Math133 Unit 5 Individual Project Version a

Student Answer and Work Form Unit 5 Ver. A

Student Name (required): _______________________________

1a. Answer: Shifted 5 units to the right

x = 5 is the vertical asymptote

(6, 0) is the x-intercept.

_______________

1a. Key work steps

g(x) = log(x – 5)

Shifted 5 units to the right

x – 5 = 0

x = 0 + 5

x = 5

x = 5 is the vertical asymptote.

Let g(x) = y

y = log(x – 5)

0 = log(x – 5)

x – 5 = 100

x – 5 = 1

x = 1 + 5

x = 6

x –intercept is (6, 0)

1b. Answer: Reflected about x-axis and vertical shift 2 units.

x = 0 is the vertical asymptote

(100, 0) is the x-intercept

_______________

1b. Key work steps

g(x) = -log(x) + 2

Reflected about x-axis and vertical shift 2 units.

x = 0 is the vertical asymptote.

Let g(x) = y

y = -log(x) + 2

0 = -log(x) + 2

Log(x) = 2

x = 102

x = 100

(100, 0) is the x-intercept.

2.a. Answer: 68% _______________

2a. Key work steps

S(t) = 68 − 20 log (t + 1), t ≥ 0

Substitute t = 0

S(0) = 68 − 20 log (0 + 1)

= 68 − 20 log (1)

= 68 – 20(0)

= 68 – 0

= 68

The average score when they initially took the test, t = 0 was 68%.

2b.Answer: 14%_______________

2b. Key work steps

S(t) = 68 − 20 log (t + 1), t ≥ 0

Substitute t = 14

S(14) = 68 − 20 log (14 + 1)

= 68 − 20 log (15)

= 68 – 20(2.70805)

= 68 – 54.161

= 13.839

= 14%

The average score after 14 months is 14% to the nearest whole number.

2c.Answer: 24 months _______________

2c. Key work steps

Average score = 40%

S(t) = 68 − 20 log (t + 1)

68 − 20 log (t + 1) = 40

− 20 log (t + 1) = 40 – 68

− 20 log (t + 1) = - 28

log (t + 1) = -28/-20

log (t + 1) = 7/5

t + 1= 10^(7/5)

t = 10^(7/5) – 1

t = 24.11886

t = 24

After 24 months, the average score was 40%.

3a. Answer: $4533.21 _______________

3a. Key work steps

P = $3000, t = 6 years, r =...