f.if.1/a1.fif.1/fa.fif.1 answers - weebly

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Algebra 1 102 Common Core Assessment Readiness

F.IF.1/A1.FIF.1*/FA.FIF.1* Answers 1. B 2. A 3. B, D, F, G 4. A function assigns each value from the

domain to exactly one value in the range. The relation is not a function because February has 28 days in a common year and 29 days in a leap year. Rubric 1 point for stating its not a function; 2 points for explanation

5. The table represents a function. The domain is {−2, −1, 0, 1, 2}. The range is {2, 6, 10, 14, 18}.

Rubric 1 point for answer; 1 point for domain; 1 point for range

6. The y-value associated with x = −2 is 4. The y-value associated with x = 0 is 3. The y-value associated with x = 2 is 2. If y = f(x), then the x-values are in the

domain of f(x), and the y-values are in the range of f(x). Rubric 1 points for each y-value; 1 point for stating the x-values are in the domain of f(x); 1 point for stating the y-values are in the range of f(x)

7. The domain of the function is the set of all real numbers.

The range is the set of real numbers greater than −3. Rubric 1 point for the domain; 2 points for the range

8. a. This is a function because no coin has more than one monetary value. The domain is the set of coins {penny, nickel, dime, quarter, half dollar}. The range is the set of monetary values assigned to each coin, {$0.01, $0.05, $0.10, $0.25, $0.50}. (Students may or may not include half dollar and dollar coins in their example. Assign full credit as long as penny, nickel, dime, and quarter are included.)

b. This is not a function because each bill is equivalent to many different combinations of coins. For example, a 1 dollar bill is equivalent to 100 pennies, but it is also equivalent to 10 dimes.

Rubric a. 1 point for answer;

1 point for explanation; 1 point for the domain; 1 point for the range

b. 1 point for answer; 1 point for explanation

Name ________________________________________ Date __________________ Class _________________


Algebra 1 Teacher Guide 102 Common Core Assessment Readiness

F.IF.2/A1.FIF.2*/FA.FIF.2* Answers 1. A 2. C 3. C 4. B, D, F, H 5. C(15) = 25(15) = 375 This value represents the production cost

of 15 graphing calculators. So, it costs $375 to produce 15 graphing calculators. Rubric 1 point for value; 1 point for interpretation

6. f(−2) = 13(−2) − (−2)2 = −26 − 4 = −30 f(−1) = 13(−1) − (−1)2 = −13 − 1 = −14 f(0) = 13(0) − 02 = 0 − 0 = 0 f(1) = 13(1) − 12 = 13 − 1 = 12 f(2) = 13(2) − 22 = 26 − 4 = 22 The range of the function is

{−30, −14, 0, 12, 22}. Rubric 0.5 point for each value in the range; 0.5 point for work

7. a. The domain of the function is {2, 3, 4, 5, 6, 7}.

b. V(2) = 23 = 8 cubic feet V(3) = 33 = 27 cubic feet V(4) = 43 = 64 cubic feet V(5) = 53 = 125 cubic feet V(6) = 63 = 216 cubic feet V(7) = 73 = 343 cubic feet

Rubric a. 1 point b. 0.5 point for each value 8. a. The profit function is:

P(n) = 149.99n − 25(99.50)

= 149.99n − 2487.50

b. The domain of the function is all whole numbers between 0 and 25, inclusive. The store cannot sell a negative number of televisions and they can

only sell up to the number in stock, which is 25.

c. The store will make a profit of $1,262.25.

Rubric a. 1 point b. 1 point for domain; 1 point for

explanation c. 1 point 9. a. Set p(t) equal to zero and solve for t.

0 = −8t +100−100 = −8t

t = 12.5

It would take 12.5 minutes for the printer to use all 100 sheets.

b. The domain of the function is all values of t, where 0 ≤ t ≤ 12.5. The printer takes 12.5 minutes to print all 100 pages, so the upper bound on t is 12.5. The printer starts printing at 0 minutes, so the lower bound is 0.

c. The range of the function is all values of p(t), where 0 ≤ p(t) ≤ 100. There are 100 sheets of paper in the printer at the start, so the upper bound on p(t) is 100. There cannot be a negative number of sheets, so the lower bound on p(t) is 0.

d. The printer will have p(7) = −8(7) + 100 = −56 + 100 = 44 sheets of paper left. So, Tanya’s report is 100 − 44 = 56 pages long.

Rubric a. 1 point for answer; 1 point for

explanation b. 1 point for domain; 1 point for

explanation c. 1 point for range; 1 point for

explanation d. 1 point for answer; 1 point for work



F.IF.4/A1.FIF.4*/FA.FIF.4*/IA.FIF.4* Answers 1. C 2. A, C, G 3.

The C(t)-intercept is $15, which is the

cost for sending up to 1000 texts. Rubric 2 points for graph; 1 point for intercept; 1 point for interpretation

4. The profit increases as the orchard goes from 0 to 40 trees. Then, the profit decreases from 40 to 80 trees. The owner of the orchard earned the least profit when there were no trees planted and when there were 80 trees planted. The most profit was earned when there were 40 trees planted. Rubric 2 points for description of relationship; 2 points for stating where the orchard owner earned the least profit; 1 point for stating where the orchard owner earned the most profit

5. a.

b. The function is decreasing for x < 0 and increasing for 0 < x.

c. f(x) is positive for all values of x except 0.

d. f(x) will never be negative because absolute value can never be negative.

Rubric a. 1 point b. 1 point for stating where f(x)

decreases; 1 point for stating where f(x) increases

c. 1 point d. 2 points



F.IF.5/A1.FIF.5*/FA.FIF.5*/IA.FIF.5* Answers 1. C 2. D 3. C 4. a. The domain is t such that 0 ≤ t ≤ 7. b. The domain represents the time that

the object is in the air. Rubric 1 point for each part

5. The domain is the real numbers. The range is the real numbers greater

than −2. Rubric 1 point for the domain; 1 point for the range

6. a. P(c) = 350c − 1800 b. The domain of P(c) is the whole

numbers. The company cannot sell a negative number of the tablet computers and they cannot sell a fractional number of tablet computers.

c. −1800, −1450, −1100, −750, −400, −50, 300, 650

Rubric a. 2 points for the function b. 1 point for the domain;

1 point for the explanation; c. 2 points for range values

7. Brand A A(h) = 4.19h The domain of A(h) is the nonnegative

real numbers since brand A can be purchased in any nonnegative amount from the deli counter.

Brand B B(h) = 4.79h The domain of B(h) is {0, 0.5, 1, 1.5, …}

since brand B can only be purchased in increments of either 0.5 pound or 1 pound. Rubric 1 point for each function rule; 1 point for each domain; 1 point for each explanation



F.IF.6/A1.FIF.6*/IA.FIF.6* Answers 1. B 2. A 3. C 4. C, D, E 5. a. No b. Yes c. No d. Yes e. No 6. a. Anchorage: about −5.6 minutes of

daylight per day Los Angeles: about −2.0 minutes of

daylight per day b. On average, Anchorage loses about

5.6 minutes of daylight each day during the month of October, while Los Angeles loses about 2 minutes of daylight each day.

c. Since October 17 is 16 days after October 1, evaluate the expression 711 + (−2.0)(16) to get 679 minutes, or 11 hours 19 minutes, of daylight on October 17. If the sun rises at 7:00 A.M. that day, then it sets about 11 hours and 19 minutes later, or at 6:19 P.M.

Rubric a. 0.5 point for each average rate of

change; b. 1 point c. 1 point for finding the minutes of

daylight; 1 point for the description of how to find the minutes of daylight; 1 point for the time of sunset

7. a. The average growth rate between weeks 2 and 3 is about

4000 − 2000

3 − 2= 2000

1=

2000 bacteria per week. b. The average growth rate between

weeks 3 and 4 is about

8000 − 4000

4 − 3= 4000

1=

4000 bacteria per week. c. The average growth rate between

weeks 4 and 5 is about

16000 − 8000

5 − 4= 8000

1=

8000 bacteria per week. d. The average growth rate is doubling

as each week passes. 2(2000) = 4000 bacteria per week,

2(4000) = 8000 bacteria per week e. The average growth rate between

weeks 5 and 6 will probably be 16,000 bacteria per week if the pattern continues.

Rubric a. 0.5 point b. 0.5 point c. 0.5 point d. 1 point for answer;

2 points for justification e. 0.5 point



F.IF.7a/A1.FIF.7*a/FA.FIF.7*a/IA.FIF.7*a Answers 1. B 2. A 3. a. R(n) = 5.25n b. Since Sally is selling 1 necklace at a

time and cannot sell negative necklaces, a reasonable domain for this function is the whole numbers.

c.

d. The n- and R-intercepts are both 0.

The intercept indicates that Sally will earn no revenue if she sells no necklaces.

Rubric a. 1 point b. 1 point c. 1 point d. 1 point for the intercept;

1 point for the explanation 4. a. The t-intercepts are 0 and 5. These

intercepts indicate that the object has a height of 0 meters when the object is thrown (t = 0) and again at 5 seconds after it is thrown.

The h-intercept is the same as the first t-intercept.

b. The vertex is (2.5, 30.625). c. The vertex is a maximum. This means

that the maximum height of the object is 30.625 feet.

d.

Rubric

a. 1 point for each t-intercept; 0.5 point for each interpretation

b. 1 point for the vertex c. 0.5 point for the answer;

0.5 point for the interpretation d. 2 points for correct graph 5. One side of the rectangular enclosure is

s. Let the other side of the enclosure be x. The perimeter of the combined enclosures is 1200 = 5s + 2x.

Solve for x:

1200 = 5s + 2x1200 − 5s = 2x

0.5(1200 − 5s) = x600 − 2.5s = x

The width of the rectangular enclosure is s + x = s + 600 − 2.5s = 600 − 1.5s.

The combined area of the enclosures is

A(s) = s(600 −1.5s)= 600s −1.5s2

= −1.5s2 + 600s

105–5 n

25

–25

R(n)



The square enclosure will be 200 feet by 200 feet. The rectangular enclosure will be 100 feet by 200 feet. The vertex of the function, (200, 60000), is a maximum. This means that the combined area of the enclosures is maximized when s = 200. Since s is the side length of the square, the square is 200 feet by 200 feet. Since the rectangle shares a side with the square, one of its dimensions is 200 feet. The other is given by the expression 600 − 2.5s.

600 − 2.5(200) = 600 − 500

= 100

Thus, the rectangle is 100 feet by 200 feet. Rubric 2 points for the function; 1 point for the graph; 1 point for each set of dimensions; 2 points for explanation involving the vertex



F.IF.8a/A1.FIF.8*a/FA.FIF.8*a Answers 1. B 2. B 3. B, C, E 4. a. False b. True c. False d. True e. True 5. a.

f (x) = 4x2 + 4x −15= 4x2 +10x − 6x −15= 2x(2x + 5)− 3(2x + 5)= (2x − 3)(2x + 5)

The zeros of the function are x = 3

2

and x = − 5

2.

b. The vertex is halfway between the zeros of the function, so the

x-coordinate is − 1

2. The value of the

function at x = − 1

2 is

4 − 1

2⎛⎝⎜

⎞⎠⎟

2

+ 4 − 12

⎛⎝⎜

⎞⎠⎟−15 = −16.

The vertex of f(x) is − 1

2, −16

⎛⎝⎜

⎞⎠⎟. The

coefficient of x2 is positive, so the parabola opens up, and the vertex is the minimum value.

Rubric a. 1 point for factoring; 0.5 point for

each zero b. 1 point for coordinates of the vertex;

1 point for stating that the vertex is a minimum; 1 point for explanation

6. a. Miguel is correct in saying that the function has no x-intercepts. However, the axis of symmetry can still be found by completing the square and finding the vertex. The axis of symmetry passes through the vertex.

b. Complete the square:

f (x) = −2x2 −16x − 34= −2 x2 + 8x( )− 34

= −2 x2 + 8x +16 −16( )− 34

= −2 x2 + 8x +16( ) + 32− 34

= −2 x + 4( )2 − 2

The vertex of the function is (−4, −2), so the axis of symmetry is x = −4.

Rubric a. 1 point b. 1 point for work; 1 point for answer

7. a.

h(x) = − 9125

x2 + 45

= − 9125

x2 − 625( )= − 9

125(x − 25)(x + 25)

The zeros of the function occur where the sides of the arch are at the water level. They are 25 feet to the left and right of the center of the bridge, so the bridge is 50 feet long.

b. The coefficient of x2 is negative, so the vertex is a maximum value of the function. The vertex is halfway between the zeros of the function, at

x = 0. h(0) = − 9

125(0)2 + 45 = 45

feet,

so the sailboat will not be able to pass under the bridge.

Rubric a. 1 point for answer; 1 point for

explanation involving the zeros b. 1 point for answer; 1 point for

explanation involving the vertex as a maximum



F.IF.9/A1.FIF.9*/FA.FIF.9*/IA.FIF.9* Answers 1. C 2. D 3. B, E 4. Both f(x) and g(x) are defined on the

domain from 1 to 4. However, f(x) is only defined for the whole numbers 1, 2, 3, and 4, while g(x) is defined for all real numbers between 1 and 4, inclusive. The range of f(x) is the set {4, 6, 10, 18}. The range of g(x) is g(1) ≤ g(x) ≤ g(4), or 4 ≤ g(x) ≤ 19. The initial value for both functions is f(1) = 4 = 12 + 3 = g(1). Rubric 1 point for comparing domains; 1 point for comparing ranges; 1 point for comparing initial values

5. The maximum value of the function shown in the graph is f(6) = 4. The maximum value of the function in the table is g(0) = 5. Since the maximum known value for g(x) is greater than the maximum value of f(x), g(x) has a greater maximum value on the domain −6 ≤ x ≤ 6. Rubric 1 point for answer; 1 point for explanation

6. a. Plan A costs more for 3 months. The cost of plan A for 3 months is 70(3) = $210, and the cost of plan B for 3 months is $200.

b. 5 months. The cost of plan A for 1 through 6 months is A(1) = $70, A(2) = $140, A(3) = $210, A(4) = $280, A(5) = $350, and A(6) = $420. Comparing these values to the corresponding values of B(t), the first time the difference is greater than 50 is when t = 5.

Rubric a. 1 point for answer; 1 point

for explanation

b. 1 point for answer; 2 points for explanation comparing function values



F.BF.3/A1.FBF.3*/FA.FBF.3*/IA.FBF.3* Answers 1. B 2. C 3. Possible answer: The graph of f(x) is

reflected about the y-axis, shifted left 1 unit, vertically stretched by a factor of 2, reflected about the x-axis, and shifted up 3 units to obtain g(x).

Rubric 5 points for transformations; 1 point for graph

4. The slope m acts as a vertical stretch or shrink factor. If 0 < |m| < 1, then the graph of g(x) is a vertical shrink of the graph of f(x) by a factor of m. If |m| > 1, then the graph of g(x) is a vertical stretch of the graph of f(x) by a factor of m. The sign of the slope can also transform the graph of g(x). If m < 0, then the graph of g(x) is a reflection about the x-axis of the graph of f(x). If m > 0, no reflection occurs. Rubric 2 points for description of how m can be a shrink or a stretch (either vertical or horizontal); 1 point for description of how m can be a reflection

5. a. f(x) = x; vertical shift up; g(x) = x + 3 Alternate answer: f(x) = x; horizontal

shift left; g(x) = x + 3

b. f(x) = 2x; vertical shift down; h(x) = 2x − 2

Rubric a. 0.5 point for parent function; 1 point for

transformation; 1 point for rule b. 0.5 point for parent function; 1 point for

transformation; 1 point for rule

6. a.

g(x) = − 12

x2 − 2x + 2

= − 12

(x2 + 4x)+ 2

= − 12

(x2 + 4x + 4 − 4)+ 2

= − 12

(x + 2)2 − 4⎡⎣ ⎤⎦ + 2

= − 12

(x + 2)2 + 2+ 2

= − 12

(x + 2)2 + 4

b. Possible answer: The graph of f(x) is shifted left 2 units, vertically shrunk by

a factor of 12

, reflected across the

x-axis, and shifted up 4 units. c.


1 point for reasonable work b. 2 points for transformations c. 1 point



F.LE.1a/A1.FLQE.1*a/FA.FLQE.1*a Answers 1. B 2. C 3. A, D, E 4. A 5. G 6. F 7. H 8. E 9.

x2 − x1 = x4 − x3 Given

m(x2 − x1) = m(x4 − x3 ) Multiplication property of equality

mx2 − mx1 = mx4 − mx3 Distributive property

mx2 + b − mx1 − b = mx4 + b − mx3 − b Addition and subtraction properties

(mx2 + b) − (mx1 + b) = (mx4 + b) − (mx3 + b) Distributive property

f (x2) − f (x1) = f (x4 ) − f (x3 ) Definition of f (x)

Rubric 1 point for each correctly completed part

10. a. Time, t (years)

Salary, S (dollars)

1 39,500 2 41,000 3 42,500 4 44,000 5 45,500

b. Sandra’s salary increases by $1500 each year. S(2) − S(1) = 41,000 − 39,500 = 1500 S(3) − S(2) = 42,500 − 41,000 = 1500 S(4) − S(3) = 44,000 − 42,500 = 1500 S(5) − S(4) = 45,500 − 44,000 = 1500

Rubric a. 2 points b. 1 point for answer; 1 point for appropriate work



11. a. P(0) = 3500(0.97)0 = 3500 people P(1) = 3500(0.97)1 = 3395 people

P(1)P(0)

=33953500

= 0.97

P(2) = 3500(0.97)2 ≈ 3293 people

P(2)P(1)

≈32933395

≈ 0.97

The population changes by the same factor over each 1 year interval.

b.

P(2)P(0)

=32933500

≈ 0.94

P(3) = 3500(0.97)3 ≈ 3194 people

P(3)P(1)

≈31943395

≈ 0.94

The population changes by the same factor over each 2 year interval. Rubric

a. 1 point for the factor between 2000 and 2001; 1 point for the factor between 2001 and 2002; 1 point for work

b. 1 point for the factor between 2000 and 2002; 1 point for the factor between 2001 and 2003; 1 point for work



F.LE.2/A1.FLQE.2*/IA.FLQE.2* Answers 1. B, C 2. C 3. D 4. B1 = 100(1.0025)t B2 = 100 + 0.25t

Rubric 1 point for each function

5. a. Exponential. The population each day is 25% greater than the population on the previous day.

b. P(t) = 1000(1.25)t Rubric

a. 1 point for answer; 1 point for explanation

b. 1 point

6. r = 16.20

18= 0.9

V (t) = a(0.9)t

18 = a(0.9)0

18 = aV (t) = 18(0.9)t

Rubric 1 point for answer; 2 points for work

7. s5 = 36, s12 = 64

d = 64 − 36

12− 5= 28

7= 4

sn = s1 + d(n −1)64 = s1 + 4(12−1)64 = s1 + 4420 = s1

sn = 20 + 4(n −1)

Rubric 1 point for correct sequence; 2 points for work

8. a. m = 150 − 60

30 − 20= 90

10= 9

y = mx + b60 = 9(20)+ b60 = 180 + b

−120 = b

The art club’s profit is modeled by the function P(c) = 9c − 120.

b. The slope is 9, which means that each comic book costs $9.

The P-intercept is −120, which could mean that the art club spends $120 on supplies to make the comic books.

c. The art club needs to sell 55 comic books to make $375.

375 = 9c −120495 = 9c55 = c

Rubric a. 1 point b. 1 point for interpretation of slope;

1 point for interpretation of intercept c. 1 point for answer; 1 point for work 9. a. Geometric. The ratio of consecutive

distances is

34.3136.35

≈ 32.3934.31

≈ 30.5732.39

≈ 0.944.

b.

d(n) = d(1) i (0.944)n−1

= 36.35 i (0.944)n−1

c.

d(n) = 36.35 i(0.944)n−1

d(20) = 36.35 i(0.944)20−1

= 36.35 i(0.944)19

≈12.16

The 20th fret is about 12.16 mm from the 19th fret.

Rubric a. 1 point for answer; 1 point for ratio b. 1 point for answer; 1 point for work c. 1 point for answer; 1 point for work



F.LE.3/A1.FLQE.3*/FA.FLQE.3* Answers 1. A, B, E 2. C 3. C 4. a. A(t) < B(t) b. A(t) < B(t) c. A(t) > B(t) d. A(t) > B(t) e. A(t) > B(t) 5. a. A(t) = 300t + 1500 B(t) = 1500(1.15)t b.

Time, t (years)

Town A population,

A(t)

Town B population,

B(t) 0 1500 1500 1 1800 1725 2 2100 1984 3 2400 2281 4 2700 2624 5 3000 3017 6 3300 3470 7 3600 3990 8 3900 4589

c. Town B will have a larger population than town A. A increases by the same amount (300) each year. B increases by the same percent (15%) each year. 15% of 4589 is about 688, so B will continue to increase by a greater amount than A each year after 2008.

Rubric a. 1 point for each function b. 0.5 point for each value calculated c. 1 point for answer;

1 point for explanation

6. a.

b.

c. Initially, the values of h(x) are less than the values of f(x), but greater than the values of g(x). Eventually, the values of h(x) exceed the values of both f(x) and g(x).

d. The value of an exponential function may start out less than the values of a linear function or a polynomial function for the same inputs. As x increases, the value of the exponential function will eventually be greater than the values of the linear and polynomial functions for the same inputs.

Rubric a. 0.5 point for graphing each function b. 0.5 point for graphing each function c. 2 points for accurate comparison d. 2 points for accurate conjecture



F.LE.5/A1.FLQE.5*/FA.FLQE.5*/IA.FLQE.5* Answers 1. A 2. D 3. C 4. A, D 5. At t = 0, the plane begins to descend. So,

the height of the airplane, in feet, when it begins to descend is h(0) = 15,000. Rubric 1 point for correct height with correct units; 1 point for explanation

6. a. 256. The value of P(0) is 256.

b. 12

. Since the decay rate, 12

, is the

ratio of players remaining after r rounds to the ratio of players remaining after r − 1 rounds, the fraction of players eliminated in

the rth round is 1− 1

2= 1

2.


1 point for explanation b. 1 point for answer;

1 point for explanation 7. a. $1250 b. 2%. Since 0.98 is the decay factor, it

can be rewritten as 1 − 0.02, where 0.02 is the decay rate. Notice that 0.02 is equal to 2%.

Rubric a. 1 point b. 1 point for answer;


8. a. 65 miles per hour. The coefficient of t is equal to the speed, in miles per hour, that the family’s car is traveling. Notice that the coefficient of t is 65.

b. 715 miles. The distance between the family’s house and the point where they started driving on the second day is equal to d(0). Notice that d(0) = 715.



1 point for explanation 9. a. City B. The official value of city A’s

population for the first census and the official value of city B’s population for the first census is when c = 0. Since A(0) = 50,600, the value of city A’s population for the first census is 50,600. Since B(0) = 75,850, the value of city B’s population for the first census is 75,850.

b. City A. The growth factor for city A is 1.08 and the growth factor for city B is 1.069.




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