# MA107 Precalculus Algebra Exam 2 Review Solutions

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1 MA107 Precalculus Algebra Exam 2 Review Solutions February 24, The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p a. Please write a model expressing the revenue, R as a function of x. (*) Revenue is the product of price and units sold. Hence, R = xp. But we know x = 4p + 200, so I substitute 4p for x in the revenue equation. I get R = ( 4p + 200)p, or R = 4p p. b. What price should the company charge to maximize revenue? (*) Note that the formula for R is a quadratic. Hence, its graph is a parabola. A parabola has a vertex, which is either a min point (if the parabola opens up), or max (if the parabola opens down). The a coefficient of the quadratic (ax 2 + bx + c) is 4. means the parabola opens down, so the vertex is a max. That it is negative value The vertex is the point (p, R), where p is the price and R is the revenue at that price. The formula for the vertex is ( b 2a, f( b 2a ). So p = b 2a = = 25. This means the maximum revenue is obtained when p = 25. c. What is the maximum revenue? (*) We found the price, p, which generates maximum revenue. Revenue at any price is given by R = 4p p. So to find the maximum revenue, plug in the price which gives the maximum revenue. R(25) = 4(25) (25) =

2 d. What quantity x corresponds to the maximum revenue? (*) Recall that x = 4p Then when p = 25, x = 4(25) = 100. e. How do you know there is a maximum and not a minimum point on the graph of this function? (*) Because the a coefficient is negative, as explained in the solution to question b. 2. Write a polynomial f(x) which has the roots 2, 2, 4 and such that the graph of f(x) touches the x axis at 2 and 2, and crosses the x axis at 4, and has a degree of 9. (*) If f(x) has the roots 2, 2, and 4, then it has the factors (x+2), (x 2), (x 4) respectively. So start by letting f(x) be the producto of these factors. f(x) = (x + 2)(x 2)(x 4) Now the powers of the factors (x + 2) and (x 2) must be even, as f(x) touches the x axis at their respective roots. The power of the (x 4) factor must be odd, as f(x) crosses the x axis at its respective root. Also, the degree of the polynomial must be 9. This means the sum of the three powers must be 9. One possibility: f(x) = (x + 2) 2 (x 2) 2 (x 4) 5 3. Write a polynomial f(x) which has the roots 1, 1 and such that the graph of f(x) touches the x axis at both 1 and 1, and has a degree of 6. (*) One possibility: f(x) = (x 1) 4 (x + 1) What are the roots of the polynomial (2x 3)(x 3 + 7x 2 )? (*) I want to find the values of x such that (2x 3)(x 3 + 7x 2 ) = 0. (In other words, the x-intercepts). So (2x 3) = 0 and (x 3 + 7x 2 ) = 0. If 2x 3 = 0, then 2x = 3 and so x = 3 2. There is one solution. To find the other solutions, solve (x 3 + 7x 2 ) = 0. I can factor out an x 2 from the LHS. Then I get x 2 (x + 7) = 0. So x 2 = 0 means x = 0. x + 7 = 0 means x = 7. 2

3 5. What is the domain of the rational function f(x) = x2 +5x+6 x 2 +6x+8? (*) The domain of a rational function is all real numbers except those values of x for which the denominator is 0. I solve x 2 + 6x + 8 = 0. x 2 + 6x + 8 = (x + 4)(x + 2), so x + 4 = 0 and x + 2 = 0. x = 2. Then x = 4 and So the denominator is all real numbers except x = 4 and x = 2. {x x 4, x 2}. 6. For the rational function above, please state the vertical asymptotes. (*) The locations of the vertical asymptotes are precisely the roots of the denominator which are not roots of the numerator. So there s two ways to go about this solution. We already know the roots of the denominator. So we can find the roots of the numerator, and compare the lists. We could also write f(x) in reduced form, and solve the denominator for zero. Looking at the next couple of questions, we ll need to know the roots of the numerator. So let s go that route. I solve x 2 + 5x + 6 = 0. x 2 + 5x + 6 = (x + 3)(x + 2), so x + 3 = 0 and x + 2 = 0. x = 2. Then x = 3 and The numerator has roots 3 and 2. The denominator has roots 4 and 2. 4 is the only root of the denominator which is not a root of the numerator. there is a vertical asymptote at x = 4. So 3

4 7. For the rational function above, please state the horizontal asymptotes. (*) Recall the rule for rational functions: 1. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. 2. If the degree of the numerator is less than the degree of the denominator, the line y = 0 is a horizontal asymptote. 3. If the degrees are the same, throw out all but the dominant term of the numerator and denominator. Simplify to get a real number L. Then the line y = L is the horizontal asymptote The degrees of the numerator and denominator are both 0. The dominant term of the numerator is x 2. The dominant term of the denominator is x 2. x 2 x 2 = 1. So y = 1 is the horizontal asymptote. 8. For the rational function above, please state the zeroes. (*) The roots of a rational function are the zeroes of the numerator which are not zeroes of the denominator. We found the roots of the numerator to be 2, 3. But 2 is a zero of the denominator. So the only remaining zero is 3. Hence, x = 3 is the zero of f(x). 9. Let p(x) = x 3 + x 2 6x. What is the degree of p(x)? (*) What are the roots of p(x)? What is the multiplicity of each root? (*) p(x) = x 3 + x 2 6x = x(x 2 + x 6) = x(x 2)(x + 3) The roots are 0, 2, 3. Each has multiplicity For each root, state if the graph of p(x) crosses or touches the x-axis at the intercept. (*) Because each root has multiplicity 1, the graph of p(x) crosses the x-axis at each one. 12. Please state the power function that the graph of f(x) resembles for large values of x. (*) Take the dominant term. The power function is x 3. 4

5 13. What is the remainder when p(x) is divided by (x 5)? (*) In general, if p(x) is divided by (x c), then the remainder is p(c). Take c = 5. Then the remainder is p(5) = = Consider q(x) = 7x x 4 22x 2 + 5x + 21 List all the possible rational roots. (*) Factors of the constant coefficient, p : ±1, ±3, ±7, ±21 Factors of the leading coefficient, q : ±1, ±7 All the possible ratios: p q : ±1, ±3, ±7, ±21, ± 1 7, ± Use the Intermediate Value Theorem to find an open interval which contains at least one real root of q(x). (*) q( 4) = 95. q( 3) = 456. Because the sign changes as x moves from 4 to 3, there must be a real root in the interval ( 4, 3). 16. Solve the rational inequality (x 4)(x+2)2 (x 3) > 0. (*) Let f(x) = LHS = (x 4)(x+2)2 (x 3). First I find the roots of the numerator. These are: 4, 2. Next I find the roots of the denominator. These is one: 3. These roots divide the real number line into four intervals. I have: (, 2), ( 2, 3), (3, 4), (4, ). I pick a value of x in the interval (, 2). Say, x = 3. f( 3) > 0, so for all x in this interval, f(x) > 0. I pick a value of x in the interval ( 2, 3). Say, x = 0. f(0) > 0, so for all x in this interval, f(x) > 0. I pick a value of x in the interval (3, 4). Say, x = 3.5. f(3.5) < 0, so for all x in this interval, f(x) < 0. I pick a value of x in the interval (4, ). Say, x = 5. f(5) > 0, so for all x in this interval, f(x) > 0. We wanted intervals for which f(x) > 0. criteria. They are: (, 2), ( 2, 3), (4, ) There are three intervals which satisfy this 5

6 17. Sketch the graph of f(x) = (x 2) 2 (x + 1) 3 x(x 1). (*) The important features are: The graph touches the x axis at x = 2. The graph crosses the x axis at x = 1. The graph crosses the x axis at x = 0. The graph crosses the x axis at x = 1. The y-intercept is I wish to enclose a rectangular yard with a fence. I have 400 feet of fencing available. a. Express the area A of the yard as a function of w, the width of the rectangle. (*) I know: A = lw. I also know: 2l + 2w = 400. Solving for l, I have l = 200 w. Then A = w(200 w) = w w. b. Express the area A of the yard as a function of l, the length of the rectangle. (*) Instead of solving 2l + 2w = 400 for l, solve it for w. Then w = 200 l. So, I can get A = l l. c. Find the maximum area. Find the dimensions of the rectangle which give this area. (*) Given A = l l is a quadratic, its graph is a parabola. So it has a vertex. The parabola opens down because the leading coefficient is negative. So the vertex is a maximum. The vertex is the point (l, A), where l = b 2a = = 100. Knowing l = 100 and w = 200 l, then w = = 100. The dimensions are w = 100, l = 100. When l = 100, the area is l l = (100) (100) =

7 19. A cylindrical box is to the constructed such that the sum of the height and radius is 100 inches. Construct a function which states the volume of the box as a function of its radius. (*) Knowing the volume of a cylinder is V = πr 2 h, I have a model which takes r (radius) as input and gives me the volume. But what is h? This is an unknown. To eliminate it, I need more information. The sum of height and radius is 100. So h + r = 100. Solving for h, I get h = 100 r. So now I can substitute 100 r for h in my volume formula. V = πr 2 (100 r). Now I have a formula for volume whch is a function of r, as was requested. 20. Which polynomial below is that whose graph is depicted to the right? Choice A x2 (x 3)(x 5) 3 Choice B x3 (x 3)(x 5) 3 Choice C x2 (x + 3) 3 (x 5) 3 Choice D x3 (x 3) 2 (x 5) 2 Choice E x3 (x 3) 2 (x 5) 3 (*) The correct answer is A. The graph has roots 0, 3 and 5 (eliminating choice C). The graph crosses the x axis at x = 3 (eliminating choices D, E). The graph crosses the x axis at x = 5 (also eliminating choice D). Choice B is eliminated because the graph touches the x axis at x = 0, but the power of x is odd. 21. Which polynomial below is that whose graph is depicted to the right? Choice A. (x 1) 2 (x + 2) 3 Choice B. (x + 1) 2 (x + 2) 3 Choice C. (x 1) 2 (x 2) 4 Choice D. (x 1) 3 (x 2) 3 Choice E. (x 1) 2 (x 2) 3 (*) The correct answer is E. The graph has roots 1 and 2 (eliminating choices A, B). The graph touches the x axis at x = 1 (eliminating choice D). The graph crosses the x axis at x = 2 (eliminating choice C). 7

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