Modeling: Functions
- Theheight,in meters,ofarockas itfalls ata giventime(x),in seconds,can be foundusing theexpression–5x2+h0,whereh0is the startingheightwheretherockfalls.
PartA
Arockfallsfrom astarting heightof 80 meters.Writea function,f(x),thatmodelsthe heightoftherock as itfalls.Makesuretouse properfunctionnotation.
PartB
Using the functionfrompart A,whatis the value off(3)?
PartC
Ahmedstatesthatthe domain for thisfunctioninthegivencontextisx ≤ 4.ExplainwhyAhmedisincorrect,and provide acorrectdomain.
Reasoning: Algebra
2. ShannonandJermainearesolvingquadraticequations.Thistableshowstheirwork.
Steps / Shannon / JermaineInitialequation / 2
x–6x+5=12 / 2
x+2x–29=2x+7
Step1 / 2
x–6x=7 / 2
x +2x–36=2x
Step2 / 2
x–6x+9=16 / 2
x–36=0
Step3 / 2
(x–3) =16 / (x– 18)(x+18)=0
Step4 / x–3 =±4 / x–18=0orx+ 18 =0
Step5 / x=±7 / x=18or -18
Both Shannon and Jermainehaveerrorsintheir work.Writeaclearexplanation ofeachstudent’serror.Providethe correctsolutionsfor both equations.
Shannon
Correctsolution(s):
Explanation oferror:
Jermaine
Correctsolution(s):
Explanation oferror:
Modeling: Geometry
3. Mr. Fontenot planted fourtypesofsoybeanson hisland inorderto compareoverallcost(forplantingandharvesting)andcropharvest. Thetable showsthenumberofacresplanted,the costperacre,andthenumberof bushelsofsoybeansproducedforthedifferenttypesof soybeans.
TypeofSoybean / NumberofAcresPlanted / Cost(peracre)toHarvest / NumberofBushelsProducedA / 200 / $174.70 / 9,000
B / 150 / $180.90 / 7,500
C / 100 / $192.40 / 5,900
D / 75 / $204.00 / 4,500
PartA
RegulationsspecifythatMr.Fontenotcannotdevote more than80%ofafieldtoone particular typeofsoybean.He wants to designa fieldso thathecan harvestthe mostsoybeansfor thelowest cost.
Whatisthebestdesignplan for Mr.Fontenot’s 525 acres? Includespecificdetailsaboutwhichsoybeansyouchose,howmanyacresofeachtypeshould be planted,and whyyouchose thosesoybeans.
PartB
This table shows the profitMr.Fontenotcan earnper bushelfor eachtypeofsoybean.
TypeofSoybean / Profit perBushelA / $4.50
B / $3.88
C / $3.96
D / $4.24
Determineifthe designplancreatedinpartAis the mostprofitable 80/20 design.Includespecificdetailsaboutthe profitabilityof the planfrom part Acomparedto other possible designplans.
Reasoning: Numbers & Quantity
4. Sarah stated that if the complex number 2 + 4i is one of the roots to the quadratic equation x2 + bx + c = 0, then the second root is -2- 4i.
Part A
Write an argument thatdisproves Sarah's statement.
Part B
State the second root and check your solution to justify your answer.
Scoring Rubrics and Correct Answers
Question 1ScoringRubric
3 / Student response includes each of the following 3 elements:
1 modeling point: Correct function in Part A
1 computation point: Correct height in Part Bforf(3)fromequationinpart A
1 modeling point: Correct and complete explanation ofwhyAhmed’sdomainisincorrect, including correct domain
2 / Student response includes 2 of the 3 elements
1 / Student response includes 1 of the 3 elements
0 / Thestudent’sresponseisincorrect,irrelevant totheskill or conceptbeing measured,or blank
Note: Apointmaybe awardedinpart Bifthestudentcorrectlysolves an incorrectfunctionprovidedinpartAfor thevalueoff(3).
SampleAnswer
PartA.
f(x) = -5x^2+80
Part B.
35 meters
Part C.
Ahmedisincorrectbecause the timehas to start at0. Youcannot haveanegativetime.Thecorrectdomainis0<= x<= 4.
Question 2
ScoringRubric
3 / Student response includes each of the following 3 elements:
1 reasoning point: Correct and complete explanation ofShannon'scalculationerror
1 reasoning point: Correct and complete explanation ofJermaine'scalculationerror
1 computation point: Correct calculation of 7 and -1 as solutions to Shannon’s equation and 6 and -6 as solutions to Jermaine’s equation
2 / Student response includes 2 of the 3 elements
1 / Student response includes 1 of the 3 elements
0 / Thestudent’sresponseisincorrect,irrelevant totheskill or conceptbeing measured,or blank
SampleAnswer
Shannon
CorrectSolutions:x=7orx= -1
Explanation oferror:
Shannon'serror isafter step4.Sheshould haveseparated the equationsoutsuch that x-3=4orx-3= -4. Thensolvebothforx: x=7orx=-1.
Jermaine
CorrectSolutions:x=6orx= -6
Explanation oferror:
Jermaine'serror isafter step2.Heshouldhavetakenthesquarerootof 36 instead ofdividing it by2.Step3should be (x-6)(x+6)=0whichgives x -6=0orx+6=0.Therefore,x=6orx= -6.
Question 3
ScoringRubric
3 / Student response includes each of the following 3 elements:
1 modeling point: Correct designof420 acresof C and105 acresof D in Part A
1 modeling point: Correct designof420 acresof D and105 acresof C in Part B
1 modeling point: Correctsupportfortheir selections in Parts A and B
2 / Student response includes 2 of the 3 elements
1 / Student response includes 1 of the 3 elements
0 / Thestudent’sresponseisincorrect,irrelevant totheskill or conceptbeing measured,or blank
SampleAnswer
PartA.
Mr.Fontenotshouldplant420 acresofsoybeanC and 105 acresofsoybeanD.
525*0.8= 420
Priceper bushelA= $3.88
B= $3.62C = $3.26D = $3.40
SoybeanChasthe lowest cost per bushel to produceand therefore shouldbe plantedonthemaximum80%.SoybeanD hasthe nextlowest costper bushelto produceandshould be used forthe other
20%.
Part B.
Thedesignplan inpartAisnotthe mostprofitable 80/20design.Mr.Fontenotshouldplant420 acresofsoybeanD and 105acresofsoybean C.Based onthe numbersofbushelsper acre andthe profitper bushel,SoybeanDhas the highestlevelofprofitfor thelarger sectionof420 acres.SoybeanCgives thenexthighestprofitbased on bushels per acre andprofit,andshould be used fortheother20%.
Question 4
ScoringRubric
3 / Student response includes each of the following 3 elements:
1 reasoning point: Correct and complete explanation disprovingSarah’sstatement
1 reasoning point: Correct and complete justification of solution
1 computation point: Correct calculations disproving Sarah’s statement and providing 2 – 4iasacorrectsolution
2 / Student response includes 2 of the 3 elements
1 / Student response includes 1 of the 3 elements
0 / Thestudent’sresponseisincorrect,irrelevant totheskill or conceptbeing measured,or blank
SampleAnswer
Part A.
Substitute known solution in equation: (2 + 4i)2 + b(2 + 4i) + c = 0
Expand terms in equation and rewrite as: (-12 + 2b + c) + (16 + 4b)i = 0
Real part and imaginary part equal zero.
-12 + 2b + c = 0 and 16 + 4b = 0
Solve for b: b = -4, substitute and solve for c: c = 20
Part B.
Since the given equation has real numbers, the second root is the complex conjugate of the given root: 2 - 4i is the second solution.
Check: (2 - 4i)2 - 4 (2 - 4i) + 20
(Expand) = 4 - 16 - 16i - 8 + 16i + 20
= (4 - 16 - 8 + 20) + (-16 + 16)i = 0