WCCUSD Algebra II
1
Benchmark 1 Study Guide
Linear Functions, Tables and Graphs.
1´
Graphing a Function’s Rule:
Method #1:
Given a function’s rule, you can make a table of values
and select values for x to substitute into your equation.
Example: f (x) = 2x + 3
f (−2) = 2(−2) + 3
f (−2) = −1
x
y
-2
-1
-1
1
0
3
You try:
A. Make a table of values for the function
2
f (x) = − x + 4 . (Hint: Select x-values that
3
are multiples of 3 to facilitate calculations.)
etc…
1
5
2
7
Plot these points to graph the function.
Method #2:
Identify the rate of change and the y-intercept. Then
plot your y-intercept and use a slope triangle to
determine other points on your line.
B. Determine the graphed function’s rule:
For f (x) = 2x + 3 , the rate of change (slope) is
2, y-intercept is (0, 3) .
5
x − 4 by
2
identifying the y-intercept and rate of change.
C. Graph the function f (x) =
A-CED.2
Page 1 of 13
A-CED.2
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
2
Benchmark 1 Study Guide
Inverse Functions
Given a relation or a function its inverse can
be determined by reversing the order of its xand y-coordinates such that (x, y) → (y, x) .
Example 1:
Find the inverse for the relation
2´
You try:
A. Find the inverse for the relation
(
+
" 1%
)(−4, 7), $10, ', 3, −2 ,
# 2&
*
-
(
)
{(2, 3), (−4,1), (9, − 7 )}
Solution: Reverse the order of the x- and ycoordinates:
B. Find the inverse function for
5
h(x) = − x +1 .
4
Example 2:
2
x −5.
3
Solution: Consider g(x) as the output value y:
2
y = x−5
3
Reverse the order of x and y. Then solve for y:
2
x = y−5
3
2
x+5= y
3
3
3 2
( x + 5) = ⋅ y
2
2 3
3
( x + 5) = y
2
3
15
x+ = y
2
2
Find the inverse function for g(x) =
∴ g−1 (x) =
C. Find the inverse function for the following
graphed function, f(x).
3
15
x+
2
2
or
3x +15
g−1 (x) =
2
F-BF.4
F-BF.4
Page 2 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
3
Benchmark 1 Study Guide
Writing Linear Equations
Using Two Data Points
3´
Here are three ways to express a linear function:
1) Slope-Intercept Form: y = mx + b
y − y1 = m(x − x1 )
f (x) = m(x − x1 ) + y1
3) Standard Form: Ax + By = C
2) Point-Slope Form:
or
You try: After 5 minutes of climbing, Juan
reached an elevation of 300 feet. After 9
minutes, he was at an altitude of 500 feet.
Assuming a constant elevation gain, express
algebraically the height he will reach, h(m),
after m minutes. Show your equation in three
different forms.
Example: During the third consecutive week of saving
money, Ciera had $100. After 5 weeks, she had $160.
Assuming she saved consistently every week, find the
equation that would represent her savings, s(w), after w
weeks.
For Slope Intercept Form: Set up your two data
points: (3,100) and (5,160) .
---Find the rate of change (slope):
Δy
Δx
160 −100
m=
5−3
60
m=
2
m = $30 per week
m=
---Find the initial condition by substituting either point
and the rate of change into the slope-intercept equation:
y = mx + b
100 = 30(3) + b
100 = 90 + b
10 = b
---Rewrite information:
y = mx + b
y = 30x +10
For Standard Form: Rewrite the equation above in the
form ax + by = c .
y = 30x +10
−30x + y = 10
For Point-Slope Form: Find the slope (rate of change),
choose either point and substitute into the equation
y − y1 = m(x − x1 ) .
or
From here you could also rewrite to slope-intercept
form and standard form.
A-CED.2
A-CED.2
Page 3 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
4
Benchmark 1 Study Guide
Transformations of Absolute Value Functions
Example: Sketch
f (x) = 4 x + 3 − 5 .
4´
You try: Sketch the following.
A. f (x) = x − 5
B. f (x) = − x − 5
C. f (x) = x − 3
D. f (x) = x + 2 + 4
In order to graph this function, you need to know what
the graphed parent function f (x) = x looks like:
The vertex form of an absolute value function is
f (x) = a x − h + k , where a is the compression
factor, h is the amount of horizontal shift and k is the
amount of vertical shift. The vertex is at (h, k) .
Sidebar: If
a > 1 , the function’s graph narrows. If
a < 1 , the function’s graph widens. If a < 0 , the
function opens downward.
Therefore, for the function
f (x) = 4 x + 3 − 5 :
--The compression factor is 4, which narrows the
function’s graph in relation to the parent function.
--For x − h to result in x + 3 , the horizontal
transformation value of h must be −3 , meaning the
graph will shift three units left (negative x direction).
--The vertical transformation value k is −5 , meaning
the graph shifts five units down (negative x direction).
--So the vertex of the function must be at (−3, −5) .
Graph this point first.
E. f (x) =
1
x+3 −6
2
F. f (x) = −3 x −1 + 2
The solid line is f (x) = 4 x + 3 − 5 and the dashed
line is the parent function
F-IF.7 and F-BF.3
f (x) = x .
-IF.7 and F-BF.3
F-IF.7 and F-BF.3
F-IF.7 and F-BF.3
Page 4 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
5
Benchmark 1 Study Guide
Average Rate of Change
5´
The average rate of change for a function
from interval x to a, [x, a] , can be found using
the slope formula written in functional
notation:
Δy f (x) − f (a)
A=
=
Δx
x−a
where f (x) and f (a) are output values (the yvalues for Δy in the slope formula) and x and
a are the input values ( Δx in the slope
formula).
You try: Find the average rate of change on
the interval −1 ≤ x ≤ 4 for the function
f (x) = x 2 + 2x − 3 .
Example: Find the average rate of change on
the interval 1 ≤ x ≤ 4 for the function
f (x) = x 2 + 5 .
--Input values are 1 and 4. So to find the
output values, find f (1) and f (4) :
f (1) = (1)2 + 5
f (1) = 1+ 5
f (1) = 6
f (4) = (4)2 + 5
f (4) = 16 + 5
f (4) = 21
--Substitute your values:
f (x) − f (a)
A=
x−a
f (1) − f (4)
A=
1− 4
6 − 21
A=
1− 4
−15
A=
−3
A=5
F-IF.6
F-IF.6
Page 5 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
6
Benchmark 1 Study Guide
Finding Intercepts
6´
To find the x-intercept, let y = 0 (or f (x) = 0 ).
To find the y-intercept, let x = 0 .
You try: Find the intercepts for
f (x) = x 2 − 7x +10 .
Example: Find the intercepts for
f (x) = x 2 + x −12 .
Let x = 0 :
f (0) = (0)2 + (0) −12
f (0) = −12
The y-intercept
is
.
Let y or f (x) = 0 :
0 = x 2 + x −12
0 = (x + 4)(x − 3)
Factor the trinomial.
0 = x + 4 and 0 = x − 3 Zero Product Property
∴ x = −4 and x = 3
The x-intercepts are
and
.
F-IF.8
7
F-IF.8
Identifying Key Features
(domain, range, intercepts, minimum/maximum,
increasing/decreasing rate)
7´
You try:
Given the graphed function above, identify…
--The domain of this function is all real
numbers since the x-values will decrease and
increase infinitely.
--Since there are no y-values less than −5 , the
range can be expressed as −5 ≤ y < ∞ or [ − 5,∞) .
--The vertex is at (1, −5) .
--The minimum is at −5 .
-- Domain:
-- Range:
--Vertex:
-- Minimum/Maximum:
-- y-intercept:
--The y-intercept is at (0, −4) and the
x-intercepts are at (−3, 0) and (5, 0) .
-- x-intercepts:
--The function is decreasing for the interval
(−∞,1) and increasing for the interval (1,∞) .
-- Decreasing interval:
-- Increasing interval:
F-IF.4
F-IF.4
Page 6 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
Benchmark 1 Study Guide
Complex Numbers
8
8´
Complex numbers are written in the form
a ± bi where i = −1 , and therefore i 2 = −1 .
You try:
A. Simplify (−4 − i) − (−3+ 8i)
Examples:
A. Simplify (3− 2i) − (4 + 6i)
(3− 2i) − (4 + 6i)
= 3− 2i − 4 − 6i
Distribute the negative.
Commute.
= 3− 4 − 2i − 6i
= −1− 8i
Combine like terms.
B. Simplify (−4 + 5i)(4 − 3i)
B. Simplify (3− 2i)(4 + 6i)
(3− 2i)(4 + 6i)
= 3(4) + 3(6i) − 2i(4) − 2i(6i)
Distribute.
= 12 +18i − 8i −12i 2
Multiply.
= 12 +18i − 8i −12(−1)
= 12 +18i − 8i +12
= 24 +10i
C. Simplify
Combine like
terms.
4
3+ i
4
3+ i
4 3− i
=
•
3+ i 3− i
4(3− i)
=
(3+ i)(3− i)
12 − 4i
=
9 + 3i − 3i − i 2
12 − 4i
=
9 − (−1)
12 − 4i
=
10
2(6 − 2i)
=
2⋅5
6 − 2i
=
5
C. Simplify
5
2 − 3i
--Multiply by conjugate.
--Distribute.
--Simplify.
--Factor, if possible.
--Simplify.
N-CN.2
N-CN.2
Page 7 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
Benchmark 1 Study Guide
Solving Quadratics
9
9´
When a quadratic doesn’t factor, there are two other
methods you can use to find the solutions: The
Quadratic Formula and Completing the Square.
x 2 + 3x = −4 .
x 2 + 3x + 4 = −4 + 4
You try:
A. Solve x 2 + 5x = −1 using the Quadratic
Formula.
Example: Solve
Set equal to zero
x 2 + 3x + 4 = 0
Determine values for a, b and c if it is in the form
ax 2 + bx + c = 0 .
∴a = 1 , b = 3 , c = 4
Substitute and solve using the Quadratic Formula:
x=
−b ± b 2 − 4ac
2a
x=
−(3) ± (3)2 − 4(1)(4)
2(1)
−3± 9 −16
2
−3± −7
x=
2
−3± −1 ⋅ 7
x=
2
−3± i 7
x=
2
x=
B. Solve x 2 + 5x = −1 by Completing the Square.
Using Completing the Square:
x 2 + 3x = −4
2
" 3%
" 3%
x 2 + 3x + $ ' = −4 + $ '
#2&
#2&
2
To “complete the square,” find
. Add to both sides.
2
"
3%
16 9
$x + ' = − +
#
2&
4 4
Create common denominator.
2
"
3%
7
$x + ' = −
#
2&
4
Combine terms on right side.
2
"
3%
7
$x + ' = −
#
2&
4
x+
3 ± −7
=
2
2
3 i 7
x=− ±
2
2
−3± i 7
x=
2
Take the square root of each
side.
Simplify. Remember:
Isolate the variable x.
Combine terms.
A-REI.4 and N-CN.7
A-REI.4 and N-CN.7
Page 8 of 13
End of Study Guide
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
Benchmark 1 Study Guide
You Try Solutions:
1´
You try:
A. Make a table of values for the function
2
f (x) = − x + 4 . (Hint: Select x-values that
3
are multiples of 3 to facilitate calculations.)
x
y
-6
8
-3
6
0
4
3
2
6
0
B. Determine the graphed function’s rule:
f (x) = 3x − 6
5
x − 4 by
2
identifying the y-intercept and rate of change.
C. Graph the function f (x) =
2´
You try:
A. Find the inverse for the relation
(
+
" 1%
)(−4, 7), $10, ', 3, −2 ,
# 2&
*
-
(
)
(
+
"1 %
)( 7, −4), $ ,10 ', −2, 3 ,
#2 &
*
-
(
)
B. Find the inverse function for
5
h(x) = − x +1 .
4
5
y = − x +1
4
5
x = − y +1
4
5
x −1 = − y
4
4
4" 5 %
− (x −1) = − $ − y '
5
5# 4 &
4
4
− x + = h −1 (x)
5
5
OR
−4x + 4
= h −1 (x)
5
C. Find the inverse function for the following
graphed function, f(x).
The equation of the
graphed line is
.
So, by a similar
calculation as in Part B,
y-intercept: (0, −4)
5
rate of change:
2
This can also be found by choosing two points on the
line, like
and
, reversing their order, plotting
the points, and determining the equation of that line.
Page 9 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
3´
Benchmark 1 Study Guide
You try: After 5 minutes of climbing, Juan
reached an elevation of 300 feet. After 9
minutes, he was at an altitude of 500 feet.
Assuming a constant elevation gain, express
algebraically the height he will reach, h(m),
after m minutes. Show your equation in three
different forms.
4´
You try: Sketch the following.
A. f (x) = x − 5
B. f (x) = − x − 5
C. f (x) = x − 3
D. f (x) = x + 2 + 4
Create data points for (minutes, feet):
(5, 300) and (9, 500)
Find the rate of change:
Δy
m=
Δx
500 − 300
m=
9−5
200
m=
4
m = 50
Write out Point-Slope Form and substitute
values:
Point-Slope Form
Slope-Intercept
Form
E. f (x) =
1
x+3 −6
2
F. f (x) = −3 x −1 + 2
Standard Form
Page 10 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
5´
Benchmark 1 Study Guide
You try: Find the average rate of change on
the interval −1 ≤ x ≤ 4 for the function
f (x) = x 2 + 2x − 3 .
6´
You try: Find the intercepts for
f (x) = x 2 − 7x +10 .
Let x = 0 :
2
f (−1) = (−1) + 2(−1) − 3
f (−1) = 1− 2 − 3
f (−1) = −4
2
f (4) = (4) + 2(4) − 3
f (4) = 16 + 8 − 3
f (4) = 21
f (0) = 0 2 − 7(0) +10
f (0) = 0 − 0 +10
f (0) = 10
The y-intercept is (0,10) .
f (x) − f (a)
x−a
−1− 21
A=
−1− 4
−22
A=
−5
22
A=
5
A=
Let f (x) = 0.
0 = x 2 − 7x +10
0 = (x − 5)(x − 2)
∴ x = 5 and x = 2
The x-intercepts are (5, 0) and (2, 0) .
Page 11 of 13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
7´
Benchmark 1 Study Guide
You try:
8´
You try:
A. Simplify
(−4 − i) − (−3+ 8i)
= −4 − i + 3− 8i
= −4 + 3− i − 8i
= −1− 9i
Given the graphed function above, identify…
-- Domain: All values of x.
-- Range:
B. Simplify
(−4 + 5i)(4 − 3i)
−∞ ≤ y < 6 or ( − ∞,6]
= −16 +12i + 20i −15i 2
= −16 + 32i −15(−1)
= −16 + 32i +15
= −1+ 32i
--Vertex: (−2, 6)
-- Minimum/Maximum: Maximum at 6
-- y-intercept: (0, 0)
C. Simplify
-- x-intercepts: (0, 0) and (−4, 0)
-- Increasing interval: −∞ < y < −2 or (-∞,-2)
-- Decreasing interval: −2 < y < ∞ or (-2,∞)
Page 12 of 13
5
2 − 3i
5 2 + 3i
=
⋅
2 − 3i 2 + 3i
5(2 + 3i)
=
(2 − 3i)(2 + 3i)
10 +15i
=
4 − 6i + 6i − 9i 2
10 +15i
=
4+9
10 +15i
=
13
MCC@WCCUSD
09/28/15
WCCUSD Algebra II
9´
Benchmark 1 Study Guide
You try:
A. Solve x 2 + 5x = −1 using the Quadratic
Formula.
x 2 + 5x = −1
x 2 + 5x +1 = 0
So,
a = 1, b = 5 , c = 1
x=
−b ± b 2 − 4ac
2a
x=
−(5) ± (5)2 − 4(1)(1)
2(1)
−5 ± 25 − 4
2
−5 ± 21
x=
2
x=
B. Solve x 2 + 5x = −1 by Completing the Square.
x 2 + 5x = −1
2
" 5%
" 5%
x + 5x + $ ' = −1+ $ '
#2&
#2&
2
2
2
"
5%
4 25
$x + ' = − +
#
2&
4 4
2
"
5 % 21
$x + ' =
#
2&
4
2
"
5%
21
$x + ' =
#
2&
4
x+
5 ± 21
=
2
2
± 21 5
x=
−
2
2
−5 ± 21
x=
2
Page 13 of 13
MCC@WCCUSD
09/28/15