Appendices ( ) ( ) Appendix A: Equations and Inequalities 13. ( ) 1. Solve the equation 2x+ 7 = x + 8= x + 15.

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1 Appendices Appendi A: Equations and Inequalities. Solve the equation + = + = + = + = + = = 8 Moreover, replacing with 8 in + = yields a true statement. Therefore, the given statement is true.. The equations = and + = are not equivalent. The first has a solution set {, } while the solution set of the second is { }. Since the equations do not have the same solution set, they are not equivalent. The given statement is false. 5. B cannot be written in the form a + b =. A can be written as 5 = or 5 + ( ) =, C can be written as = or + =, and D can be written as.. or. +. =. =. 5+ = + = = 8 = 9. 6( ) = 8 ( ). 8 6 = = 8 6 = 8 = 8 = = = 5 + 8= + 8= = = Solution set: { } = = 6+ = 6+ 8 = = = 8 8. Solution set: { 8} 5. ( ) + + = + + = + = + = + = + = = +. ( ) = + ( ) = 5 ( ) = ( + ) 9. 5 = + 8 = 8 = 8 =..5 =.+ (..5) = (.+ ) 5= + 5= = = = 6+ = 6 + = 6 =. ( + ) = + + ( + ) 8+ 8= = = 8 = all real numbers identity; { } 5. ( ) 8 = 6 6= 6 6 = 6 = conditional equation; { } 8

2 Appendi A: Equations and Inequalities 9. ( + ) = ( + ) + ( + ) 9. (a) + 8= = = 6 contradiction; (c) = 5 =± 5 =± 5; E 5= = 5 =± 5; C = =± =± 5; A (d) 5= = 5; B (e) + 5= = 5; D. B is the only one set up for direct use of the square root property. ( + 5) = + 5= ± 5± = 5± = Solution set: { 5± } 5+ 6= = = = or = =, ( )( ) 5 = 5 + = ( )( ) 5 5+ = = or = = Solution set: { } 5, + = = = + ( )( ) + = = or = = Solution set: { }, = 6 =± 6 =± Solution set: { ± }. = = = ± = ± Solution set: { ± } + 5 = + 5= ± = ± = 5± Solution set: 5±. = = ± ± = ± = Solution set: { ± } + = Let a =, b =, c =. b± b ac = a ± = ( ) () ± 6 ± ± = = = = or Solution set: {, } = Let a =, b=, and c =. b± b ac = a ± = ( ) ( ) ( )( ) () ± + ± 5 = = Solution set: { ± 5 } 6 = 6+ = Let a =, b = 6,and c =. b± b ac = a ( 6) ± ( 6) ( ) 6± 6 8 = = () 6± 8 6± = = = ± Solution set: { ± }

3 Appendices = ( + ) = + = Let a =, b =, and c =. b± b ac ± = = a ± + 96 ± 9 = = Solution set: { ± 9 }. +.. = (. +..) = + = Let a =, b =,and c =. b± b ac = a ± ± 6+ = = ± ± ± = = = Solution set: { ± } ( )( + ) = + = + = Let a =, b =,and c =. b± b ac ± = = a ± 9+ ± = = 8 8 Solution set: { ± } (a) < 6 The interval includes all real numbers less than 6 not including 6. The correct interval notation is (, 6 ), so the correct choice is F. 6 The interval includes all real numbers less than or equal to 6, so it includes 6. The correct interval notation is (,6], so the correct choice is J. (c) < 6 The interval includes all real numbers from to 6, not including, but including 6. The correct interval notation is (, 6], so the correct choice is A. (d) The interval includes all real numbers between and, including and. The correct interval notation is [, ], so the correct choice is H. (e) 6 The interval includes all real numbers greater than or equal to 6, so it includes 6. The correct interval notation is 6,, so the correct choice is I. [ ) (f) 6 The interval includes all real numbers greater than or equal to 6, so it includes 6. The correct interval notation is [ 6, ), so the correct choice is D. (g) The interval shown on the number line includes all real numbers between and 6, including, but not including 6. The correct interval notation is [, 6), so the correct choice is B. (h) The interval shown on the number line includes all real numbers between and 8, not including or 8. The correct interval notation is (, 8), so the correct choice is G. (i) The interval shown on the number line includes all real numbers less than, not including, and greater than, not including. The correct interval notation is (, ) (, ), so the correct choice is E. (j) The interval includes all real numbers less than or equal to 6, so it includes 6, The correct interval notation is (, 6 ], so the correct choice is C , Solution set: ( ]

4 Appendi A: Equations and Inequalities , Solution set: [ ) 65. ( ) Solution set: (,6] < ( + ) < + 5+ < + + < + < < < < Solution set: (,) ( ) ( )( + 5) Solution set: 5, ) ( + ) 5 + 5( + ) ( ) Solution set: (,. 5< 5+ < 5 5< 5+ 5< 5 < < 6 6 < < 5< < 5, Solution set: Solution set: [, 6 ]. > + > > + > > > 8 8 < < < < 6, 6 9. Solution set: ( ) ( 5) ,9 Solution set: [ ]

5 Appendices Appendi B: Graphs of Equations.. 9. ( 5,.5) has a negative -coordinate and a positive y-coordinate. The point lies in quadrant II.. (.,.8) has a negative -coordinate and a negative y-coordinate. The point lies in quadrant III.. (a) Since (a, b) lies in quadrant II, a < and b >. Therefore, a > and ( a, b) lies in quadrant I. Since (a, b) lies in quadrant II, a < and b >. Therefore, a > and b <, so ( a, b) lies in quadrant IV. (c) Since (a, b) lies in quadrant II, a < and b >. Therefore, b < and (a, b) lies in quadrant III. (d) Since (a, b) lies in quadrant II, a < and b >. The point (b, a) has a positive - coordinate and a negative y-coordinate. Therefore, (b, a) lies in quadrant IV. 5. False. The epression should be ( ) ( y y ) +.. True. The midpoint has coordinates a+ a b+ ( b) a b,, = = ( a, b). 9. ( abc,, ) = ( 9,,5) a + b = 9 + = 8+ = 5 = 5 = c The triple is a Pythagorean triple since a + b = c.. ( abc,, ) = ( 5,,5) a + b = 5 + = 5 + = 5 5 = 5 = c The triple is not a Pythagorean triple since a + b c.. Let (, y ) = ( 5, 6 ). Let (, y ) = ( 5,) since we want the point on the -ais with -value 5. d = = + 6 = + 6 = 6 = 6 Thus, the point (5, 6) is 6 units from the -ais. 5. P( 5, ), Q(, ) (a) d( P, Q ) = [ ( 5)] + [ ( )] = = 8 = 8 The midpoint M of the segment joining points P and Q has coordinates 5 + ( ) + 8 6, =, = 9,.. P(8, ), Q (, 5) (a) d( P, Q ) = ( 8) + (5 ) = 5 + = = The midpoint M of the segment joining points P and Q has coordinates 8 + 5, + =,. 9. P ( 8, ), Q (, 5) (a) d( P, Q) = ( 8) + ( 5 ) = + ( 9) = + 8 = The midpoint M of the segment joining points P and Q has coordinates ( 5) 5, =,.

6 Appendi B: Graphs of Equations. P (, 5 ), Q (, 5) (a) d( P, Q) ( ) ( 5 5) ( ) ( 5 5) = + = + = 8+ 5 = The midpoint M of the segment joining points P and Q has coordinates ( 5), 5 5 =, =,.. Find such that the distance between (, ) and (, ) is 5. Use the distance formula and solve for : 5= + = + = = + 6 9= ± = = ± = or = 5 5. Find y such that the distance between (, y) and (, 9) is. Use the distance formula and solve for y: = + 9 y ( 5) ( 9 y) = + 5 ( 9 y) ( y) ( y) = + = = 9 ± 9 = 9 y y = 9± 9 y = or = 9 9. Let P (, y) = be a point 5 units from (, ). Then, by the distance formula, 5= + y 5= + y 5 = + y This is the required equation. The graph is a circle with center (, ) and radius The endpoints of the segment are (99,.) and (6, 8.) M =, = ( 998,.5) The estimate is.5%. This is close to the actual figure of.%. In eercises 5, other ordered pairs are possible.. (a) y y-intercept: = 6y = 6y = y = -intercept: y = 6 = = = = additional point. (a) y - and y-intercept: = additional point additional point

7 Appendices 5. (a) y - and y-intercept: = additional point 8 additional point. (a) Center (, ), radius 6 ( ) ( y ) ( ) ( y ) + = 6 + = 6 + y = 6 5. (a) Center (, ), radius ( ) ( y ) ( y ) + = + = 6 9. (a) Center (, ), radius 6 ( ) ( y ) ( ) ( y ) + = 6 + = 6 ( ) + y = 6 5. (a) Center (, 5), radius ( y ) + 5 = [ ( )] + ( y 5) = ( + ) + ( y 5) = The center of the circle is located at the midpoint of the diameter determined by the points (, ) and (5, ). Using the midpoint formula, we have C =, = (,). The radius is one-half the length of the diameter: r = ( 5 ) + ( ) = The equation of the circle is + y = 5. The center of the circle is located at the midpoint of the diameter determined by the points (, ) and (, ). Using the midpoint formula, we have + ( ) + C =, = (, ). The radius is one-half the length of the diameter: r = ( ) + = The equation of the circle is + + y =

8 Appendi C: Functions 5 Appendi C: Functions. The relation is a function because for each different -value there is eactly one y-value. This correspondence can be shown as follows.. Two ordered pairs, namely (, ) and,6, have the same -value paired with different y-values, so the relation is not a function. 5. The relation is a function because for each different -value there is eactly one y-value. This correspondence can be shown as follows.. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of corresponds to two values of y., ;, 9. domain: [ ] range: [ ] y = represents a function since y is always found by squaring. Thus, each value of corresponds to just one value of y. can be any real number. Since the square of any real number is not negative, the range would be zero or greater.. Two sets of ordered pairs, namely (, ) and (, ) as well as (, ) and (, ), have the same -value paired with different y-values, so the relation is not a function.,,,, domain: {,, } ; range: { } 9. The relation is a function because for each different -value there is eactly one y-value. domain: {,,5,, } ; range: {,, } domain: (, ); range: [, ). The ordered pairs (, ) and (, ) both satisfy = y 6. This equation does not represent a function. Because is equal to the sith power of y, the values of are nonnegative. Any real number can be raised to the sith power, so the range of the relation is all real numbers.. The relation is a function because for each different -value there is eactly one y-value. This correspondence can be shown as follows. Domain: {,, }; range: {,, }. This graph represents a function. If you pass a vertical line through the graph, one -value corresponds to only one y-value., ;, domain: ( ) range: 5. This graph does not represent a function. If you pass a vertical line through the graph, there are places where one value of corresponds to two values of y., ;, domain: [ ) range: domain: [, ) range: (, ). y = 5 represents a function since y is found by multiplying by and subtracting 5. Each value of corresponds to just one value of y. can be any real number, so the domain is all real numbers. Since y is twice, less 5, y also may be any real number, and so the range is also all real numbers. (continued on net page)

9 6 Appendices (continued from page 5) domain: (, ); range: (, ) 5. For any choice of in the domain of y =, there is eactly one corresponding value of y, so this equation defines a function. Since the quantity under the square root cannot be negative, we have. Because the radical is nonnegative, the range is also zero or greater. domain: [, ); range: [, ). For any choice of in the domain of y = + there is eactly one corresponding value of y, so this equation defines a function. Since the quantity under the square root cannot be negative, we have +. Because the radical is nonnegative, the range is also zero or greater. domain:, ) ; range: [, ) 9. Given any value in the domain of y =, we find y by subtracting, then dividing into. This process produces one value of y for each value of in the domain, so this equation is a function. The domain includes all real numbers ecept those that make the denominator equal to zero, namely =. Values of y can be negative or positive, but never zero. Therefore, the range will be all real numbers ecept zero.. B domain: (,) (, ); range: (,) (, ). f f = + = + = + = 5. g = + + g. 9. = + + = = f = + f = + = + = g = + + g = + +. ( ) ( ) ( ) = + + = f = + f p = p+. f = + f ( ) ( ) + = + + = 6+ = 5. (a) f =. (a) f = 5 9. (a) f = f ( ) f = f = =

10 Appendi D: Graphing Techniques 5. (a) f ( ) = f = (c) f ( ) = (d) f = 5. (a) f = (c) Since f =, a = f 6 = 9 (d) Since f f f = = 8 =, =,,8 (e) We have ( 8, ( 8) ) ( 8,) (, ( ) ) (, ). f = and f = Using the distance formula, we have d = 8 + = + = + = 55. (a) [, ) (, ] (c) [, ] 5. (a) (,] [, ) (c) none 59. (a) none (, ]; [, ) (c) (,) 6. (a) Yes, it is the graph of a function. [, ] (c) When t = 8, y = from the graph. At 8 A.M., approimately megawatts is being used. (d) The most electricity was used at hr or 5 P.M. The least electricity was used at A.M. (e) f ( ) 9; At noon, electricity use is about 9 megawatts. (f) increasing from A.M. to 5 P.M.; decreasing from midnight to A.M. and from 5 P.M. to midnight Appendi D: Graphing Techniques. (a) B; y = ( ) is a shift of y = units to the right. (c) F; y = is a vertical stretch of y = by a factor of. (d) A; y = ( + ) is a shift of y = units to the left. (e) D; y = + is a shift of y = units upward. (f) C; y = is a vertical shrink of y =.. (a) B; y = + is a shift of y = units upward. 5. A; y = is a shift of y = units downward. (c) G; y = ( + ) is a shift of y = units to the left. (d) C; y = ( ) is a shift of y = units to the right. (e) F; y = is a vertical stretch of y = by a factor of. (f) D; y = is a reflection of y = across the -ais. (g) H; y = ( ) + is a shift of y = units to the right and unit upward. (h) E; y = ( + ) + is a shift of y = units to the left and unit upward. (i) I; y = ( + ) is a shift of y = units to the left and unit down. y = y = y = 6 6 E; y = is a shift of units downward. y = (continued on net page)

11 8 Appendices (continued from page ). y = y =. y = y = y = y = y = 9 y = 9. (a) y = f is a vertical shrinking of f, by a factor of. The point that corresponds to ( 8, ) on this translated function is ( ) = 8, 8,. y f = is a vertical stretching of f, by a factor of. The point that corresponds 8, on this translated function is to ( 8, ) = ( 8, 8 ). 5. (a) The point that is symmetric to (5, ) with respect to the -ais is (5, ). The point that is symmetric to (5, ) with respect to the y-ais is ( 5, ). (c) The point that is symmetric to (5, ) with respect to the origin is ( 5, ).. (a) The point that is symmetric to (, ) with respect to the -ais is (, ). The point that is symmetric to (, ) with respect to the y-ais is (, ). (c) The point that is symmetric to (, ) with respect to the origin is (, ).

12 Appendi D: Graphing Techniques 9 The result is the same as the original equation, so the graph is symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the origin only. 9.. y = + Replace with to obtain y = ( ) + = +. The result is the same as the original equation, so the graph is symmetric with respect to the y-ais. Since y is a function of, the graph cannot be symmetric with respect to the -ais. Replace with and y with y to obtain y = ( ) + y = + y =. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph is symmetric with respect to the y-ais only. + y = Replace with to obtain ( ) + y = + y =. The result is the same as the original equation, so the graph is symmetric with respect to the y-ais. Replace y with y to obtain + ( y) = + y = The result is the same as the original equation, so the graph is symmetric with respect to the -ais. Since the graph is symmetric with respect to the -ais and y-ais, it is also symmetric with respect to the origin y = + Replace with to obtain y = ( ) ( ) + y = + +. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-ais. Since y is a function of, the graph cannot be symmetric with respect to the -ais. Replace with and y with y to obtain y = ( ) ( ) + y = + + y =. The result is not the same as the original equation, so the graph is not symmetric with respect to the origin. Therefore, the graph has none of the listed symmetries. y = graph of y = unit downward. y = + graph of y = units upward.. y = Replace with to obtain y = ( ) y = ( ) y =. The result is not the same as the original equation, so the graph is not symmetric with respect to the y-ais. Replace y with y to obtain y = y =. The result is not the same as the original equation, so the graph is not symmetric with respect to the -ais. Replace with and y with y to obtain y = ( ) y = ( ) y = y =.

13 5 Appendices y =. graph of y = unit to the right. f = ( ) graph of y =, units to the right and units down. It is then stretched vertically by a factor of. 9. y = + graph of y = units to the left.. y = + graph of y = units to the left, and then units down. 5.. y = graph of y =, unit down. It is then stretched vertically by a factor of.. It is the graph of f = translated unit to the left, reflected across the -ais, and translated units up. The equation is y = It is the graph of f = translated units left, stretched vertically, and then translated units down. The equation is y = f() = + 5: Translate the graph of f( ) up units to obtain the graph of t = (+ 5) + = +. Now translate the graph of t() = + left units to obtain the graph of g = ( + ) + = + 6+ = +. (Note that if the original graph is first translated to the left units and then up units, the final result will be the same.). Answers will vary. There are four possibilities for the constant, c. i) c > c > The graph of F( ) is stretched vertically by a factor of c. ii) c > iii) c < iv) c < c < The graph of F is shrunk vertically by a factor of c. F is stretched vertically by a factor of c and reflected over the -ais. c > The graph of c < The graph of F is shrunk vertically by a factor of c and reflected over the -ais.