Week 1

1. Given the function f (x) = x2 − 2Mx , where M is someparameter, find f (3).Answer: f (3) = 9− 6M

2. Find the domain of the function f (x) =1

x − 5.

Answer: x 6= 5

3. Find the slope of the line connecting the points (2, 3) and(4, 8).

Answer: m =y0 − y1x0 − x1

=3− 8

2− 4=−5

−2=

5

2

Week 2

1. The function f (x) = .02x + .01 models the blood alcoholpercent of a 180 pound male as a function of the number ofdrinks he has had (where x is the number of drinks). Whatcould you predict about the slope of a function that describesthe BAC for a 120 pound male in comparison to the modelabove?Answer: The slope would be greater since a person whoweighs less would probably be affected more by each drink.

2. What is the slope, m, and the y -intercept, b, of the line2y + 6x = 8?Answer: 2y = −6x + 8, so y = −3x + 4 and m = −3, b = 4.

Week 2, cont.

3. The number of cell phone subscribers (in millions) between1995 and 1999 can be modeled by N(x) = 11.75x + 32.95,where x is the number of years after 1995. Which of thefollowing is true?

A) N(x) is a linear function.B) N(0) represents the number of subscribers in 1995, in millions.C) There were 32.95 subscribers in 1995.D) All of the above.E) Both (A) and (B).

Answer: (E), both (A) and (B) are true, but (C) is falsebecause there were 32.95 million subscribers in 1995.

Week 3

1. Solve the following inequality: −3(x + 4) > 53.

Answer: x + 4 <53

−3so x <

−65

32. A jeweler sells necklaces with a revenue function (in $) of

R(x) = 90.75x and a cost function (in $) ofC (x) = 24.50x + 4770, where x is the number of necklacessold. How many necklaces must she sell in order to breakeven?Answer: Set R(x) = C (x)90.75x = 24.50x + 477066.25x = 4770 so x = 72 necklaces

Week 3, cont.

3. Solve |3x − 6| > 9Answer: First solve |3x − 6| = 9. Either 3x − 6 = 9 andx = 5, or −(3x − 6) = 9 and x = −1.Now plug in other numbers from the number line.

x −1 5

Plugging in x = −2, 0 and 6, we see|3(−2)− 6| = | − 12| = 12 > 9|3(0)− 6| = | − 6| = 6 6> 9|6(6)− 6| = 30 > 9So the final answer is x > 5 or x < −1.

Week 4

1. Given the quadratic function f (x) = x2 + 2x + 5 what is thex-coordinate of the vertex of the graph of y = f (x).Answer: Use completing the square.f (x) = x2 + 2x + 1− 1 + 5 = (x + 1)2 + 4 so the vertex is(−1, 4) and the x-coordinate is h = −1.

2. Find the roots of f (x) = x2 + 9x + 8.Answer: x2 + 9x + 8 = 0(x + 8)(x + 1) = 0x = −1, −8

3. Find the roots of f (x) = x2 + 4x + 2 and simplify your answer!Answer: Quadratic formula:−4±

√42 − 4(1)(2)

2(1)=−4± 2

√2

2= −2±

√2

Week 5

1. Given f (x) = 3x − 5 and g(x) = x2 + 5, find f (x)g(x).(Simplify your answer.)Answer: (3x − 5)(x2 + 5) = 3x3 − 5x2 + 15x − 25

2. Find the inverse of the function f (x) = 3x − 5.Answer: Solve x = 3y − 5

y =x + 5

3.

Week 5, cont.

3. Based on the table below, find f (g(2)).

x f (x) g(x)

−2 −3 1

−1 2 0

0 1 −1

1 −2 −3

2 0 −2

Answer: Since g(2) = −2, f (g(2)) = f (−2) = −3.

Week 6

1. Find g(4) if f (x) = x2 − 3x and g(x) = f (x − 2).Answer: g(4) = f (4− 2) = f (2) = −2

2. Find f (3) if f (x) = 2.456x0.245. Round your answer to thenearest thousandth.Answer: 3.215

3. Find f (3) if f (x) is the piecewise function

f (x) =

{2x + 1, if x > 5

x − 5, if x ≤ 5

Answer: 3 ≤ 5 so f (3) = 3− 5 = −2

Week 7

1. Solve the quadratic inequality x2 − 5x < 6.Answer: First solve x2 − 5x − 6 = 0.(x − 6)(x + 1) = 0 so x = 6,−1.Now plug in other numbers from the number line.

x −1 6

Plugging in x = −2, 0 and 7, we see(−2)2 − 5(−2) = 14 6< 602 − 5(0) = 0 < 672 − 5(7) = 14 6< 6So the final answer is −1 < x < 6.

Week 7, cont.

2. Solve the quadratic inequality (x − 3)(x + 4) > 0.Answer: First solve (x − 3)(x + 4) = 0, so x = 3,−4

x −4 3

Plugging in x = −5, 0 and 4, we see(−5− 3)(−5 + 4) = 8 > 0(0− 3)(0 + 4) = −12 6> 0(4− 3)(4 + 4) = 8 > 0So the final answer is x < −4 or x > 3.

3. Solve x =√x + 2. Remember to CHECK your potential

answers!Answer: Square both sides so x2 = x + 2x2 − x − 2 = 0(x − 2)(x + 1) = 0 so the potential answers are x = 2,−1CHECK:

√2 + 2 = 2 but

√−1 + 2 = 1 6= −1, so the final

answer is x = 2.

Week 8

1. Find the horizontal asymptote of the function f (x) = 3x − 4.Answer: y = −4

2. Which of the following statements are true for the functionf (x) = 3x4?

A) As x approaches ∞, f (x) approaches 0.B) As x approaches −∞, f (x) approaches 0.C) As x approaches ∞, f (x) approaches −4.D) As x approaches −∞, f (x) approaches −4.E) None of the above

Answer: (D)

3. Find the domain of f (x) = log2(5− 4x)Answer: 5− 4x > 0, so x < 5

4 .

Week 9

1. Solve ex = 5.Answer: Take ln of both sides to get x = ln(5).

2. Convert the following logarithmic equation into an exponentialequation: y = ln(x + 2)Answer: ey = x + 2

3. Expand log(x2y) using the properties of logarithms.Answer: By the multiplication rule,log(x2y) = log(x2) + log(y). Then by the exponent rule weget the final answer, 2 log(x) + log(y).

Week 10

1. What are the degree, leading coefficient and constant term ofthe polynomial y = 5x4 − 3x2 + 2x − 7?Answer: Degree = 4, leading coefficient = 5 and constantterm = −7.

2. Consider the polynomial f (x) = −7x4 − 5x . What can yousay about the number of turning points for f (x)?Answer: f (x) has either 3 or 1 turning point(s).

Week 10, cont.

3. Consider the graph of the polynomial y = f (x) shown below,and assume that everything“interesting” can be seen on thegraph. Which of the following is true about f (x)?

y = f (x)

A) Degree is even and leading coefficient is positive.B) Degree is even and leading coefficient is negative.C) Degree is odd and leading coefficient is positive.D) Degree is odd and leading coefficient is negative.

Answer:(C)

Week 11

1. Which of the following is a linear factor off (x) = x3 − 6x2 + 21x − 26?

A) x − 2B) x + 2C) xD) Both (A) and (B).E) All of the above.

Answer: f (2) = 0, f (−2) = −100 and f (0) = −26 so x = 2is the only one of these which is a root of f (x), and x − 2 isthe only linear factor.

Week 11, cont.

2. Consider the cubic polynomial f (x) = x3 − 6x2 + 21x − 26.Since x − 2 is a linear factor of f (x), f (x) = (x − 2)p(x).Find the polynomial p(x).

Answer:

1 −6 21 −262 2 −8 26

1 −4 13 0

p(x) = x2 − 4x + 13

3. Which of the following is a polynomial with a single root atx = −4 and a double root at x = 7?

A) y = (x − 4)(x + 7)2

B) y = (x + 4)(x − 7)(x − 7)C) y = 12(x + 4)(x − 7)2

D) Both (A) ad (B).E) Both (B) and (C).

Answer: (E)

Week 12

1. What are the poles of the rational function

f (x) =2x + 1

x2 + 4x + 3?

Answer: Set the denominator x2 + 4x + 3 = 0(x + 3)(x + 1) = 0 so the poles are x = −3,−1.

2. Find the horizontal or slope asymptote of the rational function

f (x) =x2 − 5x + 4

2x2 + 4x + 3.

Answer: Since the degree of the top and bottom are thesame, we divide the leading coefficients and f (x) has ahorizontal asymptote at y = 1

2 .

Week 12, cont.

3. What are the zeros of the function f (x) =x2 − 5x + 4

x2 + 6x − 7?

Answer: f (x) =(x − 4)(x − 1)

(x + 7)(x − 1)The potential zeros of f (x) are the zeros of the numerator,x = 4, 1. But x = 1 is also a zero of the denominator, so it isnot in the domain of f (x). So the final answer is x = 4.

Week 13

1. Find a system of equations corresponding to the augmentedmatrix below. [

2 3 5−1 2 1

]Answer: 2x + 3y = 5 and −x + 2y = 1

2. What is the size of the matrix below? 2 3−1 2

0 7

Answer: 3x2

Week 13, cont.

3. Solve the system of equations corresponding to the followingmatrix, assuming that the first column corresponds to x andthe second column corresponds to y .[

5 3 40 2 6

]Answer: 5x + 3y = 4 and 2y = 6.Solving the second equation, y = 3.Plugging this into the first equation, 5x + 9 = 4, so x = −1.

Week 14

Thanksgiving

Week 15

1. Find the product[

2 3 5] 2

34

.

Answer: 2(2) + 3(3) + 5(4) = 33

2. What is the system of equations that corresponds to thefollowing matrix equation?[

5 30 2

] [xy

]=

[5−1

]Answer: 5x + 3y = 5, 2y = −1

3. Find the inverse of

[1 41 3

].

Answer:

1

1(3)− 4(1)

[3 −4−1 1

]=

[−3 4

1 −1

]