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Multiple-Choice Questions 1. 3 5 D The equation of line l is y x , so the equation of any line perpendicular to 2 2 2 l is given by y x b . Since the line passes through B , the coordinates of B (6,1) 3 2 must satisfy this equation. This yields b 3 , so the equation of the line is y x 3 , 3 or 2x 3y 9 in standard form. 2. B The equation of the circle centered at A passing through B is given by x 3 2 y 2 82 since the center is A (3,2) and the radius is the distance 2 between A and B , which is x 2 y 2 6x 4 y 69 0 . 82 . Expanding this equation yields the general form 1 1 3 , or y x3 3 , has no symmetries since x 2 x2 replacing x by x in the equation does not return y or y . 3. D The graph of y x 3 4x is given by Df x | x 4, x 1,0,1 x3 x since 4 x 0 implies that x 4 and x3 x 0 implies that x( x 2 1) 0 , or x( x 1)( x 1) 0 , or x 0, 1 . The domain of the function f (x) 4. C 5. B If x 4 is a zero of f , then f (4) 0 . This implies that 2(4) A 0 , or A 8 . Now, if x 1 a vertical asymptote of the graph of f , then B 1 . 6. A The graph of the quadratic function f (x) x 2 6x a , where a is some real coefficient, is a parabola whose vertex is the point 3, f (3) 3,9 a and whose vertex is also a global maximum (i.e. the parabola opens down). Therefore, the range of f is given by R f y | y 9 a. 7. D The graph of y 3 x 3 1 x 3 1 7 is also the graph of y 3x 6 shifted vertically 4 units upwards since y 3 x3 1 x3 1 7 3 x 6 1 7 3x 6 4 . 8. B If a parabola passes through the origin and its line of symmetry is the vertical line x 4 , then the parabola must be described by an equation of the form Only the equation y a( x 4)2 16a , where a is some real coefficient. 2 (x 4) y 8 does not satisfy this form. [You can check that in this equation if 2 0 . Hence, this parabola does not pass through the origin.] x 0 , then y 16 9. B If a polynomial function f has a zero at x 3 of multiplicity 2 , another zero at x 2 of multiplicity 3, and a third zero at x 1 of multiplicity 5, then the function f has the form f ( x) a( x 3)2 ( x 2)3 ( x 1)5 , where a is some real coefficient. Now if the graph of f intersects the y-axis at 216, then f (0) 216 . This implies that a 3 and so f ( x) 3( x 3)2 ( x 2)3 ( x 1)5 . 10. D The graph of R( x) horizontal R( x) asymptote, 2 x 2 x has three vertical asymptotes, one x 3 x 3 4 x 2 4 x and no x-intercept. To check this, note that ( x 2) 1 for x 2 . From this 2 x x 3 x 2 x x 3 x 2 x x 3 x 2 reduced form we see that the graph has 3 vertical asymptote at x 0 , x 3 , and x 2 . Since the numerator is 1, the graph has no x-intercept. Since R is proper, the graph has one horizontal asymptote: y 0 (the x-axis). 2 Problem 1 Start with the graph of y #1 y 1 , then use the following transformations: x2 1 1 1 y . This is a horizontal shift of the graph of y 2 4 units to the 2 2 x x x 4 1 , we see x2 that these are now changed to, respectively, points (3,1) and (5,1) on the graph of 1 1 y . Note that the vertical asymptote x 0 on the graph of y 2 is now 2 x x 4 right. Using test points ( 1,1) and (1,1) , for example, on the graph of y moved to x 4 on the graph of y #2 y 1 x 4 2 y 1 x 4 2 1 x 4 2 . . This is a reflection of the graph of y 1 x 4 about 2 the x-axis. Our test points are now moved to, respectively, points (3, 1) and (5, 1) . #3 y 1 x 4 2 y 3 1 x 4 2 . This is a vertical shift of the graph of y 1 x 4 2 3 units upwards. Our test points are now moved to, respectively, points (3, 2) and (5, 2) . Note that the horizontal asymptote is now the line y 3 . The range of the function y 3 1 x 4 2 is then given by y | y 3 , or (,3) . Problem 2 a) One possibility is to place the vertex of the arch at (0, 25) and the bases of the two legs at (60, 0) and ( 60, 0) . Then the equation of the parabola is given by y ax 2 25 , for some positive coefficient a . Using the point (60, 0) , for example, we get a 25 25 1 x2 . So the equation of the parabolic arch is y 25 . 602 3600 144 144 [Note: you could also place the vertex at the origin, or come up with a different configuration. The equation would then change accordingly] b) x2 25 . We then need to solve for x in the equation f ( x) 6 to find 144 out how far a 6-feet tall man would have to walk from the midpoint of the arch legs Let f ( x) before hitting his head on the arch. approximately 52.3 feet. This yields a distance of 12 19 feet, or Problem 3 If f is an odd function, then f ( x) f ( x) for any x in the domain of f . Since y f ( x) f ( x) f ( x) , we conclude that the function defined by y f ( x) is necessarily even (so its graph is symmetrical with respect to the y-axis) whenever f is odd. Problem 4 1 2 4 7 14 28 are 1, 2, 4, 7, 14, 28, , , , , , 3 3 3 3 3 3 according to the Rational Zero Theorem. a) The potential zeros of b) Factoring by grouping, we have: f f ( x) 3x3 7 x 2 12 x 28 x 2 3x 7 4 3x 7 3 x 7 x 2 4 . Therefore, irreducible. 7 is the only real zero of f . Note that the quadratic factor 3 x 2 4 is Problem 5 a) x 1 x 2 x 1 x3 1 Note that R( x) 2 is already in lowest terms. x 9 x 3 x 3 DR , the domain of R , is then given by DR x | x 3 (, 3) (3,3) (3, ) . b) The vertical asymptotes of R are the vertical lines x 3 and x 3 . GR , the graph of R , also has the oblique asymptote y x since R( x) c) x3 1 9x 1 2 x. 2 x 9 x 9 [Check this result using long division.] To find the point where GR intersects its non-vertical asymptote, solve for x in the equation R( x) x , or x3 1 x . We then have x3 1 x x 2 9 x 3 9 x , which 2 x 9 1 1 1 gives 1 9x , or x . The intersection point is then , . 9 9 9 [Check this result graphically!] Bonus Problem 1 The two centers of the circles are (2, 3) and ( 3, 2) . The equation of the line passing 1 13 through both of these points is then given by y x , or x 5 y 13 in standard form. 5 5 Bonus Problem 2 Let C be the point on the line passing through O and A (the line y 2 x ) which is located in Quadrant I such that OC OB 10 . Then C 2, 2 2 . Now, let M be the midpoint of 2 1 2 2 3 , line segment BC . Then M . The line passing through O and M is then 2 2 the bisector of angle AOB . This line has slope approximately equal to 14.1.