Problem 1 · 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 2 · 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 3 · 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 4 · Unit Circle and Trigonometric Functions · 9 points
Locate an exact angle An angle is θ = 5 π 6 \displaystyle \theta=\frac{5\pi}{6} θ = 6 5 π
A. Convert to degrees and identify the reference angle and quadrant. B. Find sine, cosine, and tangent exactly. C. Give all angles in [0, 2π) with the same sine. Show worked solution and rubric A. θ = 150 ∘ \displaystyle \theta=150^\circ θ = 15 0 ∘ π 6 \displaystyle \frac\pi6 6 π B. sin θ = 1 2 \displaystyle \sin\theta=\frac12 sin θ = 2 1 cos θ = − 3 2 \displaystyle \cos\theta=-\frac{\sqrt3}{2} cos θ = − 2 3 tan θ = − 3 3 \displaystyle \tan\theta=-\frac{\sqrt3}{3} tan θ = − 3 3 C. The angles are π 6 \displaystyle \frac\pi6 6 π 5 π 6 \displaystyle \frac{5\pi}6 6 5 π Scoring rubric 3 points: θ = 150 ∘ \displaystyle \theta=150^\circ θ = 15 0 ∘ π 6 \displaystyle \frac\pi6 6 π 3 points: sin θ = 1 2 \displaystyle \sin\theta=\frac12 sin θ = 2 1 cos θ = − 3 2 \displaystyle \cos\theta=-\frac{\sqrt3}{2} cos θ = − 2 3 tan θ = − 3 3 \displaystyle \tan\theta=-\frac{\sqrt3}{3} tan θ = − 3 3 3 points: The angles are π 6 \displaystyle \frac\pi6 6 π 5 π 6 \displaystyle \frac{5\pi}6 6 5 π Review this unit Problem 5 · Trigonometric Graphs and Models · 9 points
Model a Ferris wheel A rider starts at the lowest point, 2 meters above ground. The wheel radius is 10 meters and one revolution takes 40 seconds. Let t be seconds after starting.
Problem diagram 5 10 15 20 25 30 35 40 2 4 6 8 10 12 14 16 18 20 22 24 x y A. Write a height model and explain each parameter. B. Find the height at t = 10 and all times in the first revolution at height 12. C. State the domain and range for one revolution and sketch the cycle. Show worked solution and rubric A. h ( t ) = 12 − 10 cos ( π 20 t ) \displaystyle h(t)=12-10\cos(\frac\pi{20}t) h ( t ) = 12 − 10 cos ( 20 π t ) 2 π 40 = π 20 \displaystyle \frac{2\pi}{40}=\frac\pi{20} 40 2 π = 20 π B. h ( 10 ) = 12 \displaystyle h(10)=12 h ( 10 ) = 12 cos ( π 20 t ) = 0 \displaystyle \cos(\frac\pi{20}t)=0 cos ( 20 π t ) = 0 t = 10 , 30 \displaystyle t=10,30 t = 10 , 30 C. For one revolution, 0 ≤ t ≤ 40 \displaystyle 0\leq t\leq40 0 ≤ t ≤ 40 2 ≤ h ≤ 22 \displaystyle 2\leq h\leq22 2 ≤ h ≤ 22 ( 0 , 2 ) , ( 10 , 12 ) , ( 20 , 22 ) , ( 30 , 12 ) , ( 40 , 2 ) \displaystyle (0,2),(10,12),(20,22),(30,12),(40,2) ( 0 , 2 ) , ( 10 , 12 ) , ( 20 , 22 ) , ( 30 , 12 ) , ( 40 , 2 ) Solution diagram 5 10 15 20 25 30 35 40 2 4 6 8 10 12 14 16 18 20 22 24 x y Scoring rubric 3 points: h ( t ) = 12 − 10 cos ( π 20 t ) \displaystyle h(t)=12-10\cos(\frac\pi{20}t) h ( t ) = 12 − 10 cos ( 20 π t ) 3 points: h ( 10 ) = 12 \displaystyle h(10)=12 h ( 10 ) = 12 3 points: For one revolution, 0 ≤ t ≤ 40 \displaystyle 0\leq t\leq40 0 ≤ t ≤ 40 2 ≤ h ≤ 22 \displaystyle 2\leq h\leq22 2 ≤ h ≤ 22 Review this unit Problem 6 · Identities and Trigonometric Equations · 9 points
Solve without losing roots Solve 2 sin 2 x − sin x = 0 \displaystyle 2\sin^2x-\sin x=0 2 sin 2 x − sin x = 0 0 ≤ x < 2 π \displaystyle 0\leq x<2\pi 0 ≤ x < 2 π
A. Factor and find all solutions. B. Explain why dividing by sin(x) loses solutions. C. Verify the identity (1-cos²x)/sin(x) = sin(x), stating its domain. Show worked solution and rubric A. sin x ( 2 sin x − 1 ) = 0 \displaystyle \sin x(2\sin x-1)=0 sin x ( 2 sin x − 1 ) = 0 x = 0 , π , π 6 , 5 π 6 \displaystyle x=0,\pi,\frac\pi6,\frac{5\pi}6 x = 0 , π , 6 π , 6 5 π 2 π \displaystyle 2\pi 2 π B. Dividing by sin x \displaystyle \sin x sin x x = 0 , π \displaystyle x=0,\pi x = 0 , π C. Using 1 − cos 2 x = sin 2 x \displaystyle 1-\cos^2x=\sin^2x 1 − cos 2 x = sin 2 x sin 2 x / sin x = sin x \displaystyle \sin^2x/\sin x=\sin x sin 2 x / sin x = sin x sin x ≠ 0 \displaystyle \sin x\ne0 sin x = 0 x ≠ k π \displaystyle x\ne k\pi x = k π k \displaystyle k k Scoring rubric 3 points: sin x ( 2 sin x − 1 ) = 0 \displaystyle \sin x(2\sin x-1)=0 sin x ( 2 sin x − 1 ) = 0 3 points: Dividing by sin x \displaystyle \sin x sin x x = 0 , π \displaystyle x=0,\pi x = 0 , π 3 points: Using 1 − cos 2 x = sin 2 x \displaystyle 1-\cos^2x=\sin^2x 1 − cos 2 x = sin 2 x sin 2 x / sin x = sin x \displaystyle \sin^2x/\sin x=\sin x sin 2 x / sin x = sin x sin x ≠ 0 \displaystyle \sin x\ne0 sin x = 0 x ≠ k π \displaystyle x\ne k\pi x = k π k \displaystyle k k Review this unit Problem 7 · Analytic and Applied Trigonometry · 9 points
Survey an oblique triangle Triangle ABC has sides a = 7 \displaystyle a=7 a = 7 b = 10 \displaystyle b=10 b = 10 C = 60 ∘ \displaystyle C=60^\circ C = 6 0 ∘
A. Find side c exactly using a named theorem. B. Find the area exactly. C. Explain why these data determine one triangle up to reflection, unlike some side-side-angle data. Show worked solution and rubric A. Law of Cosines: c 2 = 7 2 + 10 2 − 2 ( 7 ) ( 10 ) cos 60 ∘ = 79 \displaystyle c^2=7^2+10^2-2(7)(10)\cos60^\circ=79 c 2 = 7 2 + 1 0 2 − 2 ( 7 ) ( 10 ) cos 6 0 ∘ = 79 c = 79 \displaystyle c=\sqrt{79} c = 79 B. K = 1 2 a b sin C = 35 sin 60 ∘ = 35 3 2 \displaystyle K=\frac12ab\sin C=35\sin60^\circ=\frac{35\sqrt3}{2} K = 2 1 ab sin C = 35 sin 6 0 ∘ = 2 35 3 C. Two sides and the included angle give SAS uniqueness. A nonincluded angle can admit two placements with different third angles, which is the SSA ambiguity.Scoring rubric 3 points: Law of Cosines: c 2 = 7 2 + 10 2 − 2 ( 7 ) ( 10 ) cos 60 ∘ = 79 \displaystyle c^2=7^2+10^2-2(7)(10)\cos60^\circ=79 c 2 = 7 2 + 1 0 2 − 2 ( 7 ) ( 10 ) cos 6 0 ∘ = 79 c = 79 \displaystyle c=\sqrt{79} c = 79 3 points: K = 1 2 a b sin C = 35 sin 60 ∘ = 35 3 2 \displaystyle K=\frac12ab\sin C=35\sin60^\circ=\frac{35\sqrt3}{2} K = 2 1 ab sin C = 35 sin 6 0 ∘ = 2 35 3 3 points: Two sides and the included angle give SAS uniqueness Review this unit Problem 8 · Polar Coordinates · 9 points
Compare polar representations Point P has polar coordinates ( 4 , 2 π 3 ) \displaystyle (4,\frac{2\pi}{3}) ( 4 , 3 2 π )
A. Convert P to rectangular coordinates. B. Give one representation with negative radius. C. Convert r = 4cos(θ) to rectangular form and identify its center and radius. Show worked solution and rubric A. x = 4 cos ( 2 π 3 ) = − 2 \displaystyle x=4\cos(\frac{2\pi}3)=-2 x = 4 cos ( 3 2 π ) = − 2 y = 4 sin ( 2 π 3 ) = 2 3 \displaystyle y=4\sin(\frac{2\pi}3)=2\sqrt3 y = 4 sin ( 3 2 π ) = 2 3 B. ( − 4 , 5 π 3 ) \displaystyle (-4,\frac{5\pi}3) ( − 4 , 3 5 π ) π \displaystyle \pi π C. Multiplying by r \displaystyle r r r 2 = 4 r cos θ \displaystyle r^2=4r\cos\theta r 2 = 4 r cos θ x 2 + y 2 = 4 x \displaystyle x^2+y^2=4x x 2 + y 2 = 4 x ( x − 2 ) 2 + y 2 = 4 \displaystyle (x-2)^2+y^2=4 ( x − 2 ) 2 + y 2 = 4 ( 2 , 0 ) \displaystyle (2,0) ( 2 , 0 ) Scoring rubric 3 points: x = 4 cos ( 2 π 3 ) = − 2 \displaystyle x=4\cos(\frac{2\pi}3)=-2 x = 4 cos ( 3 2 π ) = − 2 y = 4 sin ( 2 π 3 ) = 2 3 \displaystyle y=4\sin(\frac{2\pi}3)=2\sqrt3 y = 4 sin ( 3 2 π ) = 2 3 3 points: ( − 4 , 5 π 3 ) \displaystyle (-4,\frac{5\pi}3) ( − 4 , 3 5 π ) π \displaystyle \pi π 3 points: Multiplying by r \displaystyle r r r 2 = 4 r cos θ \displaystyle r^2=4r\cos\theta r 2 = 4 r cos θ x 2 + y 2 = 4 x \displaystyle x^2+y^2=4x x 2 + y 2 = 4 x ( x − 2 ) 2 + y 2 = 4 \displaystyle (x-2)^2+y^2=4 ( x − 2 ) 2 + y 2 = 4 Review this unit Problem 9 · Sequences, Series, and Integrated Modeling · 9 points
Analyze a decaying dose An initial 80 mg dose loses 25% of its amount each hour. No new dose is taken.
A. Write a recurrence and explicit formula for the amount after n whole hours. B. Find the first whole hour at which the amount is below 20 mg. C. Evaluate the infinite sum of hourly sampled amounts and explain its units. Show worked solution and rubric A. a 0 = 80 \displaystyle a_0=80 a 0 = 80 a n + 1 = 0.75 a n \displaystyle a_{n+1}=0.75a_n a n + 1 = 0.75 a n a n = 80 ( 0.75 ) n \displaystyle a_n=80(0.75)^n a n = 80 ( 0.75 ) n B. 80 ( 0.75 ) n < 20 \displaystyle 80(0.75)^n<20 80 ( 0.75 ) n < 20 n > ln ( 1 / 4 ) ln ( 3 / 4 ) ≈ 4.819 \displaystyle n>\frac{\ln(1/4)}{\ln(3/4)}\approx4.819 n > ln ( 3/4 ) ln ( 1/4 ) ≈ 4.819 a 4 = 25.3125 \displaystyle a_4=25.3125 a 4 = 25.3125 a 5 = 18.984375 \displaystyle a_5=18.984375 a 5 = 18.984375 C. ∑ n = 0 ∞ 80 ( 0.75 ) n = 320 \displaystyle \sum_{n=0}^{\infty}80(0.75)^n=320 n = 0 ∑ ∞ 80 ( 0.75 ) n = 320 Scoring rubric 3 points: a 0 = 80 \displaystyle a_0=80 a 0 = 80 a n + 1 = 0.75 a n \displaystyle a_{n+1}=0.75a_n a n + 1 = 0.75 a n a n = 80 ( 0.75 ) n \displaystyle a_n=80(0.75)^n a n = 80 ( 0.75 ) n 3 points: 80 ( 0.75 ) n < 20 \displaystyle 80(0.75)^n<20 80 ( 0.75 ) n < 20 n > ln ( 1 / 4 ) ln ( 3 / 4 ) ≈ 4.819 \displaystyle n>\frac{\ln(1/4)}{\ln(3/4)}\approx4.819 n > ln ( 3/4 ) ln ( 1/4 ) ≈ 4.819 3 points: ∑ n = 0 ∞ 80 ( 0.75 ) n = 320 \displaystyle \sum_{n=0}^{\infty}80(0.75)^n=320 n = 0 ∑ ∞ 80 ( 0.75 ) n = 320 Review this unit 