College Algebra MA 134Midterm Exam-Take Home Portion100 Points

  1. (E04a) Graph y =- x3 + 4 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: (-∞,∞)

(b)Table

  1. (E04b) Graph f(x) = 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.

(d) Y-Intercept: (0,2)

  1. (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.
x / y
0 / 0.0
-1 / (+/-)3.5
-2 / (+/-)4.9
-3 / (+/-)6.0
-4 / (+/-)6.9
-5 / (+/-)7.7
-6 / (+/-)8.5

  1. (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.

(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)

  1. (E10a) Use the Quadratic Formula to find the exact values of the x-intercepts for the function in #4.

= = =

= = = -1

  1. (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 fact that (, f()can define this point for you.)

(a)Has a maximum point since is a quadratic and leading term is (-)

(b)Xv = = = = - 1,500

g(-1500) = -0.002(-1500)2 – 6(-1500) + 100 = 4600

  1. (E11f) For the equation y = 5x2+10x+5
  1. What is the Degree? Answer: 2
  2. What is the y-intercept if it exists? Answer: 5
  3. What are the Zeros of this function?(Hint: Factor it)

Y = 5(x2+2x+1) = 5(x + 1)(x+1)

Answer: Zeros are {-1)

  1. Graph this function accurately plotting the zero(s) and labeling them (it) as well as any maximum or minimum point.

  1. (E11b) For the equation y = 2x2(x+4)2(x-3)3
  1. What is the Degree of this polynomialANSWER: Degree = 7
  2. What are the Zeros of this function?ANSWER: (0, -4, 3)
  3. Tell the multiplicity of each Zero.ANSWER: 0 (Mult2), -4(Mult 2), 3(Mult 3)
  4. Graph this function accurately plotting the zeros and labeling them as well as any maximum or minimum points.

  1. (E07ad) Find the domain for each of the following:
  1. b. f(x) = x2 – 4x

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:(-∞,∞)

  1. (E20abc) An object is thrown vertically up and its height in feet after t seconds is given by the formula h(t) = 96t – 16t2.
  1. 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

  1. After how many seconds does it attain its maximum height? [Hint: at tv]

ANSWER: We answered that in [part A above: 3 seconds

  1. 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.

  1. (E25a-h) Each of the following is arithmetic or geometric:
  1. Find the nth term of 10,6,2,-2,-6, . . .

ANSWER: an = a1 +(n-1)d = 10 + (n-1)(-4) = 14 -4n

  1. 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

  1. Find the 20th term of 2,6,18,54,. . .

Answer: An = a1r(n-1) = (2)(3)19 = 2,324,522,934

  1. Evaluate Use Calculator: 20,503
  2. Evaluate Use Calculator: 122
  3. Find the sum of the first 10 terms of + + + + . . .

This is an infinite sequence that converges:

ANSWER: = = Sn = = = 1.332

  1. Find the sum : + + + + . . .

ANSWER: = : Sn = = = =

  1. Evaluate r = .4 , a1 = 2.4

ANSWER: Sn = = = = = 4

  1. (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.

  1. (E26) Use the graph of the function to answer questions a – h:

a) =-8

b) For what value(s) of x is ? -3
c) The domain of f is (-7,5]
d) The range of f is [-9,7]
e) For what interval(s) is decreasing? (-4,1)
f) For what interval(s) is 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. The points total to 110.

As I marked the papers I counted off and after this total was determined. I looked at all the papers to see who missed the least points. That number was -10. I added that amount back to all papers. So your score was determined like this:

100 - (The points I counted off) + (10 points) .