C Challenge
Using the graph of the function f(x) shown, sketch the graph of each transformed function.
-
f
(
x
+
1
)
f open x plus 1 close
-
f
(
x
)
−
2
f open x close minus 2
-
f
(
x
+
2
)
+
1
f open x plus 2 close plus 1
-
−
2
f
(
x
)
negative 2 f open x close
- Graph all of the following functions in the same viewing window. After you enter each new function, view its graph.
-
y
=
x
2
y equals , x squared
-
y
=
x
2
+
2
y equals , x squared , plus 2
-
y
=
(
x
+
2
)
2
y equals . open x plus 2 close squared
-
y
=
(
x
−
1
)
2
−
4
y equals . open x minus 1 close squared . minus 4
-
y
=
−
x
2
−
2
y equals , negative x squared , minus 2
Based on your results, make a sketch of the graph of
f
(
x
)
=
−
(
x
+
2
)
2
+
1
f open x close equals negative . open x plus 2 close squared . plus 1 and check your prediction on your calculator.
Standardized Test Prep
SAT/ACT
What is an equation for each vertical translation of
y
=
2
x
−
1
?
y equals 2 x minus 1 question mark
- 3 units down
-
y
=
2
x
−
7
y equals 2 x minus 7
-
y
=
2
x
+
2
y equals 2 x plus 2
-
y
=
2
x
+
5
y equals 2 x plus 5
-
y
=
2
x
−
4
y equals 2 x minus 4
-
3
5
units up
3 fifths . unitsup
-
y
=
2
x
−
2
5
y equals 2 x minus , 2 fifths
-
y
=
2
x
−
11
5
y equals 2 x minus , 11 over 5
-
y
=
2
x
−
8
5
y equals 2 x minus , 8 fifths
-
y
=
2
x
+
1
5
y equals 2 x plus , 1 fifth
-
What is the slope of the line in the graph below?
-
−
5
2
negative , 5 halves
-
−
2
5
negative , 2 fifths
-
2
5
2 fifths
-
5
2
5 halves
Short Response
- The weight of a gold bar varies directly with its volume. If a
40
cm
3
40 , cm cubed bar weighs 772 grams, how much will a
100
cm
3
100 . cm cubed bar weigh?
Mixed Review
See Lesson 2-5.
-
A musician's manager keeps track of the ticket prices and the attendance at recent performances. Use a graphing calculator to determine the equation of the line of best fit for the given data.
Ticket Prices($) |
41.00 |
41.50 |
42.00 |
43.00 |
43.50 |
44.00 |
44.50 |
45.00 |
45.00 |
47.00 |
Number Sold |
256 |
276 |
250 |
241 |
210 |
235 |
195 |
194 |
205 |
180 |
Get Ready! To prepare for Lesson 2-7, do Exercises 56–58.
See Lesson 1-6.
Solve each absolute value equation.
-
|
x
−
3
|
+
2
=
7
vertical line x minus 3 vertical line plus 2 equals 7
-
|
2
x
+
1
|
−
14
=
9
vertical line 2 x plus 1 vertical line negative 14 equals 9
-
1
3
|
5
x
−
3
|
=
6
1 third , vertical line 5 x minus 3 vertical line equals 6