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Solved Test 2 for College Algebra | MATH 121, Exams of Algebra

Material Type: Exam; Class: COLLEGE ALGEBRA; Subject: MATHEMATICS; University: La Sierra University; Term: Fall 2004;

Typology: Exams

Pre 2010

Uploaded on 08/16/2009

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Math 121, Test 2, October 19, 2004
Name: Hints and Answers
Instructions. Part I: Do each of the following 10 problems. Each problem is worth 5pts.
Show all appropriate details in your solutions. Calculators are not allowed. Good Luck! When
you turn this in, please pick up Part II which has 2problems.
1. Find (a) the distance between (โˆ’1,3) and (โˆ’5,6) and (b) the midpoint of (โˆ’1,3) and
(โˆ’5,6).
Answer. (a) d=p(โˆ’5โˆ’(โˆ’1))2+ (6 โˆ’3)2=p(โˆ’4)2+ 32=โˆš16 + 9 = โˆš25 = 5
(b) The midpoint is ๎˜’โˆ’1+(โˆ’5)
2,3+6
2๎˜“=๎˜’โˆ’3,9
2๎˜“.
2. Write the equation of the circle x2+ 2x+y2โˆ’6y+ 1 = 0 in standard form, and find the
center and radius of the circle.
Answer. Complete the square(s) to write the circle in standard form:
x2+ 2x+1+y2โˆ’6y+ 9 + 1 = 1 + 9 and so
(x+ 1)2+ (yโˆ’3)2= 9.
Therefore, the center is (โˆ’1,3) and the radius is 3.
3. Let f(x) = โˆšxโˆ’2 and g(x) = โˆš8โˆ’x. Find the domains of (a) f(b) g(c) fg (d) f /g.
Write your answers in interval form.
Answer. (a) xโˆ’2โ‰ฅ0, thus xโ‰ฅ2, i.e. [2,โˆž).
(b) 8 โˆ’xโ‰ฅ0, thus xโ‰ค8, i.e. (โˆ’โˆž,8].
(c) The intersection of the domains of fand gwhich is [2,8].
(d) The intersection of the domains of fand g, except where g(x) = 0, which is [2,8).
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Download Solved Test 2 for College Algebra | MATH 121 and more Exams Algebra in PDF only on Docsity!

Math 121, Test 2, October 19, 2004

Name: Hints and Answers

Instructions. Part I: Do each of the following 10 problems. Each problem is worth 5pts. Show all appropriate details in your solutions. Calculators are not allowed. Good Luck! When you turn this in, please pick up Part II which has 2 problems.

  1. Find (a) the distance between (โˆ’ 1 , 3) and (โˆ’ 5 , 6) and (b) the midpoint of (โˆ’ 1 , 3) and (โˆ’ 5 , 6).

Answer. (a) d =

(โˆ’ 5 โˆ’ (โˆ’1))^2 + (6 โˆ’ 3)^2 =

(โˆ’4)^2 + 3^2 =

(b) The midpoint is

  1. Write the equation of the circle x^2 + 2x + y^2 โˆ’ 6 y + 1 = 0 in standard form, and find the center and radius of the circle.

Answer. Complete the square(s) to write the circle in standard form:

x^2 + 2x + 1 + y^2 โˆ’ 6 y + 9 + 1 = 1 + 9 and so (x + 1)^2 + (y โˆ’ 3)^2 = 9.

Therefore, the center is (โˆ’ 1 , 3) and the radius is 3.

  1. Let f (x) =

x โˆ’ 2 and g(x) =

8 โˆ’ x. Find the domains of (a) f (b) g (c) f g (d) f /g. Write your answers in interval form.

Answer. (a) x โˆ’ 2 โ‰ฅ 0, thus x โ‰ฅ 2, i.e. [2, โˆž).

(b) 8 โˆ’ x โ‰ฅ 0, thus x โ‰ค 8, i.e. (โˆ’โˆž, 8].

(c) The intersection of the domains of f and g which is [2, 8].

(d) The intersection of the domains of f and g, except where g(x) = 0, which is [2, 8).

  1. Find the equation of the line that passes through the point (2, 3) and is perpendicular to the line x โˆ’ 2 y = 4. Write your answer in slope-intercept form.

Answer. The orginal line in slope intercept form is y = 12 x โˆ’ 2, thus the slope of the original line is m 1 = โˆ’ 1 /2. The slope of the perpendicular line is m 2 = โˆ’ 1 /m 1 = โˆ’2. Therefore, the equation of the perendicular line is y โˆ’ 3 = โˆ’2(x โˆ’ 2) which simplifies to y = โˆ’ 2 x + 7 in slope-intercept form.

  1. Consider the function f (x) = โˆ’ 2 x^2 + 6x โˆ’ 5. (a) Find the range of f. (b) Does f have a minimum? If so, what is it? (c) Does f have a maximum? If so, what is it?

Answer. The x-coordinate of the vertex is โˆ’b/ 2 a = โˆ’ 6 / โˆ’ 4 = 3/2. The y-coordinate is y = โˆ’2(3/2)^2 + 6(3/2) โˆ’ 5 = โˆ’ 1 /2. The parabola opens downward because a < 0. Therefore:

(a) The range is (โˆ’โˆž, โˆ’ 1 /2].

(b) There is no minimum because the graph opens downward.

(c) The maximum is f (3/2) = โˆ’ 1 /2.

  1. Compute the difference quotient

f (x + h) โˆ’ f (x) h

for the function f (x) = 2x^2 โˆ’ 3 x.

Answer.

f (x + h) โˆ’ f (x) h

2(x + h)^2 โˆ’ 3(x + h) โˆ’ (2x^2 โˆ’ 3 x) h

=

2(x^2 + 2xh + h^2 ) โˆ’ 3 x โˆ’ 3 h โˆ’ 2 x^2 + 3x h

=

4 xh + 2h^2 โˆ’ 3 h h = 4 x + 2h โˆ’ 3.

Name:

Part II. You may use a calculator on this part of the test. Show all appropriate work in your solutions.

  1. A magazine company had a profit of $300,000 per year when it had 100,000 subscribers and a profit of $800,000 per year when it had 150,000 subscribers. Assume the profit P is a linear function of x the number of subscribers.

(a) Find a formula for P (x).

Answer. The formula for P is the formula of a line going through (100000, 300000) and (150000, 800000). Thus the slope of P is

m =

Thus P (x) = 10x + b, plug in one of the points to find b, i.e. 300, 000 = 10(100, 000) + b and so b = โˆ’ 700 , 000. Therefore, the equation is P (x) = 10x โˆ’ 700 , 000.

(b) How many subscribers are needed to make a profit of $1,000,000 per year?

Answer. Solve 1, 000 , 000 = 10x โˆ’ 700 , 000, to get that x = 170, 000 subscribers.

(c) What would the profit be if there were 300,000 subscribers?

Answer. The profit would be P (300, 000) = 10(300, 000) โˆ’ 700 , 000 = $2, 300 , 000.

  1. A farmer has $20,000 to spend to fence a rectangular corral. Because extra reinforcement is needed on one side, the corral costs $30 per foot along that side. It costs $10 per foot to fence the remaining sides. What dimensions of the corral will maximize the area of the corral?

Answer. Let x be length of the rectangle along the expensive side, and y be the length of the other dimension of the rectangle. Then the cost of the fence is 30x + 10x + 10y + 10y = 20, 000 and we wish to maximize area A = xy. Now solve the cost equation for x to get y = 1000 โˆ’ 2 x. Then A = xy = x(1000 โˆ’ 2 x) = โˆ’ 2 x^2 + 1000x.

This is a quadratic function whose maximum occurs at the vertex when

x = โˆ’b/ 2 a = โˆ’ 1000 / โˆ’ 2(2) = 250.

Thus y = 1000 โˆ’ 2(250) = 500. Therefore, the dimensions are 250 feet by 500 feet (where 250 feet is the length of the expensive side).