1. (5 pts) Which of these graphs represent a onetoone function? Answer(s): ____________
(no explanation required.) (There may be more than one graph that qualifies.)
2. (5 pts) Convert to a logarithmic equation: 7^{x} = 2401. (no explanation required) 2. ______
A.
B.
C.
D.
3. (10 pts) Based on data about the growth of a variety of ornamental cherry trees, the following logarithmic model about these trees was determined:
h(t) = 6.47 ln(t) + 2.83, where t = age of tree in years and h (t) = height of tree, in feet.
(Note that “ln” refers to the natural log function) (explanation optional)
Using the model,
(a) At age 3 years, how tall is this type of ornamental cherry tree, to the nearest tenth of a foot?
(b) At age 12 years, how tall is this type of ornamental cherry tree, to the nearest tenth of a foot?
4. (5 pts) Solve the equation. Check all proposed solutions. Show work in solving and in checking, and state your final conclusion.
5. (10 pts)
(a) _______ (fill in the blank)
(b) Let State the exponential form of the equation.
(c) Determine the numerical value of , in simplest form. Work optional.
6. (10 pts) Let f (x) = 2x^{2} – x – 10 and g(x) = 3x + 1
(a) Find the composite function and simplify the results. Show work.
(b) Find . Show work.
7. (15 pts) Let
(a) Find f ^{–}^{ 1 }, the inverse function of f. Show work.
(b) What is the domain of f ? What is the domain of the inverse function?
(c) What is f (2) ? f (2) = ______ work/explanation optional
(d) What is f ^{–}^{ 1}( ____ ), where the number in the blank is your answer from part (c)? work/explanation optional
8. (15 pts) Let f(x) = e^{ x – 1 } + 4.
Answers can be stated without additional work/explanation.
(a) Which describes how the graph of f can be obtained from the graph of y = e^{x} ? Choice: ________
A. Shift the graph of y = e^{x } to the left by 1 unit and up by 4 units.
B. Shift the graph of y = e^{x } to the right by 1 unit and up by 4 units.
C. Reflect the graph of y = e^{x } across the xaxis and shift up by 4 units.
D. Reflect the graph of y = e^{x } across the yaxis and shift up by 4 units.
(b) What is the domain of f ?
(c) What is the range of f ?
(d) What is the horizontal asymptote?
(e) What is the yintercept? State the approximation to 2 decimal places (i.e., the nearest hundredth).
(f) Which is the graph of f ?
GRAPH A GRAPH B GRAPH C GRAPH D
NONLINEAR MODELS – For the latter part of the quiz, we will explore some nonlinear models.
9. (15 pts) QUADRATIC REGRESSION
Data: On a particular summer day, the outdoor temperature was recorded at 8 times of the day, and the following table was compiled. A scatterplot was produced and the parabola of best fit was determined.
t = Time of day (hour)

y = Outdoor
Temperature (degrees F.)

7

52

9

67

11

73

13

76

14

78

17

79

20

76

23

61



Quadratic Polynomial of Best Fit:
y = –0.3476t^{2} + 10.948t – 6.0778 where t = Time of day (hour) and y = Temperature (in degrees)
REMARKS: The times are the hours since midnight. For instance, 7 means 7 am, and 13 means 1 pm.
(a) Using algebraic techniques we have learned, find themaximum temperaturepredicted by the quadratic model and find thetime when it occurred. Report the time to the nearest quarter hour (i.e., __:00 or __:15 or __:30 or __:45). (For instance, a time of 18.25 hours is reported as 6:15 pm.) Report the maximum temperature to the nearest tenth of a degree. Show algebraic work.
(b) Use the quadratic polynomial to estimate the outdoor temperature at 7:30 am, to the nearest tenth of a degree. (work optional)
(c)Use the quadratic polynomial y = –0.3476t^{2} + 10.948t – 6.0778 together with algebra toestimate the time(s) of day when the outdoor temperature y was 75 degrees.
That is, solve the quadratic equation 75 = –0.3476t^{2} + 10.948t – 6.0778 .
Show algebraic work in solving. State your results clearly; report the time(s) to the nearest quarter hour.
10. (10 pts) EXPONENTIAL REGRESSION
Data:A cup of hot coffee was placed in a room maintained at a constant temperature of 69 degrees, and the coffee temperature was recorded periodically, in Table 1.
t = Time Elapsed
(minutes)

C = Coffee
Temperature (degrees F.)

0

166.0

10

140.5

20

125.2

30

110.3

40

104.5

50

98.4

60

93.9

TABLE 1

REMARKS:
Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees.
So, the temperature difference between the coffee temperature and the room temperature will decrease to 0.
We will fit the temperature difference data (Table 2) to an exponential curve of the form y=Ae^{–}^{bt}.
Notice that as tgets large, ywill get closer and closer to 0, which is what the temperature difference will do.
So, we want to analyze the data wheret= time elapsed andy=C – 69, the temperature difference between the coffee temperature and the room temperature.

TABLE 2
t = Time Elapsed (minutes)

y= C – 69 Temperature
Difference
(degrees F.)

0

97.0

10

71.5

20

56.2

30

41.3

40

35.5

50

29.4

60

24.9


Exponential Function of Best Fit (using the data in Table 2):
y = 89.976 e ^{–}^{ 0.023}^{t} where t = Time Elapsed (minutes) and y = Temperature Difference (in degrees)
(a) Use the exponential function to estimate the temperature difference y when 25 minutes have elapsed. Report your estimated temperature difference to the nearest tenth of a degree. (explanation/work optional)
(b) Sincey=C– 69, we have coffee temperatureC=y+ 69. Take your difference estimate from part (a) and add 69 degrees. Interpret the result by filling in the blank:
When 25 minutes have elapsed, the estimated coffee temperature is ________ degrees.
(c)Suppose the coffee temperature C is 100 degrees.Theny=C– 69 = ____ degrees is the temperature difference between the coffee and room temperatures.
(d) Consider the equation _____ = 89.976 e ^{– 0.023t} where the ____ is filled in with your answer from part (c).