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WCCUSD Algebra II Benchmark 1 Study Guide
Page 1 of 13 MCC@WCCUSD 09/28/15
1 Linear Functions, Tables and Graphs. 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)+3f (−2) = −1
etc…
x -2 -1 0 1 2 y -1 1 3 5 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. For f (x) = 2x +3 , the rate of change (slope) is 2, y-intercept is (0,3) .
A-CED.2
1´ You try: A. Make a table of values for the function
f (x) = − 23x + 4 . (Hint: Select x-values that
are multiples of 3 to facilitate calculations.) B. Determine the graphed function’s rule:
C. Graph the function f (x) = 52x − 4 by
identifying the y-intercept and rate of change.
A-CED.2
WCCUSD Algebra II Benchmark 1 Study Guide
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2 Inverse Functions Given a relation or a function its inverse can be determined by reversing the order of its x- and y-coordinates such that (x, y)→ (y, x) . Example 1: Find the inverse for the relation 2,3( ), −4,1( ), 9,− 7( ){ }
Solution: Reverse the order of the x- and y-coordinates:
Example 2:
Find the inverse function for g(x) = 23x − 5 .
Solution: Consider g(x) as the output value y:
y = 23x − 5
Reverse the order of x and y. Then solve for y:
x = 23y− 5
x + 5= 23y
32x + 5( ) = 3
2⋅23y
32x + 5( ) = y
32x +15
2= y
∴g−1(x) = 32x +15
2
or
g−1(x) = 3x +152
F-BF.4
2´ You try: A. Find the inverse for the relation B. Find the inverse function for
h(x) = − 54x +1 .
C. Find the inverse function for the following graphed function, f(x).
F-BF.4
−4, 7( ), 10, 12
"
#$
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&', 3,−2( )
()*
+,-
WCCUSD Algebra II Benchmark 1 Study Guide
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3 Writing Linear Equations Using Two Data Points
Here are three ways to express a linear function:
1) Slope-Intercept Form: y =mx + b 2) Point-Slope Form: y− y1 =m(x − x1) or
f (x) =m(x − x1)+ y1 3) Standard Form: Ax +By =C
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):
m =ΔyΔx
m =160−100
5−3
m =602
m = $30 per week
---Find the initial condition by substituting either point and the rate of change into the slope-intercept equation:
y =mx + b100 = 30(3)+ b100 = 90+ b 10 = b
---Rewrite information: y =mx + by = 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) .
From here you could also rewrite to slope-intercept form and standard form.
A-CED.2
3´ 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.
A-CED.2
or
WCCUSD Algebra II Benchmark 1 Study Guide
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4 Transformations of Absolute Value Functions
Example: Sketch .
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. The solid line is f (x) = 4 x +3 − 5and the dashed
line is the parent function f (x) = x .
F-IF.7 and F-BF.3
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
E. f (x) = 12x +3 − 6 F. f (x) = −3 x −1 + 2
F-IF.7 and F-BF.3
-IF.7 and F-BF.3
f (x) = 4 x +3 − 5
F-IF.7 and F-BF.3
WCCUSD Algebra II Benchmark 1 Study Guide
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5 Average Rate of Change
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:
A = ΔyΔx
=f (x)− f (a)x − a
where f (x) and f (a) are output values (the y-values for Δy in the slope formula) and x and a are the input values (Δx in the slope formula). Example: Find the average rate of change on the interval 1≤ x ≤ 4 for the function f (x) = x2 + 5 .
--Input values are 1 and 4. So to find the output values, find f (1) and f (4) : f (1) = (1)2 + 5f (1) =1+ 5f (1) = 6
f (4) = (4)2 + 5f (4) =16+ 5f (4) = 21
--Substitute your values:
A = f (x)− f (a)x − a
A = f (1)− f (4)1− 4
A = 6− 211− 4
A = −15−3
A = 5
F-IF.6
5´ You try: Find the average rate of change on the interval −1≤ x ≤ 4 for the function f (x) = x2 + 2x −3 .
F-IF.6
WCCUSD Algebra II Benchmark 1 Study Guide
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6 Finding Intercepts
To find the x-intercept, let y = 0 (or f (x) = 0 ). To find the y-intercept, let x = 0 .
Example: Find the intercepts for f (x) = x2 + x −12 .
Let x = 0 : f (0) = (0)2 + (0)−12f (0) = −12
Let y or f (x) = 0 : 0 = x2 + 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
F-IF.8
7 Identifying Key Features
(domain, range, intercepts, minimum/maximum, increasing/decreasing rate)
--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 .
--The y-intercept is at (0,−4) and the x-intercepts are at (−3,0) and (5, 0) .
--The function is decreasing for the interval (−∞,1) and increasing for the interval (1,∞) .
F-IF.4
6´ You try: Find the intercepts for f (x) = x2 − 7x +10 .
F-IF.8
7´ You try:
Given the graphed function above, identify…
-- Domain: -- Range: --Vertex: -- Minimum/Maximum: -- y-intercept: -- x-intercepts: -- Increasing interval: -- Decreasing interval:
F-IF.4
The y-intercept is .
The x-intercepts are and .
WCCUSD Algebra II Benchmark 1 Study Guide
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8 Complex Numbers Complex numbers are written in the forma± bi where i = −1 , and therefore i2 = −1 . Examples:
A. Simplify (3− 2i)− (4+ 6i) (3− 2i)− (4+ 6i)= 3− 2i− 4− 6i= 3− 4− 2i− 6i= −1−8i
B. Simplify (3− 2i)(4+ 6i)
(3− 2i)(4+ 6i)= 3(4)+3(6i)− 2i(4)− 2i(6i)=12+18i−8i−12i2
=12+18i−8i−12(−1)=12+18i−8i+12= 24+10i
C. Simplify 43+ i
43+ i
=4
3+ i•
3− i3− i
=4(3− i)
(3+ i)(3− i)
=12− 4i
9+3i−3i− i2
=12− 4i9− (−1)
=12− 4i
10
=2(6− 2i)
2 ⋅5
=6− 2i
5
N-CN.2
8´ You try: A. Simplify (−4− i)− (−3+8i) B. Simplify (−4+ 5i)(4−3i)
C. Simplify 52−3i
N-CN.2
Distribute the negative.
Commute.
Combine like terms.
Distribute. Multiply.
Combine like terms.
--Multiply by conjugate. --Distribute. --Simplify. --Factor, if possible. --Simplify.
WCCUSD Algebra II Benchmark 1 Study Guide
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9 Solving Quadratics
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. Example: Solve x2 +3x = −4 .
x2 +3x + 4 = −4+ 4x2 +3x + 4 = 0
Determine values for a, b and c if it is in the form ax2 + bx + c = 0 .
∴a =1 , b = 3 , c = 4
Substitute and solve using the Quadratic Formula:
x = −b± b2 − 4ac2a
x = −(3)± (3)2 − 4(1)(4)2(1)
x = −3± 9−162
x = −3± −72
x = −3± −1 ⋅ 72
x = −3± i 72
Using Completing the Square: x2 +3x = −4
x2 +3x + 32"
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2
= −4+ 32"
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2
x + 32
"
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2
= −164+
94
x + 32
"
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&'
2
= −74
x + 32
"
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&'
2
= −74
x + 32=± −7
2
x = − 32±i 7
2
x = −3± i 72
A-REI.4 and N-CN.7
9´ You try: A. Solve x2 + 5x = −1 using the Quadratic Formula. B. Solve x2 + 5x = −1 by Completing the Square.
A-REI.4 and N-CN.7
End of Study Guide
Set equal to zero
To “complete the square,” find . Add to both sides.
Create common denominator. Combine terms on right side. Take the square root of each side. Simplify. Remember: Isolate the variable x. Combine terms.
WCCUSD Algebra II Benchmark 1 Study Guide
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You Try Solutions: 1´ You try:
A. Make a table of values for the function
f (x) = − 23x + 4 . (Hint: Select x-values that
are multiples of 3 to facilitate calculations.)
x -6 -3 0 3 6 y 8 6 4 2 0
B. Determine the graphed function’s rule:
f (x) = 3x − 6
C. Graph the function f (x) = 52x − 4 by
identifying the y-intercept and rate of change. y-intercept: (0,−4)
rate of change: 52
2´ You try: A. Find the inverse for the relation
−4, 7( ), 10, 12
"
#$
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&', 3,−2( )
()*
+,-
B. Find the inverse function for
h(x) = − 54x +1 .
y = − 54x +1
x = − 54y+1
x −1= − 54y
−45
(x −1) = − 45−
54y
"
#$
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&'
−45x + 4
5= h−1(x)
OR
−4x + 45
= h−1(x)
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,
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.
7,−4( ), 12,10
"
#$
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&', −2, 3( )
()*
+,-
WCCUSD Algebra II Benchmark 1 Study Guide
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3´ 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. Create data points for (minutes, feet): (5,300)and (9, 500) Find the rate of change:
m =ΔyΔx
m =500−3009− 5
m =2004
m = 50
Write out Point-Slope Form and substitute values:
4´ You try: Sketch the following. A.
B.
C. D.
E. F.
f (x) = x − 5 f (x) = − x − 5
f (x) = x −3 f (x) = x + 2 + 4
f (x) = 12x +3 − 6 f (x) = −3 x −1 + 2
Point-Slope Form Slope-Intercept Form Standard Form
WCCUSD Algebra II Benchmark 1 Study Guide
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5´ You try: Find the average rate of change on the interval for the function
. f (−1) = (−1)2 + 2(−1)−3f (−1) =1− 2−3f (−1) = −4
f (4) = (4)2 + 2(4)−3f (4) =16+8−3f (4) = 21
A = f (x)− f (a)x − a
A = −1− 21−1− 4
A = −22−5
A = 225
6´ You try: Find the intercepts for .
Let x = 0 : f (0) = 02 − 7(0)+10f (0) = 0− 0+10f (0) =10
The y-intercept is (0,10) . Let f (x) = 0. 0 = x2 − 7x +100 = (x − 5)(x − 2)
∴x = 5 and x = 2
The x-intercepts are (5, 0) and (2, 0) .
−1≤ x ≤ 4f (x) = x2 + 2x −3
f (x) = x2 − 7x +10
WCCUSD Algebra II Benchmark 1 Study Guide
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7´ You try: Given the graphed function above, identify…
-- Domain: All values of x. -- Range: −∞≤ y < 6 or (−∞,6] --Vertex: (−2,6) -- Minimum/Maximum: Maximum at 6 -- y-intercept: (0, 0) -- x-intercepts: (0, 0) and (−4, 0) -- Increasing interval: −∞ < y < −2 or (-∞,-2) -- Decreasing interval: −2 < y <∞ or (-2,∞)
8´ You try: A. Simplify
(−4− i)− (−3+8i)= −4− i+3−8i= −4+3− i−8i= −1− 9i
B. Simplify
(−4+ 5i)(4−3i)= −16+12i+ 20i−15i2
= −16+32i−15(−1)= −16+32i+15= −1+32i
C. Simplify
52−3i
=5
2−3i⋅2+3i2+3i
=5(2+3i)
(2−3i)(2+3i)
=10+15i
4− 6i+ 6i− 9i2
=10+15i
4+ 9
=10+15i
13
WCCUSD Algebra II Benchmark 1 Study Guide
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9´ You try: A. Solve x2 + 5x = −1 using the Quadratic Formula. x2 + 5x = −1x2 + 5x +1= 0 So, a =1 , b = 5 , c =1
B. Solve x2 + 5x = −1 by Completing the Square.
x2 + 5x = −1
x2 + 5x + 52"
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2
= −1+ 52"
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2
x + 52
"
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&'
2
= −44+
254
x + 52
"
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&'
2
=214
x + 52
"
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2
=214
x + 52=± 21
2
x = ± 212
−52
x = −5± 212
x = −b± b2 − 4ac2a
x = −(5)± (5)2 − 4(1)(1)2(1)
x = −5± 25− 42
x = −5± 212