Pg The perimeter is P = 3x The area of a triangle is. where b is the base, h is the height. In our case b = x, then the area is


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1 Pg. 9. The perimeter is P = The area of a triangle is A = bh where b is the base, h is the height 0 h= btan 60 = b = b In our case b =, then the area is A = = 0. By Pythagorean theorem a + a = d a a = d d = a = d a = d a = d The area of the square is d d A= a = = = d
2 . From the triangle ABC by Pythagorean theorem AC = AB + BC = a + a = a From the triangle ACD by Pythagorean theorem AD = AC + CD = a + a = a Thus, d = a d a = d a = d a = d a = The surface area is d d A 6a 6 = = = 6 = d The volume is d d d V = a = = = 7 9. (a) y = This equation can be rewritten as y, y< 0 = y, y 0 The first line passes through the points (, ) and (5, 5). The second line passes through the points (, ) and (5, 5). The graph is not graph of a function because we can draw a vertical line that intersects it more than once.
3 (b) y = This equation can be rewritten as y = 0 (y )(y + ) = 0 y = 0 or y + = 0 y = or y = The graph is not graph of a function because we can draw a vertical line that intersects it more than once.. (a) + y = In the Quadrant I > 0 and y > 0, so we have + y = y = + This is a line with the slope m = and the yintercept (0, ). In the Quadrant II < 0 and y > 0, so we have + y = y = + This is a line with the slope m = and the yintercept (0, ). In the Quadrant III < 0 and y < 0, so we have y = y = This is a line with the slope m = and the yintercept (0, ). In the Quadrant IV > 0 and y < 0, so we have y = y = This is a line with the slope m = and the yintercept (0, ). The graph is not graph of a function because we can draw a vertical line that intersects it more than once.
4 (b) + y = We have two equations: + y = and + y = or y = + and y = The first graph is a line with the slope m = and the yintercept (0, ). The second graph is a line with the slope m = and the yintercept (0, ). The graph is not graph of a function because we can draw a vertical line that intersects it more than once. 7., F( ) = +, > When then F() =, so the part of the graph of F that lies to the left of the vertical line = must coincide with the parabola y =. The coordinate of the verte is b 0 = = = 0 a ( ) Find y for = 0: y = 0 =. The verte is (0, ). Find few points: y = 0 5 When > then F() = +, so the part of the graph of F that lies to the right of the vertical line = must coincide with the parabola y = +. The coordinate of the verte is b = = = a Find y for = : y = + () =
5 The verte is (, ). Find few points: y = , < 0 G ( ) =, > 0 When < 0 then G() = /, so the part of the graph of F that lies to the left of the vertical line = 0 must coincide with the graph of y = /. Find few points: y = / When > 0 then G() =, so the part of the graph of F that lies to the right of the vertical line = 0 must coincide with the graph of y =. This line have slope m = and yintercept (0, 0).
6 9. (a) On the interval [0, ] the line passes through the point (0, 0) and (, ), so the equation of the graph on this interval is y =. On the interval [, ] the line passes through the point (, ) and (, 0). The slope is y y 0 m = = = The equation of the line is y y = m( ) y = ( ) y = + y = + The equation of the function is, 0 < f( ) = +, (b) The equation of the function is, 0 < 0, < f( ) =, < 0,
7 0. (a) On the interval (0, ] the line passes through the point (0, ) and (, 0). The slope is y y 0 m = = = 0 The equation of the line is y y = m( ) y = ( 0) y = + On the interval (, 5] the line passes through the point (, ) and (5, 0). The slope is y y 0 m = = = 5 The equation of the line is y y = m( ) y 0 = ( 5) 5 y = + The equation of the function is +, 0 < f( ) = 5 +, < 5 (b) On the interval (, 0] the line passes through the point (, 0) and (0, ). The slope is y y 0 m = = = 0 ( ) The equation of the line is y y = m( ) y 0 = ( ( )) y = On the interval (0, ] the line passes through the point (0, ) and (, ). The slope is y y m = = = 0 The equation of the line is y y = m( ) y = ( 0) y = + The equation of the function is, 0 f( ) = < +, 0 <
8 Pg.9. f() =, g( ) =, h() =, j() = (a) y = y = ( f o g)( ) (b) y = y = ( jo g)( ) (c) y = y = ( go g)( ) (d) y = y = ( jo j)( ) (e) y = ( ) ( ho f)( ) = ( ) y = ( goho f)( ) (f) y = ( 6) ( jo f)( ) = ( 6) y = ( ho jo f)( ). (a) y = y = ( f o j)( )
9 (b) y = y = ( ho g)( ) (c) (d) y = 9 y = ( ho h)( ) y = 6 h ( ) = f( ) = 6 y = ( f o h)( ) (e) y = ( go f)( ) = y = ( hogo f)( ) (f) y = ( f o h)( ) = y = ( go f oh)( ) 7. y =, left, down If c > 0 then y = f( + c) shifts the graph of y = f() a distance c units to the left y = f() c shifts the graph of y = f() a distance c units downward Therefore, y = ( + ) y = ( + ) shifts the graph of y = one unit to the left shifts the graph of y = ( + ) one unit downward Find few points: 0 y =
10 8. y =, right, down If c > 0 then y = f( c) shifts the graph of y = f() a distance c units to the left y = f() c shifts the graph of y = f() a distance c units downward Therefore, y = ( ) shifts the graph of y = ( ) shifts the graph of Find few points: y = one unit to the right y= y = ( ) one unit downward
11 5. y = The graph of y = is the graph of y = moved by unit downward. shifted by unit to the right and then Find few points: y = 0
12 6. y = ( + ) / + The graph of y = ( + ) / + is the graph of y = / shifted by unit to the left and then moved by unit upward. Find few points: / y = y = +, compressed vertically by a factor of If c > then y = f ( ) compress the graph of y = f( ) vertically by a factor of c. c
13 Then y = + y = y = +, stretched horizontally by a factor of If c > then y = f stretch the graph of ( ) c y = f horizontally by a factor of c. Then y = + y = y = + Pg.8 9. π cos =,,0 + = sin = cos sin cos sin =± cos Since is in Quadrant IV then sin = cos 8 8 sin = = = = = 9 9 sin tan = = = cos
14 5 π 0. cos =,, π + = sin cos sin = cos sin =± cos Since is in Quadrant II then sin = cos 5 5 sin = = = = sin tan = = = cos 5 5. tan =, π π, sin + cos = sin cos tan + = cos cos + = = cos + tan cos =± tan + cos =± tan + Since is in Quadrant III then cos = tan +
15 5 cos = = = = = = sin = tan cos = = 5 5 π 9. y = cos π The graph of y = cos is the graph of cos Find the yintercept: π π y = cos 0 = cos = 0 The yintercept is (0, 0). Find the intercepts: π cos = 0 y = shifted by π units to the right. π π = + π k k Z = π + π k The intercepts are ( π + π k, 0). Find few points: 0 π 6 y = cos π π π π π π 00 y = cos is an even function so its graph is symmetric about the yais.
16 The period is π. π 0. y = sin + 6 π The graph of y = sin + is the graph of 6 Find the yintercept: π π y = sin 0 + = sin = 6 6 The yintercept is 0,. Find the intercepts: π sin + = 0 6 π + = π k k Z 6 π = + π k 6 π The intercepts are + π k,0 6. Find few points: π y = sin shifted by units to the left. 6 0 π 6 π π π π π π
17 y = sin y = sin is an odd function so its graph is symmetric about the origin. The period is π. π. cos = sin ( ) cos α β = cosαcos β + sinαsin β Transform the left side: π π π cos = cos cos + sin sin = cos 0 + sin = sin π. cos + = sin
18 7. ( ) cos α + β = cosαcos β sinαsin β Transform the left side: π π π cos + = cos cos sin sin = cos 0 sin = sin π cos 8 + cosα cos α = π + cos + π + cos = = = 8 8. π cos 8 + cosα cos α = 5π π cos cos π π + + π cosπ cos sinπsin 5π cos = = = ( ) + 0 = = = 5. sin θ cosθ = 0
19 sin θ cosθ = 0 sinθcosθ cosθ = 0 cos θ(sinθ ) = 0 cosθ = 0 or sinθ = 0 cosθ = 0 or sinθ = π m θ = + πk k Z or θ = ( ) sin + mπ m Z π m π θ = + πk or θ = ( ) + πm 6 5. cos θ + cosθ = 0 cos θ sin θ + cosθ = 0 θ cos θ ( cos θ) + cos = 0 cos θ + cos θ + cosθ = 0 cos θ + cosθ = 0 cos θ + cosθ = 0 cos θ + cosθ cosθ = 0 cos θ(cosθ + ) (cosθ + ) = 0 (cosθ + )( cosθ ) = 0 cosθ + = 0 or cosθ = 0 cosθ = or cosθ = cosθ = or cosθ = θ = π + πk k Z or θ =± cos + πm m Z π θ = π + πk or θ =± + πm Pg.6. ht () = cott
20 The average rate of change of a function y = f() over the interval [a, b] is π π (a), π π cot cot = = = π π π π π π π (b), 6 π π cot cot 6 0 = = = π π π π π 6 6 f ( b) f( a) b a.. gt () = + cost The average rate of change of a function y = f() over the interval [a, b] is (a) [ ] 0,π + cos π (+ cos0) + ( ) (+ ) = = = π 0 π π π (a) [ π, π ] + cos π (+ cos( π )) + ( ) ( + ( )) = = = 0 π ( π) π π f ( b) f( a) b a. 8. Eample is needed 9. Eample is needed. Eample is needed. Eample is needed Pg.7
21 + lim( + ) lim = = = + 6 lim( + 6) lim s(s ) = lim lims lim(s ) s / s / s / s / = = = =.. 7. lim = = h 0 lim h 0 h+ + lim( h+ + ) lim h+ + lim h 0 h 0 h 0 = = = lim( h) + lim lim h 0 h 0 5h+ ( 5h+ )( 5h+ + ) = lim h h( 5h+ + ) h 0 h 0 5h+ 5h = lim = lim h 0 h( 5h+ + ) h 0 h( 5h+ + ) 5 5 = = lim h 0 5 h = = + t + t t + t t t( t+ ) ( t+ ) lim = lim = lim t t t t t ( t )( t+) ( t+ )( t ) t+ = lim = lim t ( t )( t+ ) t t+ + = = + 8.
22 t + t+ t + t+ t+ t( t+ ) + ( t+ ) lim = lim = lim t t t t+ t t( t ) + ( t ) t t t ( t+ )( t+ ) t+ + = lim = lim = t ( t )( t+ ) t t =. ( )( ) ( )( )( + ) u ( u )( u + u+ ) ( u )( u + u+ ) u u u + u u+ u lim = lim = lim u u u ( u+ )( u + ) (+ )( + ) lim u = u + u+ = + + =. 5. v 8 ( v )( v + v+ ) ( v )( v + v+ ) lim = lim = lim v v 6 v ( v )( v ) v + ( v )( v+ )( v + ) v + v+ + + = = lim v ( )( ) ( )( v+ v ) = = 8 lim sec = lim = = = 0 0 cos cos 0. 5 ( 5)(+ 5) lim = lim + ( + )(+ 5) ( 5) + 5 = lim = lim ( + )(+ 5) ( + )(+ 5)
23 9 ( ) ( +) = lim = lim ( + )(+ 5) ( + )(+ 5) ( ) = = lim 5 ( ) = = = + 9. lim + cos( + π ) = lim + lim cos( + π ) π π π = π + cos( π + π) = π cos0 = π = π 56. lim p ( ) =, lim r ( ) = 0, lim s ( ) = (a) lim( p ( ) + r ( ) + s ( )) = lim p ( ) + lim r ( ) + lim s ( ) = ( ) = (b) lim( p ( ) r ( ) s ( )) = lim p ( ) lim r ( ) lim s ( ) = 0 ( ) = 0 (c) ( p ( ) + 5 r ( )) lim( p( ) + 5 r( )) lim( p( )) + lim(5 r( )) lim = = s ( ) lim s ( ) lim p ( ) + 5 lim r ( ) = = = 57. f() =, = f ( + h) f( ) ( + h) + h+ h m ( ) = lim = lim = lim h 0 h h 0 h h 0 h h + h = lim = lim( + h) = h 0 h h 0 At = we have m () = () =
24 58. f() =, = f ( + h) f( ) ( + h) + h+ h m ( ) = lim = lim = lim h 0 h h 0 h h 0 h h + h = lim = lim( + h) = h 0 h h 0 At = we have m( ) = ( ) = 6. f ( ) =, = 7 f ( + h) f( ) + h ( + h )( + h+ ) m ( ) = lim = lim = lim h 0 h h 0 h h 0 h( + h+ ) + h h = lim = lim = lim h 0 h( + h+ ) h 0 h( + h+ ) h 0 + h+ = = + At = 7 we have 7 m (7) = = 7 6. f ( ) = +, = 0 f( + h) f( ) ( + h) + + m ( ) = lim = lim h 0 h h 0 h ( ( + h) + + )( ( + h) ) = lim h 0 h( ( + h) ) ( + h) + (+ ) + h+ = lim = lim h 0 h( ( + h) ) h 0 h( ( + h) ) h = lim = lim h 0 h( ( + h) ) h 0 ( + h) = = At = 0 we have m (0) = = 0+
25 Pg.8 9. It follows from the graph that 9 if < < 5 5 then < < The interval, is not symmetric about = : = = 6 6 We can choose δ to be the smaller of these numbers, that is 7 δ = It follows from the graph that if.6 < <. then.8 < + <. The interval (.6,.) is not symmetric about = :. = 0..6 = 0.9 We can choose δ to be the smaller of these numbers, that is δ = 0.9
26 0. f( ) = 7, L =, 0 =, ε = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : 7 < < 7 < < 7 < 5 9< 7< 5 6 < < Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 6 < < : = 9 6 = 7 The distance from 0 = to the nearer endpoint of (6, ) is 7. If we take δ = 7 or any smaller positive number, then 0 < < 7 7 <. f( ) =, L =, 0 =, ε = 0.05 Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : < < < < < 0. 0 < < 5 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 0 < < 5 : 5 = 0 =
27 The distance from 0 = to the nearer endpoint of 0,5 is. If we take δ = or any smaller positive number, then 0 < < < f( ) = 5, L =, 0 =, ε = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : 5 < < 6< 5 < < 7 5 < < 7 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 5 < < 7 : The distance from 0 = to the nearer endpoint of ( ) take δ = 7 or any smaller positive number, then 0 < < 7 5 = 6 < 5, 7 is 7. If we. f( ) =, 0 =, ε =0.05 ( )( + ) L= lim f( ) = lim = lim = lim( + ) = + = 0 Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : < < < 0.05
28 .95 < <.05 ( )( + ).95 < < < + < < <.05 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval.95 < <.05:.05 = = 0.05 The distance from 0 = to the nearer endpoint of (.95,.05) is If we take δ = 0.05 or any smaller positive number, then 0 < < 0.05 < f( ) = , 0 = 5, ε = ( + ) + 5( + ) L= lim f( ) = lim = lim = lim ( + )( + 5) = lim = lim( + ) = 5 + = Step : Solve the inequality f( ) L < ε to find the interval about 0 = on which the inequality holds for all 0 : ( ) < < + < ( + 5)( + ).05 < < < + < < <.95 Step : Find a value of δ > 0 that places the centered interval 0 δ < < 0 + δ inside the interval 5.05 < <.95: 5 ( 5.05) = ( 5) = 0.05
29 . lim = The distance from 0 = 5 to the nearer endpoint of ( 5.05,.95) is If we take δ = 0.05 or any smaller positive number, then < + 5 < < Given ε > 0, we want to find δ > 0 such that if 0 < < δ then < ε We have = = < ε We find a positive constant C such that < C < C and we can make ε C < ε by taking < = δ C We restrict to lie in the interval <, then < < < < < ε So C = is suitable and we should choose δ = min,. Show that δ works. ε Given ε > 0, we let δ = min,. If 0 < < δ then < < < < Also ε < ε < ε < so ε = < = ε Therefore, lim =
30 . lim = Given ε > 0, we want to find δ > 0 such that if 0 < < δ then < ε We have ( ) ( + ) ) + = = = < ε Let s assume that δ. This is a valid assumption to make since, in general, once we find a δ that works, all smaller values of δ also work. Then < δ implies that < < + < < + + < + < + Then ) + < ) (+ ) < ε ε < ( + ) ε Now choose δ = min, ( +. This guaranties that both assumptions made about δ in ) the course of this proof are taken into account simultaneously. Thus, if 0 < < δ, it follows that < ε. This completes the proof.
31 Pg (a), f( ) = 0, = Find the intercept: = 0 = 0 The intercept is (0, 0). Find few points: y = (b) lim f( ) = lim f( ) = +
32 (c) Since lim f ( ) = lim f( ) then lim f( ) = + lim f ( ) eists: 8., f( ) =, = (a) Find the intercepts: = 0 ( )( + ) = 0 = 0 or + = 0 = or = The intercept is (, 0) (the point (, 0) is not on the graph of f). Find the yintercept: f(0) = 0 = The yintercept is (0, ). The coordinate of the verte is b 0 = = = 0 a ( ) The verte is (0, ). Find few points: y =
33 (b) lim f( ) = 0 lim f( ) = 0 + (c) Since lim f( ) = 0 lim f ( ) = lim f( ) then + lim f ( ) eists: lim = = = lim As, we have ( ) 0 and ( + ),so lim = (+ 5) (+ 5) lim lim lim + = + = ( + )( + ) ( + )( + ) + 5 ( ) + 5 = lim = = = + ( + ) ( + ) ( ) sin tan sin sin lim lim cos = = lim = lim cos 0 cos sin = lim lim = = 0 0 cos cos 0 = 6. t t tcost cost lim = lim = lim = lim t 0 tan t t 0 sin t t 0 sin t t 0 sin t cost t limcost t 0 cos0 = = = sin t lim t 0 t
34 9. sin sin tan cos sin lim lim lim lim = 0 sin 8 0 sin 8 = 0 8 sin8 = cos 0 8 sin8 8 cos sin sin sin lim lim = = 0 8 sin8 lim cos 0 8 sin8 = cos 8 0 sin8 cos sin sin lim 0 lim lim = 8 0 sin8 = 0 cos 8 sin8 lim cos = = cos5y sin ycot 5y sin y cot 5y sin y sin 5y lim = lim = lim y 0 ycot y y 0 y cot y y 0 y cos y sin y sin y cos5y sin y sin y sin y cos5y = lim = lim y 0 y sin5y cosy y 0 y sin5y cosy sin y y sin y sin y cos5y y cos 0 = lim lim lim = lim y 0 y y 0sin 5y y 0cos y y 0 sin 5y 5y cos0 5y sin y sin y sin y y y y y = lim = lim = lim y 0 5y sin5y y 0 5 sin5y 5 y 0 sin5y 5y 5y 5y sin y lim y 0 y = = = 5 sin 5y lim 5 5 y 0 5 y
35 . sinθ tanθ sin sin sin sin lim lim cosθ θ θ θ θ = = lim = lim θ cot θ cos θ θ θ θ θ cosθ cos θ θ 0 θ θ θ cosθ cos θ sin θ sinθ sin θ sin θ = lim lim lim = lim θ 0 θ θ 0 θ θ 0cosθcosθ θ 0 θ cos 0cos 0 = = Pg. 0. y = The first term is a rational function so it is continuous everywhere ecept = (it is not defined at = ), the second term is a polynomial function so it is continuous everywhere. Therefore, y is continuous on the intervals (, ) (, ).. y = ( + ) + The first term is a rational function so it is continuous everywhere ecept = (it is not defined at = ), the second term is a constant function so it is continuous everywhere. Therefore, y is continuous on the intervals (, ) (, ) y = + Find the domain: + = 0 + = 0 ( ) ( ) = 0 ( )( ) = 0 = or = The domain is the intervals (, ) (, ) (, ). y is a rational function so it is continuous on its domain or on the intervals (, ) (, ) (, ).
36 5. π π π lim cos cos cos t 0 = = 9 sec t 9 sec 0 9 π π = cos = cos = 6 π π π π cos = cos = cos = cos = 9 sec0 9 6 Since lim f ( t) = f(0) then the function is continuous at t = 0. t 0 6. lim csc + 5 tan = + 5 = + 5 = π 6 csc + 5 tan = + 5 = + 5 = π Since lim f( ) = f π 6 then the function is continuous at π = lim sin π e sin π e sin π + = = 0 = π 0 π sin e = sin = Since lim f( ) = f 0 then the function is continuous from the right at = ( ) 9. 9 g ( ) = Find onesided limits at = : 9 ( )( + ) lim = lim = lim( + ) = + = 6 9 ( )( + ) lim = lim = lim( + ) = + = Thus, 9 lim = 6 To be continuous at = we must define g() = 6.
37 t 0. ht () = + t 0 t Find onesided limits at t = : t + t 0 t + 5t t 0 t( t+ 5) ( t+ 5) lim = lim = lim t t t t t t ( t+ 5)( t ) = lim = lim( t + 5) = + 5 = 7 t t t t + t 0 t + 5t t 0 t( t+ 5) ( t+ 5) lim = lim = lim t t t t t t ( t+ 5)( t ) = lim = lim( t + 5) = + 5 = t t t Thus, t + t 0 lim = 7 t t To be continuous at t = we must define h() = 7.. s f() s = s Find onesided limits at s = : s ( s )( s + s+ ) s + s+ + + lim = lim = lim = = s s s ( s )( s+ ) s s+ + s ( s )( s + s+ ) s + s+ + + lim = lim = lim = = s s s ( s )( s+ ) s s+ + Thus, s lim s = s To be continuous at s = we must define f () = = 0 Consider the function f() = 5 +. f( ) = ( ) 5( ) + = f( ) = ( ) 5( ) + = 9 f( ) = ( ) 5( ) + = f( ) = ( ) 5( ) + = 5 f(0) = 0 5(0) + = f() = 5() + = f() = 5() + =
38 f() = 5() + = 7 f() = 5() + = 5 f is a polynomial function, so it is continuous everywhere. f changes its sign on the intervals (, ), (0, ) and (, ). By the Intermediate Value Theorem f must have three zeros on these intervals. 56. F( ) = ( a) ( b) + Fa ( ) = ( a a) ( a b) + a= 0+ a= a Fb ( ) = ( b a) ( b b) + b= 0+ b= b a+ b The number lies between a and b, so by the Intermediate Value Theorem there eists a a+ b number c between a and b such that f() c =. Pg f( ) = (a) (b) lim lim lim = = = = lim lim lim + = = = = h ( ) = (a) 7 7 lim lim lim = = = =
39 (b) = = = lim lim lim = 7 9. g ( ) = (a) (b) lim = lim = lim = = lim = lim = lim = = h ( ) = (a) (b) lim = lim = lim = = lim = lim = lim = =
40 . h ( ) = (a) (b) lim lim = = lim = = lim lim = = lim = = lim lim lim = = = = = lim lim lim lim lim = = = = = = lim = lim = lim = lim =. lim = 7 ( 7) because ( 7) 0 as 7.. lim = 0 ( + ) because 0 and + as 0.
41 55. (a) lim = lim lim = 0 lim = (b) lim = lim lim = 0 lim = ( ) 8 (c) lim 0 = = = = = (d) ( ) lim = = + = 56. (a) (b) (c) (d) lim = lim = ( + ) + because and as. + lim = lim = + ( + ) because and as. + ( )( + ) lim = lim = because 0, + and + 6 as lim = = ( )( ) lim ( ) = lim ( + ) + 9 = lim = lim = lim = because as.
42 8. ( ) ( + 5 )( + 5+ ) lim + 5 = lim ( ) = lim = lim = lim = because as. 0. y = The domain is the intervals (, ) (, ). ( ) y( ) = = + The function is neither odd nor even. Find the intercepts: = 0 = 0 ( )( + ) = 0 = or = The intercepts are (, 0) and (, 0). Find the yintercept: 0 y(0) = = 0 The yintercept is (0, ). The denominator is zero at =, so the line = is a vertical asymptote. To find the equation of a slant asymptote we use the long division: +
43 Thus, = + The slant asymptote is y = + Find few points: y =
44 0. y = + The domain is the intervals (, ) (, ). ( ) y( ) = = ( ) + + The function is neither odd nor even. Find the intercepts: = 0 + = 0 ( )( + ) = 0 = or = The intercepts are (, 0) and (, 0). Find the yintercept: 0 y(0) = = (0) + The yintercept is 0,. The denominator is zero at =, so the line = is a vertical asymptote. To find the equation of a slant asymptote we use the long division: + + Thus, = The slant asymptote is y =. Find few points: y = +
45
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