College Algebra MA 134
10 pts
Midterm Exam-Take Home Portion
100 Points
1
1. (E04a) Graph y = - x3 + 4 You may use your graphing calculator to aid
3
you. However, you must (a) give the domain in interval notation, (b) show
a table or t-chart , (c) plot at least 5 points and (d) label any points where
the graph crosses the x or y axis.
(a) D: (-∞,∞)
(c)
(b) Table
)
(d)
10 pts
2. (E04b) Graph f(x) = √−𝑥 + 2 You may use your graphing calculator to aid
you. However, you must (a) give the domain in interval notation, (b) show
a table or t-chart , (c) plot at least 5 points and (d) label any points where
the graph crosses the x or y axis.
(a) D: (- ∞, 0]
(c)
)
(d) Y-Intercept: (0,2)
(b) Table of Values
College Algebra MA 134
10 pts
Midterm Exam-Take Home Portion
100 Points
3. (E11h) Graph y2 = -12x You may use your graphing calculator to aid you.
However, you must (a) give the domain in interval notation, (b) show a
table or t-chart , (c) plot at least 5 points and (d) label any points where the
graph crosses the x or y axis.
(a) D: (- ∞, 0]
𝑇𝑜 𝑔𝑟𝑎𝑝ℎ, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑦: 𝑌 = +/−√−12𝑥
(c)
)
(b) Table of Values
x
0
-1
-2
-3
-4
-5
-6
y
0.0
(+/-)3.5
(+/-)4.9
(+/-)6.0
(+/-)6.9
(+/-)7.7
(+/-)8.5
(d) x and y –int at (0,0)
10 pts
4. (E10a) Graph f(x) = 3x2 + 6x + 1 You may use your graphing calculator to aid
you. However, you must (a) give the domain in interval notation, (b) show
a table or t-chart , (c) plot at least 5 points and (d) label any points where
the graph crosses the x or y axis AND (e) give the coordinates of the
maximum or minimum point.
(a) D: (-∞, ∞)
(b) Table of Values
(c)
)
(d) Y-Intercept (0,1) X-Intercepts (-1.8, 0), (-0.2, 0) {approximate}
(e) X = -b/2a = -6/2(3) = -1 and y= 3(-1)2 +6(-1) +1 = -1 Min at (-1,-2)
College Algebra MA 134
0 pts
Midterm Exam-Take Home Portion
5. (E10a) Use the Quadratic Formula to find the exact values of the xintercepts for the function in #4. (I did not grade this-original instructions were
wrong)
𝑥=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−6 ± √4(6)
6
=
=
−6 ± √ 62 − 4(3)(1)
−6 ± 2√(6)
6
=
2(3)
−3 ± √(6)
3
=
= -1 ±
−6 ± √36−12
6
√6
√6
3
=
−6 ± √24
6
3
X-Intercepts at (-1 + , 0) and (-1 10 pts
100 Points
√6
,
3
0)
6. (E10b) For the equation g(x) = -0.002x2 – 6x + 100, (a) tell if this function has
a maximum or minimum point and (b) Calculate the Max/Min pt. (Use the
−𝑏
−𝑏
fact that ( , f( )) can define this point for you.)
2𝑎
2𝑎
(a) Has a maximum point since is a quadratic and leading term is (-)
(b) Xv =
−𝑏
2𝑎
=
−(−6)
2(−0.002)
=
6
−0.004
= - 1,500
g(-1500) = -0.002(-1500)2 – 6(-1500) + 100 = 4600
Maximum Point at: (-1500, 4600)
10 pts
7. (E11f) For the equation y = 5x2+10x+5
a. What is the Degree? Answer: 2
b. What is the y-intercept if it exists? Answer: 5
c. What are the Zeros of this function?(Hint: Factor it)
Y = 5(x2+2x+1) = 5(x + 1)(x+1)
Answer: Zeros are {-1)
d. Graph this function accurately plotting the zero(s) and labeling them (it)
as well as any maximum or minimum point.
Min at:
(-1,0)
y-int at:
(5,0)
College Algebra MA 134
10 pts
8.
Midterm Exam-Take Home Portion
100 Points
For the equation y = 2x2(x+4)2(x-3)3
a. What is the Degree of this polynomial ANSWER: Degree = 7
b. What are the Zeros of this function? ANSWER: (0, -4, 3)
c. Tell the multiplicity of each Zero. ANSWER: 0 (Mult 2), -4(Mult 2), 3(Mult 3)
d. Graph this function accurately plotting the zeros and labeling them as
well as any maximum or minimum points. Max: (-4, 0), (0,0)
Min: Approx (-2.3,-4650), (1.2, -500)
(E11b)
0
-4
3
10 pts
9. (E07ad) Find the domain for each of the following:
𝑥−5
a. 2
b. f(x) = x2 – 4x
𝑥 −7𝑥−18
a) Find domain by finding zeros of denominator (x+2)(x-9) Zeros (-2, 9)
D:(-∞,-2) U (-2,9) U( 9,∞)
b) x2 – 4x = x(x-4) Zeros (0,4)
D:(-∞,∞)
College Algebra MA 134
6 pts
Midterm Exam-Take Home Portion
100 Points
10. (E20abc) An object is thrown vertically up and its height in feet after t
seconds is given by the formula h(t) = 96t – 16t2.
A. Find the maximum height attained by the object. [Hint: Xv= -b/2a
actually in this case its called tv]
Rewrite the equation to appear like this: h(t) = -16t2 +96t + 0 ( y = ax2 + bx + c format)
Since sign or square term is negative it has a maximum at t = to some value -b/2a
t=-b/2a = -96/2(-16) = 3 That’s 3 seconds by the way. To find the maximum height substitute 3
into h(t) and find h(3)= -16(3)2 + 96()3) = 144
ANSWER: 144 Feet
B. After how many seconds does it attain its maximum height? [Hint: at
tv]
ANSWER: We answered that in [part A above: 3 seconds
C. After how many seconds does it return from initially leaving the
person’s hand does it take to return to its starting position?
You can solve the equation for the x-intecepts. That’s at ground level
, when y = 0 so you set equation = 0 and solve it:
Ie. 96t – 16t2 = 0 16t(6 – t) = 0 Set each factor equal to zero
And you find that occurs at 0 and 6 seconds. So it takes 6 seconds for
the entire trip.
8 pts
11. (E25a-h) Each of the following is arithmetic or geometric:
a. Find the nth term of 10,6,2,-2,-6, . . .
ANSWER: an = a1 +(n-1)d = 10 + (n-1)(-4) = 14 -4n
b. Find the 100th term and the sum of the first 100 terms of 7,10,13,16, . . .
ANSWER: a100 = 7+(100-1)3 = 304
Sn = n/2(a1 + an) = 100/2(7 + 304) = 50(311) = 15,550
c. Find the 20th term of 2,6,18,54,. . .
Answer: An = a1r(n-1)
= (2)(3)19 = 2,324,522,934
d. Evaluate ∑100
𝑘=0 4𝑘 + 3 Use Calculator: 20,503
College Algebra MA 134
Midterm Exam-Take Home Portion
100 Points
e. Evaluate ∑6𝑘=3 𝑘(𝑘 + 2) Use Calculator: 122
2
1
1
1
f. Find the sum of the first 10 terms of
+ + + +...
3 3
6 12
This is an infinite sequence that converges:
ANSWER: =
2
∑90
3
g. Find the sum :
ANSWER: =
1
. ( )𝑛 =
2
2 1
1
3
+
2
∑∞
0 3
3
Sn =
+ +
6
1
. ( )𝑛
2
1
𝑎1(1−𝑟 𝑛
(1−𝑟)
(2/3) (1 −
=
1
2
110
)
2
= 1.332
+...
12
: Sn =
𝑎1
(1−𝑟)
=
2
3
=
1
1−
2
2
3
1
2
=
𝟒
𝟑
𝑘
h. Evaluate ∑∞
r = .4 , a1 = 2.4
𝑘=1 6(0.4)
ANSWER: Sn =
5 pts
𝑎1
(1−𝑟)
=
2.4
1− .4
=
=
2.4
.6
=4
12. (E23) The average weight of a baby born in 1900 was 6.25 pounds. In 2000,
the average weight of a newborn was 6.625 pounds. We will assume form
our purposes that the relationship is linear. Find the equation that relates
the year to the average weight of a newborn. Using this equation, predict
the average weight of a newborn in 2035.
This is a two-point slope problem. The two points are:
(1900, 6.25) and (2000, 6.625) We let the year be the independent variable
and the baby weight be the dependent variable.
W(y) = mx + b so we need to find the slope first
M = (y2 – y1)/(x2 – x1) = (6.625 – 6.25)/ (2000 – 1900) = 0.00375
Using the two –point Formula (Y – Y1) = M (X – X1) we can find the equation
(Y – 6.625) = 0.00375(X – 2000)
Y = 0.00375X – 67.5 + 6.625
Y =( 0.00375)X - 60.875
Y2035 = ( 0.00375)(2035) - 60.875 = 6.756 lbs.
College Algebra MA 134
11pts
Midterm Exam-Take Home Portion
100 Points
13. (E26) Use the graph of the function f ( x ) to answer questions a – h:
a) f (2) = -8
b) For what value(s) of x is f ( x )  3 ? -3, -5, 4.4
c) The domain of f is (-7,5]
d) The range of f is [-9,7]
e) For what interval(s) is f ( x ) decreasing? (-4,1)
f) For what interval(s) is f ( x ) increasing? (-7,4) U (1,5)
g) There is a relative maximum of 4 at -4
h) There is a relative minimum of -9 at 1.
i) Use the graph to solve f(x) < 0. (-7,-6) U (-2,4)
j) Find the real zeroes of the function. -6, -2, 4
k) f(0) = -8
Grading.
I did not count off for Problem 5. (It referred to the wrong problem)
The points total to 110.
After I found the number you missed, I subtracted from 110 then divided by
110 to get a percent. The percent score is the score I recorded in the grade
book.