Problem 1 · Functions, Transformations, and Inverses · 9 points
Undo a transformed square root Let f ( x ) = 2 x − 3 + 1 \displaystyle f(x)=2\sqrt{x-3}+1 f ( x ) = 2 x − 3 + 1
Problem diagram 2 3 4 5 6 7 8 9 10 11 12 -1 1 2 3 4 5 6 7 8 9 x y (3,1) A. State domain and range and describe transformations from the square-root parent. B. Derive the inverse and its domain. C. Verify the composition on the correct domain. Show worked solution and rubric A. The domain is [ 3 , ∞ ) \displaystyle [3,\infty) [ 3 , ∞ ) [ 1 , ∞ ) \displaystyle [1,\infty) [ 1 , ∞ ) B. From y − 1 = 2 x − 3 \displaystyle y-1=2\sqrt{x-3} y − 1 = 2 x − 3 x = ( y − 1 ) 2 4 + 3 \displaystyle x=\frac{(y-1)^2}{4}+3 x = 4 ( y − 1 ) 2 + 3 f − 1 ( x ) = ( x − 1 ) 2 4 + 3 \displaystyle f^{-1}(x)=\frac{(x-1)^2}{4}+3 f − 1 ( x ) = 4 ( x − 1 ) 2 + 3 x ≥ 1 \displaystyle x\geq1 x ≥ 1 C. f ( f − 1 ( x ) ) = 2 ( x − 1 ) 2 / 4 + 1 = ∣ x − 1 ∣ + 1 = x \displaystyle f(f^{-1}(x))=2\sqrt{(x-1)^2/4}+1=|x-1|+1=x f ( f − 1 ( x )) = 2 ( x − 1 ) 2 /4 + 1 = ∣ x − 1∣ + 1 = x x ≥ 1 \displaystyle x\geq1 x ≥ 1 Solution diagram 2 3 4 5 6 7 8 9 10 11 12 -1 1 2 3 4 5 6 7 8 9 x y (3,1) Scoring rubric 3 points: The domain is [ 3 , ∞ ) \displaystyle [3,\infty) [ 3 , ∞ ) [ 1 , ∞ ) \displaystyle [1,\infty) [ 1 , ∞ ) 3 points: From y − 1 = 2 x − 3 \displaystyle y-1=2\sqrt{x-3} y − 1 = 2 x − 3 x = ( y − 1 ) 2 4 + 3 \displaystyle x=\frac{(y-1)^2}{4}+3 x = 4 ( y − 1 ) 2 + 3 3 points: f ( f − 1 ( x ) ) = 2 ( x − 1 ) 2 / 4 + 1 = ∣ x − 1 ∣ + 1 = x \displaystyle f(f^{-1}(x))=2\sqrt{(x-1)^2/4}+1=|x-1|+1=x f ( f − 1 ( x )) = 2 ( x − 1 ) 2 /4 + 1 = ∣ x − 1∣ + 1 = x x ≥ 1 \displaystyle x\geq1 x ≥ 1 Review this unit Problem 2 · Quadratic and Polynomial Functions · 9 points
Interpret a repeated zero Let p ( x ) = ( x − 1 ) 2 ( x + 2 ) \displaystyle p(x)=(x-1)^2(x+2) p ( x ) = ( x − 1 ) 2 ( x + 2 )
Problem diagram -3 -2 -1 1 2 3 -8 -6 -4 -2 2 4 6 8 10 12 14 x y (-2,0) (1,0) A. Find the zeros, multiplicities, and y-intercept. B. Determine the sign on each interval separated by its zeros. C. State end behavior and whether the graph crosses or touches at each zero. Show worked solution and rubric A. The zeros are 1 \displaystyle 1 1 − 2 \displaystyle -2 − 2 p ( 0 ) = 2 \displaystyle p(0)=2 p ( 0 ) = 2 B. The squared factor is positive away from 1. The polynomial is negative on ( − ∞ , − 2 ) \displaystyle (-\infty,-2) ( − ∞ , − 2 ) ( − 2 , 1 ) \displaystyle (-2,1) ( − 2 , 1 ) ( 1 , ∞ ) \displaystyle (1,\infty) ( 1 , ∞ ) C. The leading term is x 3 \displaystyle x^3 x 3 p ( x ) → − ∞ \displaystyle p(x)\to-\infty p ( x ) → − ∞ x → − ∞ \displaystyle x\to-\infty x → − ∞ p ( x ) → ∞ \displaystyle p(x)\to\infty p ( x ) → ∞ x → ∞ \displaystyle x\to\infty x → ∞ − 2 \displaystyle -2 − 2 Solution diagram -3 -2 -1 1 2 3 -8 -6 -4 -2 2 4 6 8 10 12 14 x y (-2,0) (1,0) Scoring rubric 3 points: The zeros are 1 \displaystyle 1 1 − 2 \displaystyle -2 − 2 3 points: The squared factor is positive away from 1 3 points: The leading term is x 3 \displaystyle x^3 x 3 p ( x ) → − ∞ \displaystyle p(x)\to-\infty p ( x ) → − ∞ x → − ∞ \displaystyle x\to-\infty x → − ∞ p ( x ) → ∞ \displaystyle p(x)\to\infty p ( x ) → ∞ x → ∞ \displaystyle x\to\infty x → ∞ Review this unit Problem 3 · Radical and Rational Functions · 9 points
Separate a hole from an asymptote Let r ( x ) = x 2 − 1 x 2 − x − 2 \displaystyle r(x)=\frac{x^2-1}{x^2-x-2} r ( x ) = x 2 − x − 2 x 2 − 1
Problem diagram -4 -3 -2 -1 1 2 3 4 5 6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y hole A. State the original domain and simplify where allowed. B. Locate the hole and all vertical and horizontal asymptotes. C. Solve r(x) = 0 and explain why a canceled factor is still excluded. Show worked solution and rubric A. Factor to get ( x − 1 ) ( x + 1 ) ( x − 2 ) ( x + 1 ) = x − 1 x − 2 \displaystyle \frac{(x-1)(x+1)}{(x-2)(x+1)}=\frac{x-1}{x-2} ( x − 2 ) ( x + 1 ) ( x − 1 ) ( x + 1 ) = x − 2 x − 1 x ≠ − 1 , 2 \displaystyle x\ne-1,2 x = − 1 , 2 B. At x = − 1 \displaystyle x=-1 x = − 1 2 3 \displaystyle \frac23 3 2 ( − 1 , 2 3 ) \displaystyle (-1,\frac23) ( − 1 , 3 2 ) x = 2 \displaystyle x=2 x = 2 y = 1 \displaystyle y=1 y = 1 C. Only x = 1 \displaystyle x=1 x = 1 Solution diagram -4 -3 -2 -1 1 2 3 4 5 6 -5 -4 -3 -2 -1 1 2 3 4 5 6 x y hole Scoring rubric 3 points: Factor to get ( x − 1 ) ( x + 1 ) ( x − 2 ) ( x + 1 ) = x − 1 x − 2 \displaystyle \frac{(x-1)(x+1)}{(x-2)(x+1)}=\frac{x-1}{x-2} ( x − 2 ) ( x + 1 ) ( x − 1 ) ( x + 1 ) = x − 2 x − 1 x ≠ − 1 , 2 \displaystyle x\ne-1,2 x = − 1 , 2 3 points: At x = − 1 \displaystyle x=-1 x = − 1 2 3 \displaystyle \frac23 3 2 ( − 1 , 2 3 ) \displaystyle (-1,\frac23) ( − 1 , 3 2 ) 3 points: Only x = 1 \displaystyle x=1 x = 1 Review this unit Problem 4 · Exponential and Logarithmic Functions · 9 points
Model a growing culture A culture has P ( t ) = 500 ⋅ 2 t / 3 \displaystyle P(t)=500\cdot2^{t/3} P ( t ) = 500 ⋅ 2 t /3 t ≥ 0 \displaystyle t\geq0 t ≥ 0
Problem diagram 1 2 3 4 5 6 7 8 9 500 1000 1500 2000 2500 3000 3500 4000 4500 x y 500 1000 2000 A. Interpret the initial value and doubling time. B. Solve exactly for the time at which there are 3000 cells. C. Write the inverse and state its domain for this physical model. Show worked solution and rubric A. P ( 0 ) = 500 \displaystyle P(0)=500 P ( 0 ) = 500 t \displaystyle t t B. 2 t / 3 = 6 \displaystyle 2^{t/3}=6 2 t /3 = 6 t = 3 log 2 6 = 3 ln 6 ln 2 \displaystyle t=3\log_2 6=\frac{3\ln6}{\ln2} t = 3 log 2 6 = ln 2 3 ln 6 C. t = 3 log 2 ( P / 500 ) \displaystyle t=3\log_2(P/500) t = 3 log 2 ( P /500 ) P ≥ 500 \displaystyle P\geq500 P ≥ 500 t ≥ 0 \displaystyle t\geq0 t ≥ 0 Solution diagram 1 2 3 4 5 6 7 8 9 500 1000 1500 2000 2500 3000 3500 4000 4500 x y 500 1000 2000 Scoring rubric 3 points: P ( 0 ) = 500 \displaystyle P(0)=500 P ( 0 ) = 500 3 points: 2 t / 3 = 6 \displaystyle 2^{t/3}=6 2 t /3 = 6 t = 3 log 2 6 = 3 ln 6 ln 2 \displaystyle t=3\log_2 6=\frac{3\ln6}{\ln2} t = 3 log 2 6 = ln 2 3 ln 6 3 points: t = 3 log 2 ( P / 500 ) \displaystyle t=3\log_2(P/500) t = 3 log 2 ( P /500 ) P ≥ 500 \displaystyle P\geq500 P ≥ 500 Review this unit Problem 5 · Sequences and Series · 9 points
Track a geometric savings plan A student deposits 100 dollars in week 1 and increases the weekly deposit by 10% each week. Ignore interest.
A. Write explicit and recursive formulas for the weekly deposit. B. Find the total deposited over the first four weeks. C. Explain whether an infinite sum exists for this model. Show worked solution and rubric A. a n = 100 ( 1.1 ) n − 1 \displaystyle a_n=100(1.1)^{n-1} a n = 100 ( 1.1 ) n − 1 a 1 = 100 \displaystyle a_1=100 a 1 = 100 a n = 1.1 a n − 1 \displaystyle a_n=1.1a_{n-1} a n = 1.1 a n − 1 n ≥ 2 \displaystyle n\geq2 n ≥ 2 B. S 4 = 100 1.1 4 − 1 1.1 − 1 = 464.1 \displaystyle S_4=100\frac{1.1^4-1}{1.1-1}=464.1 S 4 = 100 1.1 − 1 1. 1 4 − 1 = 464.1 100 + 110 + 121 + 133.1 \displaystyle 100+110+121+133.1 100 + 110 + 121 + 133.1 C. The infinite series diverges because the ratio 1.1 has magnitude greater than 1 and its terms do not approach zero.Scoring rubric 3 points: a n = 100 ( 1.1 ) n − 1 \displaystyle a_n=100(1.1)^{n-1} a n = 100 ( 1.1 ) n − 1 a 1 = 100 \displaystyle a_1=100 a 1 = 100 a n = 1.1 a n − 1 \displaystyle a_n=1.1a_{n-1} a n = 1.1 a n − 1 n ≥ 2 \displaystyle n\geq2 n ≥ 2 3 points: S 4 = 100 1.1 4 − 1 1.1 − 1 = 464.1 \displaystyle S_4=100\frac{1.1^4-1}{1.1-1}=464.1 S 4 = 100 1.1 − 1 1. 1 4 − 1 = 464.1 3 points: The infinite series diverges because the ratio 1.1 has magnitude greater than 1 and its terms do not approach zero. Review this unit Problem 6 · Modeling and Analytical Reasoning · 9 points
Choose a model and test it A data table gives ( t , P ) = ( 0 , 200 ) , ( 1 , 300 ) , ( 2 , 450 ) \displaystyle (t,P)=(0,200),(1,300),(2,450) ( t , P ) = ( 0 , 200 ) , ( 1 , 300 ) , ( 2 , 450 )
A. Compare first differences and ratios to select an exact model for these data. B. Predict P at t = 4 and solve when P = 1000. C. State one limitation and describe an observation that would challenge the model. Show worked solution and rubric A. First differences are 100 and 150, while both ratios equal 1.5. The exponential model P ( t ) = 200 ( 1.5 ) t \displaystyle P(t)=200(1.5)^t P ( t ) = 200 ( 1.5 ) t B. P ( 4 ) = 1012.5 \displaystyle P(4)=1012.5 P ( 4 ) = 1012.5 P = 1000 \displaystyle P=1000 P = 1000 t = ln 5 ln 1.5 ≈ 3.969 \displaystyle t=\frac{\ln5}{\ln1.5}\approx3.969 t = ln 1.5 ln 5 ≈ 3.969 C. Three points do not establish indefinite exponential growth. A resource limit or later measurements with consistently different ratios would challenge the model; a noninteger prediction may represent an expected count.Scoring rubric 3 points: First differences are 100 and 150, while both ratios equal 1.5 3 points: P ( 4 ) = 1012.5 \displaystyle P(4)=1012.5 P ( 4 ) = 1012.5 3 points: Three points do not establish indefinite exponential growth Review this unit 