can be written as the ratio of two other functions.
The vertical asymptote(s) and the endbehavior asymptote (horizontal, oblique, etc) of the
function are important in determining the domain and range.
(4.)
A transcendental function is a family of functions whose behavior and characteristics are
unique to that family. Thus the domain and range are dependent on the knowledge of the
function. Transcendental functions include the family of functions identified as
trigonometric functions, exponential functions and logarithmic functions.
The parent graphs are the basic graphs of particular functions. These parent graphs are useful in
determining behaviors and characteristics of other members of the same type of function. By
knowing what the basic graph looks like and how it behaves allows one to have insight to more
complicated forms of the function. Below are twelve parent graphs. Memorize these graphs.
(1.)
f (x)  x
(2.)
f (x)  x 2
1
-1
(4.)
-1
f (x)  x
(3.)
f (x)  x 3
1
1
-1
(5.)
-1
f (x)  x
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1
1
-1
(6.)
-1
1
f (x)  3 x
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
1
-1
(7.)
1
1
-1
f (x) 
1
x
-1
(8.)
(10.)
-1
1
-1
(11.)
Ex 3:
(9.)
y  ln(x)
1
-1
-1
(12.)
-2š
-š
-1
1
-1
f (x)  sin(x)
1
1
1
-1
1
f (x)  cos(x)
1
-1
-1
1
y  ex
-1
1
-1
x 2  y2  1
1
-1
1
1
š
2š
-2š
-š
-1
š
2š
To translate a graph you move the parent graph horizontally and/or vertically a prescribed
number of units.
(1.)
Translations take the form of y  c  f (x  a) , where a is the horizontal displacement and
c is the vertical displacement.
(2.)
Sketch the graph of y  (x  1)2  3 .
This is similar to the parent function f (x)  x 2 . If you evaluate the parent function at x+1
and subtract 3 from the result you would obtain y.
y  f (x  1)  3  (x  1)2  3
Draw the parent graph. Locate three easy points on the parent graph. In this case (1,1),
(0,0) and (–1,1). Translate these points by moving each point 1 unit horizontally in the
negative direction, and then follow this with a vertical movement of 3 units down. In
English: subtract 1 from each x coordinate and subtract 3 from each y coordinate. Draw
the new curve through the translated points.
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
-3
-2
-1
1
-2
Ex 4
To dilate a graph you multiply each y coordinate of the graph by a fixed constant.
(1.)
Dilations take the form of y  kf (x) .
(2.)
Sketch the graph of y  3x 2 .
Draw the parent graph, which is f (x)  x 2 in this case. Pick three reference points and
multiply each of the y coordinates by 3. Sketch the new graph.
3
2
1
-1
Ex 5:
Ex 6:
1
-1
Consider any parent function.f(x). Discuss the graph defined by:
(1.)
y  k  af (x  c)
(2.)
y  f (x)
(3.)
y fx

Conic Relations
(1.)
The general equation for a circle is defined by
x  x   y  y 
2
1
2
1
 r2
where x1 , y1  defines the center of the circle with a radius of r units.
(2.)
The general equation for an ellipse is defined by
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
x  x   y  y 
2
2
1,
1
a2
b2
1
where x1 , y1  is the relative center with x   a 2 and y   b 2 . Locate the
relative center and move x and y units. Draw the ellipse through the four points.
(3.)
The general equation of a hyperbola is defined by
x  x   y  y 
2
2
1,
a2
1
b2
1
where x1 , y1  is the relative center with x   a 2 or y   b 2 depending on the
location of the negative. The graph will be bound by the asymptotes of y  
2.
b
a
.
Calculators and Graphing
The following section deals with instructions for TI83 calculators. If you are not using a TI83, determine
where these operations, commands and keys are located on your calculator.
Ex 7:
To center a graph on a point. Place a function in Y1 and graph. Use the TRACE key to move to
a desired location on the graph. Press ENTER and the graph will center on this point.
Ex 8:
Relative Maximum and Minimum
(1.)
This refers to a value of x on the curve where the function takes on its greatest or least y
value in some small interval about x. In a parabola this would occur at the vertex.
(2.)
To find the relative maximum and/or minimum on the graph of a function, go to the
CALCULATE menu and choose either minimum or maximum. Follow the prompts.
The value of x for the relative maximum (or minimum) is temporarily stored in the X
memory. Should you need the value for a further calculation going to the home screen and
typing X can retrieve it. This will recall the stored value from the last command used. The
same applies for the Y. This will also work for other commands.
(3.)
Ex 9:
Increasing and/or decreasing behavior refers to how the range behaves relative to values
of the domain. As you move from the left of the domain to the right, the function is
increasing if the y values get larger and decreasing if the y values get smaller. The change
will occur at the relative maximum and/or minimum of the function.
A tangent (line) to a curve at a point x = c is defined as the single line that can be drawn at the
given point. It follows that if you can draw this line you can figure out the slope of the line that is
drawn tangent to the point. The concept is illustrated by the following example.
(1.)
(2.)
Draw the function f (x)  x 2  3x  5 in a standard window.
Draw the line tangent to the function f at the point x = 2.3.
OPTION 1



Push the keys 2nd PRGM
Choose Number 5: Tangent ( from the menu)
On the home screen you should see the following statement
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS



tangent(
Complete the statement so that it looks as follows tangent Y1, 2.3 
Press ENTER
On the screen the parabola should be drawn followed by a line at x = 2.3
OPTION 2







Push the keys 2nd PRGM
Choose ClrDraw. This erases the previous tangent line.
Display the graph.
Choose Tangent
Push the keys 2nd PRGM
Type 2.3
A line should now be drawn to the curve at 2.3 and the equation of the line
y  1.6x  10.29 should be displayed.
OPTION 3





Clear previous tangent line.
Push the keys 2nd TRACE
Choose dy/dx from the menu.
Type 2.3
You should see dy/dx = 1.6 which represents the slope of the tangent line.
Ex 10: A piece wise defined function is a function that is pieced together using two or more functions.
Each of the individual functions will have a restricted domain. The function below is an example
of a piece wise defined function. Note the restrictions on the individual parts.
x0
 x ;
 2
f (x)   x ; 0  x  1
1 ;
x 1

(1.)
Sketch the graph of f.
To sketch means to do without a calculator. Graph each of the three functions separately.
Use only the piece for each function that is relevant to the problem. Combine the pieces to
obtain the complete graph.
(2.)
Draw the graph of f.
You will enter each of the functions separately along with the proper restrictions. Your
entries should look as follows:
Y1  x(x  0)1
Y2  x 2 (x  0 and x  1)1
Y3  1(x  1)1
The inequality symbols and the “and” can be found in the TEST menu. Access this menu
pressing 2nd MATH . The “and” is in the logic menu. The 1 after the restrictions
prevents the graph from trying to connect with the x-axis.
Ex 11: Go to the MODE menu. At the bottom you will see three commands:
Full Horiz G-T
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
Make sure that you have a graph in Y1. Notice that the home screen changes depending on the
mode. This is useful when you want to perform multiple tasks and see the results.
3.
Combinations of Functions
Ex 12: A combination of two or more functions involves performing the operations of +, –, ÷ or  on
the range elements of the given functions for any domain elements common to all functions.
(1.)
If two functions f and g are both defined for x = c, then you could obtain a new range
value based on x = c by combining f(c) and g(c). Thus the new range element could be
defined by +, –, ÷ or  f(c) and g(c).
 c , f (c)  g(c) ,  c , f (c) – g(c) ,  c , f (c)g(c) ,  c , f (c) / g(c) 
(2.)
It is more useful to create a general rule that will work for all values of x that the two
functions have in common. This is nothing more than an algebra exercise.
(3.)
The domain for any combination will always be defined by the intersection of the
domains of the individual functions plus any natural restrictions that apply to the newly
generated rule. The key to getting the proper domain is to determine the individual
domains first and then taking the intersection of these two domains. This new domain will
apply to all combinations of the two functions less any natural restrictions. Even if the
new rule works for a given input, if that input is not in the common domain it cannot be
used.
Ex 13: Let f  {(0, 3),(3, 2),(6, 4),(1, 4)} and g  {(3, 1),(1, 5),(0, 0),(4, 6)} . Determine the
combinations of f+g, f–g, fg, f/g, and g/f.
In order to combine any functions you must first determine the common domain. In this case the
domain elements that f and g have in common are x = –1, 0, 3. Therefore, new points
(combinations) can only be created by these common values.
x
y  f (x) y  g(x)
f g
f g
fg
f /g
g/ f
1
4
5
1
9
0
3
0
3
3
4
5
20 

5
4
0
Undefined 0
3
2
1
1
3
2
2

1
2
Therefore the set defined by f+g = { (–1,1) , (0,3) , (3,1) }. The remaining sets can be defined in
a similar manner by reading the values for the combination off the chart.
Ex 14: Let f (x) 
(1.)
1
and g(x) 
x 1
x
Determine the domain for any possible combination of f and g.
The domain of any possible combination will be any values of x f and g has in common.
D f     , 1   1 ,   and Dg  [ 0 ,  )
DCOMB  D f I Dg
DCOMB  [ 0 , 1 ) ( 1 ,  )
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
The final domain for any combination will be [ 0 , 1 ) ( 1 ,  ) and any natural restrictions
that apply to the derived rule that fall in the common interval.
(2.)
Determine the rule for the combination ( f  g)(x) and the corresponding domain.
( f  g)(x)  f (x)  g(x)
1
 x
x 1
1 x x  x

x 1

The domain is restricted to [ 0 , 1 ) ( 1 ,  ) and any natural restrictions. The only natural
restriction to the new rule are x ≠ 1 and x ≥ 0 which have already been accounted for in
part (1.)
 D f g  [ 0 , 1 ) ( 1 ,  )
(3.)
Determine the rule for the combination ( fg)(x) and the corresponding domain.
( fg)(x)  f (x)g(x)
1
 x
x 1
x

; x  1 and x  0
x 1

 D fg  [ 0 , 1 ) ( 1 ,  )
(4.)
Determine the rule for the combination ( f / g)(x) and the corresponding domain.
f (x)
g(x)
1
 x  1 ; x  1 , x  0 and x  0
x
1

x x  1
( f / g)(x) 
 D f /g  ( 0 , 1 ) ( 1 ,  )
(5.)
Determine the rule for the combination (g / f )(x) and the corresponding domain.
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
(g / f )(x) 
g(x)
f (x)
x
; x  1 and x  0
1
x 1
 x x  1

 Dg/ f  [ 0 , 1 ) ( 1 ,  )
4.
Composition of Functions
Ex 15: The composition of two or more functions involves taking the output of one function and using it
as the input of a second function. The result will be a new ordered pair (thus a graph) whose
range is composed of the final output.
(1.)
Consider two functions f and g and a value of x = c. Calculating f(c) will produce an
output, call it k. Now use that output as the input of g. It in turn will produce a new
output, call it b.
x  c  f (c)  k  g(k)  d
(2.)
The notation for compositions is as follows:
(a.)
 f og (x)  f g(x)  which is read as f composed of g, where the function
f g(x)  is restricted to the domain of g. In other words f is dependent on g, thus
making g the independent variable.
(b.)
g of (x)  g  f (x)  which is read as g composed of f, where the function
g  f (x)  is restricted to the domain of f. In other words g is dependent on f, thus
making f the independent variable.
(3.)
Ex 16:
The domain of any composition is restricted to the domain of the independent function.
f  {(0, 3),(3, 2),(6, 4),(1, 4)} and g  {(3, 1),(1, 5),(0, 0),(4, 6)} . Find the sets defined
by f og and gof .
(1.)
f og is restricted to the domain of g. Thus the values of x = –1, 0, 3, and 4 apply to this
composition. That does not mean that all these values will work.
x
g(x)
f og  f g(x) 
–1
0
3
4
5
0
–1
–6
f(5) = undefined
f(0) = 3
f(–1) = –4
f(–6) = 4
f og  { (0,3) , (3,–4) , (4,4) }
(2.)
gof is restricted to the domain of f. The domain of f is x = –6, –1, 0, and 3.
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
x
f(x)
g o f  g  f (x) 
–6
–1
0
3
4
–4
3
2
g(4) = –6
g(–4) = undefined
g(3) = –1
g(2) = undefined
g o f  { (–6,–6) , (3,2) }
Ex 17: Let f (x)  x 2 & g(x)  x 1
Find the numerical value of ( f og)(4) & (g o f )(8)
( f og)(4)  f  g(4) 
(g o f )(8)  g  f (8) 
 
 f 5 
 f

Ex 18:
 f 8 2 
5
 f 64 
 64  1  65
2
5 5
f (x)  x & g(x)  1 x
(1.)
Determine the domain of f and g.
D f  [ 0 ,  ) & Dg  (   ,  1 ]
(2.)
Determine the rule and domain for ( f og)(x) .
( f og)(x)  f  g(x) 
 f
 1 x 

1 x
 4 1 x
The domain is restricted to the independent function g plus any natural restrictions on the
final rule.
D f og  Dg  Natural Restrictions
 (   , 1 ]
(3.)
Determine the rule and domain for (g o f )(x) .
(g o f )(x)  g  f (x) 
g
 x
 1 x
The domain of the gof is restricted to the domain of f. The new rule is restricted to all
values of x such that 1 x  0 .
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CALCULUS AB UNIT 2: FUNCTIONS – GRAPHING AND MANIPULATION OF FUNCTIONS
Dgof  D f  Natural Restrictions
[ 0 ,1]
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