Unit 7 – Graphs and Graphing Utilities - Classwork · 7. Graphing and Graphing Utilities - 1 - -...

7. Graphing and Graphing Utilities - 1 - - www.mastermathmentor.com - Stu Schwartz

Unit 7 – Graphs and Graphing Utilities - Classwork A. Graphing Equations Our first job is being able to graph equations. We have done some graphing in trigonometry but now need a general method of seeing a complete graph of any equation. To enter equations, you go to your

!

y = menu and enter the equation. Be aware of keys and menus on your TI-84 that allow math operations:

!

^ = exponentiation

!

x2 - squaring

!

- square root

!

x"1 - reciprocal

!

Math 3 - cube

!

Math 4 - cube root

!

Math Num 1 - absolute value When we view a graph, the first thing we do is to graph it in the calculator’s standard window. The standard window on the calculator is

!

"10 # x #10 and "10 # y #10. To get that window, you press

!

Zoom 6 : ZStandard . That sets the value of xmin = -10, xmax = 10, ymin = -10, ymax = 10. For instance, to graph the equation:

!

y = x2"10x " 7 in the

standard window: You have the ability of changing the viewing window by going

to the

!

Window menu, adjusting the values of xmin, xmax, ymin, and ymax. For instance, to view this graph in the window

!

0 " x "12 and #12 " y " 3, adjust the window and then press

!

Graph . Xscl and Yscl refer to the tick marks on the axes. In this case, both are 1. We don’t usually take the time to adjust them. We now define a complete graph: A complete graph shows all the behaviors of the graph. Behaviors include all turns of the graph – high and low points as well as intercepts, where the graph crosses the x- and y- axes. Finding a complete graph is sometimes a matter of trial and error. There is no one window that gives a complete graph. For instance, in the graph above, it is clear that the graph goes lower so we have to adjust _____. It is also clear that the graph goes further to the left so we have to adjust _____. So here is a complete graph and the window that created it. Again, let’s be clear: 1) There is no one best complete window and 2) it is a trial and error process: Here is another possible complete window graph. The better complete windows use up most of the viewing screen. The solution above is better than the one to the right.


Finally, note that the graphs above are “fatter” and thinner.” When we adjust the window, we change the size of the graph. The standard window [-10, 10], [-10, 10] has a basic problem because the screen is wider than the height giving distortion. The way of fixing that problem is going to the

!

Zoom 6 and then immediately to the

!

Zoom 5: Square view. This is the graph and the window. Note that the ratio of the screen is roughly 3 to 2. If we really wanted a complete curve drawn correctly to scale, we would have to change both the xmin, xmax and ymin, ymax so that they are in the same proportions (doubling them, tripling them, etc.) Normally, we don’t do these steps unless we want true accuracy. Mostly we are just interested in the shape of the graph and where it crosses the x-axis. 1. Graph the equations given in class according to the prescribed windows. a. Equation:

!

y = "2

Standard

!

"3,3[ ] "4,0[ ] Complete Graph [ , ] [ , ]

b. Equation:

!

y ="2x

3" 7

Standard

!

"3,3[ ] "9,2[ ] Complete Graph [ , ] [ , ] c. Equation:

!

y = "3x2" x+ 5

Standard

!

0,2[ ] 0,6[ ] Complete Graph [ , ] [ , ]


d. Equation:

!

y = x x " 4( )2

Standard

!

0,2[ ] 0,10[ ] Complete Graph [ , ] [ , ] e. Equation:

!

y = x4

+ 3x3" 9x

2" x " 2

Standard

!

"2,1[ ] "20,2[ ] Complete Graph [ , ] [ , ] f. Equation:

!

y = 2x

Standard

!

"1,2[ ] 0,3[ ] Complete Graph [ , ] [ , ]

g. Equation:

!

y = "2cos1

2x " 5

Standard

!

0,2"[ ] #7,1[ ] Complete Graph [ , ] [ , ]


h. Equation:

!

y = x x+ 3 "1

Standard

!

0,4[ ] 0,10[ ] Complete Graph [ , ] [ , ] i. Equation:

!

y = ± 25 " x2

Window to fill the screen completely Window to make it to scale

Standard [ , ] [ , ] [ , ] [ , ] B. Checking symmetry for a graph Symmetry means “mirror image.” There are three types of symmetry: x-axis symmetry, y-axis symmetry, and origin symmetry. A curve that is symmetric to the x-axis is the same on either side of the x-axis. Anything in quadrant 1 is reflected into quadrant IV and vice versa. Anything in quadrant II is reflected to quadrant III and vice versa. A curve that is symmetric to the y-axis is the same on either side of the y-axis. Anything in quadrant 1 is reflected into quadrant II and vice versa. Anything in quadrant III is reflected to quadrant IV and vice versa. A curve that is symmetric to the origin is the same on either side of the origin. Anything in quadrant 1 is reflected into quadrant III and vice versa. Anything in quadrant II is reflected to quadrant IV and vice versa.

x-axis symmetry y-axis symmetry origin symmetry


Symmetry check for x-axis symmetry: replacing every occurrence of y with –y yields an equivalent equation. Symmetry check for y-axis symmetry: replacing every occurrence of x with -x yields an equivalent equation. Symmetry check for origin symmetry: replacing every occurrence of x with –x AND y with -y yields an equivalent equation. If an equation is symmetric both to the x-axis and y-axis, it is automatically symmetric to the origin. But if an equation is symmetric to the origin, it is not necessarily symmetric to either the x or y axis. 2. Check the symmetry of the following and verify using the calculator. a.

!

y = x2" 5 b.

!

y = 4x " 2 c.

!

x3

+ y = 0 d.

!

4x " y2

= 0 e.

!

y = x " 2 f.

!

y = x " 5 g.

!

y = x " 5 h.

!

y = x3

+ 2 C. Lines in the Plane Students in Precalculus have seen lines in algebra 1 and algebra 2 so we will just review the material.

Important formulas:

The slope m of the non-vertical line passing through the points

!

x1y

1( ) and x2y

2( ) is m =y

2" y

1

x2" x

1

=Rise

Run

Slope intercept form of a line:

!

y = mx + b where m is the slope and b is the y-intercept Point slope form of a line:

!

y " y1

= m x " x1( ) where m is the slope and

!

x1y

1( ) is a point on the line Horizontal lines are in the form: y = a. Vertical lines are in the form x = a. The general form of a line is

!

Ax + By + C = 0 where A, B, and C are integers.

Intercept form: If the line passes through

!

a,0( ) and

!

0, b( ) , the equation is:

!

x

a+

y

b=1

Two lines are parallel if and only if their slopes are equal:

!

m1

= m2.

Two lines are perpendicular if and only if their slopes are negative reciprocals of each other:

!

m1

="1

m2


1. Find the slope of the line passing through the pair of points:

a.

!

"4,1( ) and 2,"7( ) b.

!

0,"7( ) and "2,"7( ) c.

!

4,"3( ) and 4,"1( ) d.

!

2

3, "3

4( ) and 4, "1

2( )

2. Find the slope and y-intercept of the given line.

a.

!

2x " y " 3 = 0 b.

!

5x + 3y " 2 = 0 c.

!

2

3x "

1

2y "1= 0

3. Find the equation of the line with the given slope passing through the given point. Verify on the calculator.

a.

!

m = 2, 4,"1( ) b.

!

m = "5, "2,6( ) c.

!

m =1

2, 4,"

1

2

#

$ %

&

' ( d.

!

m ="2

3, "

1

4,2

#

$ %

&

' (

4. Find the equation of the line passing through the given points. Verify on the calculator a.

!

3,4( ), 4,"1( ) b.

!

"2,3( ), 1,9( ) c.

!

3,4( ), 7, 1

2( ) d.

!

2

3, "3

4( ), 1

"2, 5

6( ) 5. Find the equation in general form given the intercepts: a. x-intercept

!

4,0( ) and y-intercept

!

0,"2( ) b. x-intercept

!

"2


!

0, 3

4( ) 6. Find the equation of the line through the given point a) parallel to the given line and b) perpendicular to the given line. Verify by calculator. a.

!

Point : 2,3( ), line : y = 3x "1 b.

!

Point : "2,5( ), line : 2x " 4y =1 c.

!

Point : 1

2, "3

4( ), line : x = 2 y " 2 Parallel: Parallel: Parallel: Perpendicular: Perpendicular: Perpendicular


D. Functions Functions are the mainstay of all mathematics and it is imperative you understand its definition.

relation a set of ordered pairs

!

x, y( ) domain allowable values of x in the relation range allow values of y in the relation function A relation in which for every x in the domain, there is one

and only one y in the range (no repeating x value) vertical line test If any vertical line intersects the relation in more than one

point, the relation is not a function. function notation

!

y = f x( ) even function when

!

f x( ) = f "x( ) (symmetric to y-axis) odd function when

!

" f x( ) = f "x( ) (symmetric to origin) mapping a pictorial way of illustrating an equation

!

a

b

c

1

2

3

4

a

b

c

1

2

3

4

function! not a function

Ex.

Which of the following does not represent a function?

a.

!

1

2

3

w

x

y

z b.

!

1

2

3

w

x

y

z c.

!

1

2

3

w

x

y

z d.

!

1

2

3

w

x

y

z

Which of the following does not represent a function? a.

!

students in a school{ }" locker{ } b.

!

telephone number{ }" students in a school{ } c.

!

cars in a parking lot{ }" license plate number{ } d.

!

U.S citizen who works{ }" social security #{ } Function notation: If

!

y = f x( ) = 2x " 3, and we want to find the value of y when x = 2, we simply ask for

!

f 2( ). It asks for the value of the function f when x = 2.


Using the calculator to find values of functions at specific values of x a) Graphing a curve: ex:

!

f x( ) = x2

" 4 (use

!

Y = ) b) find

!

f "1.5( ) - press

!

2nd CALC

a) b) b) alternate method of finding

!

f "1.5( ) .- press

!

VARS

c) using a table to find f(many values) - press

!

2nd TBLSET , then

!

2nd TABLE

1. If

!

f x( ) = x2" 4x + 5, find

a)

!

f 4( ) b)

!

f "5( ) c)

!

f 2a( ) d)

!

f x + 2( )

2.

!

g x( ) =2x

2" x " 3

x +1, find

a)

!

g "2( ) b)

!

g "( ) c)

!

g "x( ) d)

!

g x "1( ) E. Domain and Range

domain allowable values of x in the function. Assume the domain of a function is all real numbers with the following exceptions:

- any x value which creates a zero in the denominator - any x value which creates a square root (or even root of a negative number. Any graphs with holes, vertical asymptotes, or gaps restrict the domain. range allowable values of y in the function. Assume the range of a function is all real numbers

with the following exceptions: We usually are more interested in the domain than the range.

- squares always create positive numbers. - square (or even) roots are always positive. - absolute values always create positive numbers


When analyzing domain and range, we can examine the functions from a graphical point of view or an algebraic point of view.

1. Domain ( , ) Range ( , )





6. Domain ( , ) Range ( , ) We can also analyze domain from an algebraic point of view. The important thing to remember is that we assume the domain is all real numbers unless there is something to restrict it. Those restrictions usually fall into the categories of fractions with a variable in the denominator or the even roots of expressions. Find the domain of the following:

7.

!

y = x2" 4x + 3 8.

!

f x( ) =2x

x "1 9.

!

f x( ) =5

x2" 25


10.

!

g x( ) = 5x "10 11.

!

h x( ) = x2 + 25 12.

!

f x( ) =x + 5

x

13.

!

f x( ) = x2"10x "1

3 14.

!

f x( ) =x + 3

x2 + 2x " 3 15.

!

f x( ) =2x + 7

x2" 49

F. Piecewise Functions and Restricting the Domain

Before we examine how to restrict a domain on the calculator, we have to understand the concept of Boolean operators. First, on your calculator, find the TEST feature. To get to this screen, press, 2nd MATH . You should get the first screen. This screen opens the door to restricting domain. CLEAR and type 5 > 4. To do this, type 5 2nd MATH 3: > 4 ENTER . Your screen should look like the screen on the right. The way the calculator handles this statement is: true statements are valued as 1 while false statements are valued as 0. Since 5 > 4, you end up with 1. Now try 2 < 1. 2 2nd MATH 5 1 ENTER . Since 2 < 1 is a false statement, the calculator gives you a zero. What should this give you? 10 - 8 = 1/.5. Try it. Okay, now we are ready for a restricted domain function. Let’s examine y = 2x, x > 0. To put this into the calculator, go to your Y = screen and 2X/(X 2nd MATH 3:> 0). It looks like this: Go to TBLSET and set your table like this: Now go to TABLE. Let’s interpret this. The calculator starts with x = -3 so it evaluates Y1=2(-3)/(-3>0). Since -3>0 is false and therefore equal to zero, you are evaluating the statement -6/0 which is an error. Note that you will continue to get errors until x = 1. When x = 1, you are evaluating Y1=2(1)/(1>0). Since 1>0 is a true statement and therefore equal to one, you are finding Y1=2/1 = 2. Notice for x being positive numbers, you are always dividing by 1. Now graph the function using ZOOM 6. Values that are errors are not graphed. So you end up with y = 2x, x > 0.

So, now we are ready to try a piecewise function. Let’s try the first problem:

!

y =x + 2,x "1

x # 2,x >1

$ % &

We need to put the function in as two separate functions, Y1 and Y2. Put in the function like this and graph:


Note that you needed to enclose your expressions in parentheses. Note that this curve is not “continuous” because there is a clear break at the value of x = 1. Graph these:

1.

!

y =3x " 2, x < 2

4, x # 2

$ % &

2.

!

y =x

2 " 2, x # 0

2x " 2, x > 0

$ % &

Domain [ , ] Range [ , ] Domain [ , ] Range [ , ] Continuous YES NO Continuous YES NO

To put in a function like this:

!

y =

2x + 7, x < "2

x2 "1,"2 # x < 2

5x "1, x $ 2

%

& '

( '

, you need three functions. Y1 and

Y3 are straightforward. To put in Y2 with its compound statement, you may not do this: While this does not generate an error, it is not what you want. Do this:

You can find the operator “and” in the 2nd MATH LOGIC menu. This is the graph of that piece. So the entire function looks like this and the graph to the right. The problem of continuity needs to be done using calculus limit techniques. Do not trust your eyes using the calculator. When graphs are nearly vertical, the calculator has trouble graphing it. ZOOMING in may or may not confirm the continuity of this function at x = 2.


Finally, when TRACING, realize that there are three separate functions here and if you are tracing at an x-value in which the curve is not defined, no y-value will show. Try these.

3.

!

y =

x, x " 0

0,#2 $ x < 0

x + 2, x < #2

%

& '

( '

4.

!

f x( ) =

5 " x,x # 5

25 " x 2 ,"5 $ x < 5

x + 6,x < "5

%

& '

( '

Domain [ , ] Range [ , ] Domain [ , ] Range [ , ] Continuous YES NO Continuous YES NO 5. In problem # 4 above, find the following: a)

!

f 7( ) b.

!

f 5( ) c.

!

f "5( ) d.

!

f "8( ) 6. A new 10 week diet claims that it will take weight off a person according to the following formula. In the first 2 weeks, it will take off 7 pounds. In the next 4 weeks, it will take off 1.5 pounds a week. In the next 4 weeks, it will take off a pound a week. Complete the Chart. Week 1 2 4 6 8 10 Pounds off Write a piecewise function that describes the amount of weight taken off as a function of weeks and graph it over the length of the diet.

How long would it take to lose 14 pounds?


7. An injury lawyer charges according to the following: His partner charges this way: • First hour or part cost $50 • $500 for the first 10 hours • The next 9 hours cost $150/hour • $200 an hour for anything over that • After that, the charge is $125 an hour Complete the Chart

Hours 1 3 5 10 12 18 Lawyer 1 Lawyer 2

Write a piecewise function that describes both lawyers and graph them over 25 hours.

When is lawyer 1 cheaper to hire? When is lawyer 2 cheaper to hire?

G. Absolute Value Curves You have all learned the meaning of the absolute value. The absolute value of an expression makes the expression positive – it strips of the negative sign. However, the true definition of an absolute value is merely a piecewise function:

!

x =x, x " 0

#x, x < 0

$ % &

or expression =expression,expression " 0

#expression,expression < 0

$ % &

Absolute value curves of linear functions form the distinctive “V-shaped curve.”


So, when we graph

!

y = x , we are merely graphing

!

x, x " 0

#x, x < 0

$ % &

If we graph

!

y = x " 3 we are looking at

!

y =x " 3, x " 3 # 0

" x " 3( ), x " 3 < 0

$ % &

' & or y =

x " 3, x # 3

3" x, x < 3

$ % '

If we graph

!

y = 2x +1 , we are looking at

!

y =2x +1,2x +1" 0

# 2x +1( ),2x +1 < 0

$ % &

' & or y =

2x +1, x " #1

2

#2x #1, x < #1

2

$ % &

' &

For the following expressions, write the absolute value expressions as piecewise expressions, graph and verify on the calculator.

1.

!

y = x + 2 2.

!

y = 2x "10 3.

!

y = 5x + 3 4.

!

f x( ) = 3"3

2x

H. Translations of Graphs

Given

!

y = f x( ),a > 0

!

f x( ) + a vertical shift up - translates the graph a units upwards

!

f x( ) " a vertical shift down - translates the graph a units downwards

!

f x " a( ) horizontal shift right - translates the graph a units to the right

!

f x + a( ) horizontal shift left - translates the graph a units to the left

!

a " f x( ) Vertical stretch - if a > 1, the graph is narrower, if a < 1, it is wider

!

" f x( ) Reflection - flips the graph across the x-axis

!

f x( ) Anything below the x-axis is reflected across the x-axis.

!

y = x2

!

y = x2

+ 2

!

y = x2

"2

!

y = x " 2( )2

!

y = x + 2( )2

!

y = 2x2

!

y = .5x2

!

y = "x2


For the following problems, draw a transformation of the given graph.

1.

!

f x( )

!

f x( ) +1

!

f x( ) " 3

!

f x " 2( )

!

f x +1( )

!

2 f x( )

!

1

2f x( )

!

" f x( )

!

f x "1( ) +1

!

f x + 2( ) " 2

!

f x( )

!

f x( ) " 3

!

" f x( )

!

f x "1( ) +1

!

f x + 2( ) " 2

2.

!

f x( )

!

f x( ) "1

!

f x( ) + 2.5

!

f x " 0.5( )

!

f x +1.5( )

!

3 f x( )

!

1

2f x( )

!

" f x( )

!

f x +1( ) "1

!

" f x + 2( ) + 3

!

f x( )

!

f x( ) "1

!

" f x( )

!

f x "1( ) +1

!

f x + 2( ) " 2


3.

!

f x( )

!

f x( ) + 4

!

f x( ) " 2

!

f x " 4( )

!

f x + 3( )

!

2 f x( )

!

1

2f x( )

!

" f x( )

!

f x +1( ) "1

!

" f x + 2( ) + 3

!

f x( )

!

f x( ) + 4

!

" f x( )

!

f x +1( ) "1

!

" f x + 2( ) + 3

I. Composition of Functions The concept of composition of functions is to start with a number or variable, apply function one to that variable, and take the answer and apply that answer to function two. It can be shown pictorially. Composition of functions

!

f o g= f g x( )[ ]

x

!

g x( )

!

f g x( )[ ]

!

f o g

g f

A simple example of composition of functions is when you go to the cafeteria and select a meal: Function g is ordered pairs in the form of (student, meal). Function f is in the form of (meal, cost). So

!

f g x( )[ ] will be in the form of (student, cost). Every student who purchases a meal has a cost associated with that meal. The cashier tells the student the cost … the meal itself defined the cost, but in reality is no longer important. That is what you are doing when you find a function composition. 1.

!

f x( ) = x2 + 2x "1, g x( ) = 3x "1, find the following: a)

!

f g 3( )[ ] b)

!

g f 3( )[ ] c)

!

f f 3( )[ ] d)

!

g g 3( )[ ]


finding composite functions:

!

f x( ) = x2

+ 2x "1 and g x( ) = 3x "1, find f g 2( )( ) and g f 2( )( )

enter graphs into Y1 and Y2

!

VARS Y -VARS" 1:Function 1: Y1 (

VARS Y -VARS" 1:Function 2 : Y2 ( 2

2.

!

f x( ) = x2 + 4x " 2, g x( ) = x +1 , find the following:

a)

!

f o g( ) 3( ) b)

!

g o f( ) 3( ) c)

!

f g x( )[ ] d)

!

g f x( )[ ]

3.

!

f x( ) = 5x + 2, g x( ) =x +1

2x + 3, find the following:

a)

!

g f "2( )[ ] b)

!

f g "2( )[ ] c)

!

f g x( )[ ] d)

!

g f x( )[ ]

4. If

!

f x( ) = x2 and g x( ) =

1

x, which of the following is largest at x = 2?

a)

!

f " g b)

!

g

f c)

!

f o g d)

!

g o f J. Inverse Functions The concept of an inverse is probably the most misunderstood concept in mathematics. Try this. To your knowledge, write the inverse of the following expressions: 5 __________ Boy_________ Dog _________ Hot dog _________ . . . Write what you believe to be the definition of an inverse: ________________________________


inverse the function

!

f"1 such that

!

f"1f x( )[ ] = x

• the inverse function “undoes” function f • it is the mirror image of

!

f x( ) across the line y = x. • to find

!

f"1 , switch x and y and solve for y

• do not confuse inverse with reciprocal. “Nouns” ( expressions) have reciprocals while “verbs” (functions) have inverses.

the reciprocal to 5 is

!

1

5 while the inverse to

!

y = 5x is

!

y =1

5x .

determining if the use vertical line test on the inverse or (easier), use the horizontal line test inverse is a function on the function itself

1. You are given

!

f x( ) as a set of ordered pairs. Find

!

f "1 x( ) and determine if

!

f "1 x( ) is a function: a)

!

f x( ) = 2,3( ), 3,6( ), "1,"1( ), 0,4( )[ ] b)

!

f x( ) = 3,4( ), 4,2( ), 5,4( ), 0,0( )[ ] c)

!

f x( ) = 3,4( ), 3,5( ), 3,6( )[ ] 2. Find which of the following mappings is a function and whose inverse is also a function.

a.

!

1

2

3

x

y

z b.

!

1

2

3

x

y

z c.

!

1

2

3

x

y

z d.

!

1

2

3

x

y

z

3. For the following functions

!

f x( ), find the inverse

!

f "1 x( ) and determine if

!

f "1 x( ) is also a function.

a.

!

y = 2x "1 b.

!

y =12 " 6x c.

!

y =x "1

x + 2

d.

!

y = x2"1 e.

!

y = x3

+ 2 f.

!

y = x3" x

2" 2x "1


Unit 7 – Graphs and Graphing Utilities - Homework A1. Graph the equations given in class according to the prescribed windows. a. Equation:

!

y = 7

Standard

!

"1,1[ ] 0,10[ ] Complete Graph [ , ] [ , ]

b. Equation:

!

y =3

4x+25

2

Standard

!

0,3[ ] 0,17[ ] Complete Graph [ , ] [ , ]

c. Equation:

!

y =15 " 5x "1

3x2

Standard

!

"5,5[ ] 0,40[ ] Complete Graph [ , ] [ , ] d. Equation:

!

y = 5 x+ 43 +12

Standard

!

0,10[ ] 0,30[ ] Complete Graph [ , ] [ , ]


e. Equation:

!

y = x4" 2x

3" 2x

2+ 3x "16

Standard

!

0,5[ ] 0,40[ ] Complete Graph [ , ] [ , ] f.

!

y = 6sin x cos2x

Standard

!

0,"[ ] #10,10[ ] Complete Graph [ , ] [ , ]

g. Equation:

!

y =x 2x "1( )

2

5x+ 3( )10

Standard

!

0,1[ ] ".1,.5[ ] Complete Graph [ , ] [ , ] i. Equation:

!

y = ± 7 " x " x2 Window to make it fill screen Window to make it to scale

Standard [ , ] [ , ] [ , ] [ , ]


B1. Check the symmetry of the following and verify using the calculator. a.

!

y = 3x + 7 b.

!

x2

+ y = 6 c.

!

y = x3" x d.

!

3" 4x " y2

= 0 e.

!

4y = 3x " 2 f.

!

y = 5x "1 g.

!

x = y " 2 h.

!

y = x4" 2x

2" 6

C1. Find the slope of the line passing through the pair of points:

a.

!

4,"1( ) and 6,7( ) b.

!

"1,"1( ) and "2,"1( ) c.

!

"5,"3( ) and "5, 1

2( ) d.

!

1

3, 2

5( ) and "1, 1

2( )

C2. Find the slope and y-intercept of the given line.

a.

!

3x + y " 4 = 0 b.

!

7x " 2y + 5 = 0 c.

!

3

2x "

1

4y + 2 = 0

C3. Find the equation of the line with the given slope passing through the given point. Verify on the calculator.

a.

!

m = "3, 1,"2( ) b.

!

m = 0, "1,"5( ) c.

!

m ="2

3, 6,"2( ) d.

!

m =3

4, "

1

2,"3

8

#

$ %

&

' (

C4. Find the equation of the line passing through the given points. Verify on the calculator a.

!

5,2( ), "1,"4( ) b.

!

3,"1( ), "1,7( ) c.

!

"1,2( ), 3, 1

2( ) d.

!

1

2, "2

3( ), 1

6, "3

2( )


C5. Find the equation in general form given the intercepts: a. x-intercept

!


!

0,"6( ) b. x-intercept

!

1


!

0, 2

3( ) C6. Find the equation of the line through the given point a) parallel to the given line and b) perpendicular to the given line. Verify by calculator. a.

!

Point : 1,"3( ), line : y = 2x +1 b.

!

Point : "4,2( ), line : 3x + 2y = 3 c.

!

Point : 1, "2

3( ), line : x = 3y +1 Parallel: Parallel: Parallel: Perpendicular: Perpendicular: Perpendicular: D1. If

!

f x( ) = x3" x2 + x "1, find the following. Verify using the calculator when possible.

a)

!

f "2( ) b)

!

f 3( ) " f 1( ) c)

!

f 2x( ) d)

!

f x "1( ) D2. If

!

g x( ) = x +10 " x "10( ), find the following. Verify using the calculator when possible. a)

!

g 6( ) b)

!

g "9( ) c)

!

g 15" x( ) d)

!

g x "1( ) E1. For the following problems, find the domain and range by inspection:

a. Domain ( , ) Range ( , )


b. Domain ( , ) Range ( , )

c. Domain ( , ) Range ( , )

d. Domain ( , ) Range ( , )

e. Domain ( , ) Range ( , )

f. Domain ( , ) Range ( , )

g. Domain ( , ) Range ( , )

h. Domain ( , ) Range ( , )

i. Domain ( , ) Range ( , )


j. Domain ( , ) Range ( , ) E2. Find the domain of the following:

a.

!

y = x2

+12x "19 b.

!

f x( ) =x + 3

x + 4 c.

!

f x( ) =2x "1

x2"100

f.

!

g x( ) = 20 " 4x g.

!

h x( ) = 4x2 + 50 h.

!

f x( ) =2x +16

x "1

i.

!

f x( ) = 1" x5 j.

!

f x( ) =2x " 8

x3" 4x

k.

!

f x( ) =x2" x "1

x2" 36

F1. Graph the following:

a.

!

f x( ) =2x "1, x # 0

1" 2x, x < 0

$ % &

b.

!

f x( ) =

x2 "1, x # 2

5"1

2x, x < 2

$

% &

' &

Domain [ , ] Range [ , ] Domain [ , ] Range [ , ] Continuous: YES NO Continuous YES NO

c.

!

g x( ) =x

2 " 4x "1, x # "1

5+ x " x2, x < "1

$ % &

d.

!

h x( ) =2sin x, x " 0

4 # 4cos x, x > 0

$ % &

Domain [ , ] Range [ , ] Domain [ , ] Range [ , ] Continuous: YES NO Continuous YES NO


e. f.

Domain [ , ] Range [ , ] Domain [ , ] Range [ , ] Continuous: YES NO Continuous YES NO g. In problem f above, find the following:

1)

!

h 5( ) 2)

!

h 3( ) 3.

!

h "2( ) 4.

!

h 32( ) 5.

!

h 3( )[ ]2

F2. You have a choice of 2 phone plans. Plan 1: Plan 2: • Pay $25 a month Pay 20 cents a minute for up to 120 minutes • Pay 10 cents a minute up to 30 minutes Pay 15 cents a minute for all minutes over 120 • Pay 8 cents a minute for all minutes over 30 Write piecewise functions for plan 1

!

f x( ) and plan 2

!

g x( ) Complete the chart:

Minutes 0 10 20 30 60 100 120 180 300 Plan 1 Plan 2

What is your recommendation for the best (cheapest) plan and why?

G. Write the following absolute value expressions as piecewise expressions and graph to verify. 1.

!

y = x " 3.5 2.

!

y = 2x + 9 3.

!

y = 6x " 4 4.

!

f x( ) = "8x " 5

!

f x( ) =

4 " 2x,x # 3

2,"1$ x < 3

x,x < "1

%

& '

( '

!

h x( ) =

x2 " 4x +1,x # 4

5" x,0 $ x < 4

2x + 5,x < 0

%

& '

( '


H. For the following problems, draw the transformations on the given graphs.

1.

!

f x( )

!

f x( ) "1

!

f x( ) + 2.5

!

f x " 0.5( )

!

f x +1.5( )

!

2 f x( )

!

1

2f x( )

!

" f x( )

!

f x +1( ) "1

!

" f x + 2( ) + 3

!

f x( )

!

f x( ) "1

!

" f x( )

!

f x "1( ) +1

!

f x + 2( ) " 2

2.

!

f x( )

!

f x( ) +1

!

f x( ) " 2

!

f x " 4( )

!

f x + 2( )

!

2 f x( )

!

1

2f x( )

!

" f x( )

!

f x +1( ) "1

!

" f x + 2( ) + 3

!

f x( )

!

f x( ) +1

!

" f x( )

!

f x " 4( )

!

" f x + 2( ) + 3


I1. Given

!

f x( ) =1" 2x, g x( ) = 2x2" x "1, find the following:

a.

!

f g "1( )[ ] b.

!

g f "1( )[ ] c.

!

f f 2( )[ ] d.

!

g g 2( )[ ] I2. Given

!

f x( ) = cos x, g x( ) =1" x , find the following:

a.

!

f g 1( )[ ] b.

!

g f "( )[ ] c.

!

f g x( )[ ] d.

!

g f x( )[ ] I3. Given

!

f x( ) = sin2x, g x( ) = x , find the following:

a.

!

f g 4"2( )[ ] b.

!

g f"

6

#

$ % &

' (

)

* +

,

- . c.

!

f g x( )[ ] d.

!

g f x( )[ ]

I4. If

!

f x( ) is shown by the graph to the right . with each unit equal to 1, which is the smallest? a.

!

f f 0( )( ) b.

!

f f 1( )( ) c.

!

f f 2( )( ) d.

!

f f 4( )( ) I5. For which pairs of functions does

!

f g x( )( ) " x ?

a.

!

f x( ) = 2x, g x( ) =1

2x b.

!

f x( ) = x2, g x( ) = x ,x " 0

c.

!

f x( ) = 4x +1, g x( ) =1

4x "1 d. none of them

J1. For the following functions, find the inverse and solve for y if possible. Determine if the inverse is also a function. a.

!

y = 6x b.

!

y = 8x + 4 c.

!

y = x2" 5

d.

!

y = x3

+1 e.

!

y = x2

+ 4x + 4 f.

!

y = x , x > 0


g.

!

y =2x "1

x + 3 h.

!

y = x " 43 i.

!

y =x

2

x2"1

J2. Which of the following functions has its inverse also being a function?

a. b. c. d.

J3. Which of the following graphs is the inverse to the graph at the right?

a. b. c. d. J4. If

!

f x( ) = 2x +1 and g x( ) = x " 3 , which is the smallest value at x = 3?

a.

!

f g x( )[ ] b.

!

g f x( )[ ] c.

!

f o g( )"1 d.

!

g o f( )"1

Unit 7 – Graphs and Graphing Utilities - Classwork · 7. Graphing and Graphing Utilities - 1 - -...

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Transcript of Unit 7 – Graphs and Graphing Utilities - Classwork · 7. Graphing and Graphing Utilities - 1 - -...