MAT 124 Solutions Sample Questions for Exam 2

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1 MAT 124 Solutions Sample Questions for Exam 2 Note: Most of these results can be checked graphically. 1. a) The slope of l " is computed as follows: m " = & '(& ) * ' (* ) = +(, -(. = /, = 2. So the equation of l " is given by y = 2x + b. To determine b, plug into this equation the coordinates of A (you could aslo use the coordinates of C). This yields 2 = 2(3) + b = 6 + b b = 2 6 = 4. Therefore, we get the equation of l " : y = 2x 4. The x-intercept of l " is then the point (0, 2). b) The slope of l, is computed as follows: m, = & C(& D * C (* D = ("(E (,(((/) = ",. Since m " m, = 1, the lines l 1 and l 2 are perpendicular. 2. a) The daily cost C, in dollars, for renting a truck that drives on average x miles per day is given as follows: C(x) = 0.07x + 29 b) If a person spent $ after renting a truck for a whole week, then he/she drove, on average, 230 miles per day: 7C(x) = x = x = x = "",.ME = 230. E./N c) At the rate of 250 miles per day, this new truck rental company is cheaper since it charges a daily cost of 0.04(250) + 35 = $45 while the first company charges C(250) = 0.07(25) + 29 = $ The savings amount to $1.50 per day per truck. Since = 157.5, this implies a total savings of $ for a fleet of 15 trucks driving 250 miles per day for a whole week.

2 3. a) The average yearly population growth of the town is computed as follows:,"+.("/." = M., = 183 people/year.,eem(,ee. / a) According to this constant growth rate, the town population in the year 2000 was 1431 (3 183) = 882 people. b) The population P of the town t years after 2000 is given by the linear model below: P(t) = t. c) To predict the year in which the population of the town will reach 4,000, we solve the equation P(t) = 4000 as follows: t = t = 3118 t =.""X "X Therefore, according to this model, the town population will reach 4,000 people in the year Below is a graph of this linear model P.

3 4. a) The coefficients of f are a = 1, b = 2, c = 1. Therefore, h = ^,_ k = f(1) = 2, the vertex point of the parabola y = f(x) is V = (1, 2), and its line of symmetry is the vertical line x = 1. b) The x-intercepts of the graph of f are (1 2, 0) and (1 + 2, 0) since the two solutions of the quadratic equation f(x) = 1 + 2x x, = 0 are x = 1 ± 2. [Computation done in class] c) If h(x) = ef(x), then the domain of h is given as follows: D g = {x f(x) 0} = lx G n is above, or on, the x axisv = w1 2, 1 + 2x [ , ]. Below is the graph of f.

4 5. a) Since V = ( 1, 4), the equation of the parabola in vertex form is y = a(x + 1), + 4. Now plugging in the coordinates of the point (0, 6) into this equation yields a(0 + 1), + 4 = 6 a + 4 = 6 a = 2. Thus, the equation of the parabola is given by y = 2(x + 1) = 2x 2 + 4x + 6. b) Solving the equation y = 2(x + 1), + 4 = 0 yields (x + 1), = 2, which has no real solutions. This then implies that the parabola has no x-intercepts. Note that this result is consistent with the fact that the parabola opens up and has its vertex located in QII. 6. a) If f(x) = 3x, + 6x + c, then h = +,(.) = 1, k = f( 1) = c = c 3, and the vertex of the parabola y = f(x) is given by V = ( 1, c 3). Since the graph of f is a parabola that opens up, the absolute minimum of f occurs at its vertex point. Therefore, if the absolute minimal value of f is 8, then we have c 3 = 8 c = 5. b) If the parabola y = f(x) passes through the point ( 2, 7), then f( 2) = 7, which implies that 3( 2), + 6( 2) + c = 7 c = 7. c) By the results of part a), we conclude that the range of f is the set R f = {y y c 3} = [c 3, ). 7. The equation of the parabola that passes through the point (2, 4) and whose axis of symmetry is the vertical line x = 1 has the form y = ax 2 2ax + 4, where a 0. To see this, check that the equation of this parabola in vertex form is y = a(x 1), + k, and then plug in the coordinates of the point (2, 4) into this equation to get 4 = a(2 1), + k 4 = a + k k = 4 a. This results in the equation y = a(x 1), + 4 a = ax, 2ax + a + 4 a = ax, 2ax + 4.

5 8. Since factoring the function completely yields f(x) = 3(x, 5x + 6)(x, 8x + 12) = 3(x 2), (x 3)(x 6), the zeros of the polynomial function f are x = 2, x = 3, and x = The domain of the function f(x) = x 3x, is given by D f = x 0 x 1 3 = Š0, 1 3. To see why, note that the parabola y = x 3x, = x(1 3x) (the radicand of the function f) has x-intercepts at the origin and Œ0, 1 3 and has the vertex in Quadrant I. It is therefore above the x-axis on the interval Š0, 1. The graph of this parabola is shown 3 below. 10. a) Since f(0) = 243 (0 2) - = 243 ( 32) = 275, the y-intercept is the point (0, 275). To find any x-intercepts, we solve the equation y = f(x) = 0 as follows: 243 (x 2) - = 0 (x 2) - Ž = 243 x 2 = 243 Therefore, the only x-intercept of the graph of f is the point (5, 0). = 3 x = 5.

6 b) Start with the graph of the basic power function y = x -. Then, a horizontal shift by 2 units to the right transforms the graph of this basic power function to the graph of y = (x 2) -. Now, a reflection about the x-axis transforms the graph of y = (x 2) - into the graph of y = (x 2) -. Finally, a vertical shift 243 units upwards transforms the graph of y = (x 2) - into the graph of y = 243 (x 2) - = f(x). Here are the respective transformations for three points: Point #1: (0, 0) (2, 0) (2, 0) (2, 243) Point #2: (2, 32) (4, 32) (4, 32) (4, 211) Point #3: (3, 243) (5, 243) (5, 243) (5, 0) [Check all these results graphically.] 11. a) Below is the graph of this arch as a parabola.

7 b) Setting up the vertex point of the arch at (0, 630) and the two x-intercepts at (315, 0) and ( 315, 0), so that the center (the midpoint of the arch legs) coincides with the origin, we get the following equation for this parabolic arch: y = f(x) = Œ +.E."- x, = Œ,."- x, x, Note that there are other ways to set up this arch. Another possibility is to place the vertex point at the origin. c) Sincef(200) = f( 200) = Œ,."- (200), , we conclude that the arch height 200 feet away from the center is approximately 376 feet (rounded to the nearest feet). d) Solving the equation f(x) = 20 yields x So the 20-feet-tall statue can be placed 310 feet away from the center. 12. a) Factoring the function f completely yields f(x) = x / x. + 6x, = x, (x, + x 6) = x, (x 2)(x + 3) Thus the x-intercepts are the points (0, 0), (2, 0), and ( 3, 0). S Since f(0) = 0, the y-intercept is also the origin (the point (0, 0)). b) For large values of x (either positive or negative), the graph of f resembles the graph of the power function y = x 4. As a result,, the graph of f will end in QIII on the left and in QIV on the right. Below is the graph of f.

8 Bonus Problem Suppose there is a parabola y = ax, + bx + c, where a 0, that passes through these three collinear points. This would then imply that the coefficients a, b, c satisfy the following equations: 4a + 2b + c = 1 (1) 9a + 3b + c = 2 (2) 16a + 4b + c = 3 (3) Subtracting (1) from (2) yields the equation 5a + b = 1. On the other hand, subtracting (2) from (3) yields the equation 7a + b = 1. Solving for both of these equations results in a clear inconsistency. As a result, the initial system of equations in a, b, c has no solution and no such parabola exists.