wccusd algebra ii benchmark 1 study guide · 2015. 11. 15. · wccusd algebra ii benchmark 1 study...

WCCUSD Algebra II Benchmark 1 Study Guide

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



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

"

#$

%

&', 3,−2( )

()*

+,-



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



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



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



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 .



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.



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"

#$%

&'

2

= −4+ 32"

#$%

&'

2

x + 32

"

#$

%

&'

2

= −164+

94

x + 32

"

#$

%

&'

2

= −74

x + 32

"

#$

%

&'

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.



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

"

#$

%

&', 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

"

#$

%

&'

−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( )

()*

+,-



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



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



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



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"

#$%

&'

2

= −1+ 52"

#$%

&'

2

x + 52

"

#$

%

&'

2

= −44+

254

x + 52

"

#$

%

&'

2

=214

x + 52

"

#$

%

&'

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

wccusd algebra ii benchmark 1 study guide · 2015. 11. 15. · wccusd algebra ii benchmark 1 study...

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