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Math Algebra ISE Intermediate Algebra

To draw the graph. Solution: The graphical points of the equation y 2 − 4 x 2 − 16 = 0 is ( 0 , 4 ) a n d ( 0 , − 4 ) Explanation: Given: The equation is y 2 − 4 x 2 − 16 = 0 Formula used: The equation is in its standard form of the hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 Calculation: Resolving equation in form of hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 We get , y 2 − 4 x 2 − 16 = 0 ⇒ y 2 − 4 x 2 = 16 ⇒ y 2 16 − 4 x 2 16 = 1 D i v i d i n g e a c h t e r m b y 16 ⇒ y 2 16 − x 2 4 = 1 − − − − − − − ( 1 ) ⇒ y 2 4 2 − x 2 2 2 = 1 W r i t t e n i n t h e s tan d a r d f o r m o f h y p e r b o l a . As compared with equation y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 , We get value of a = 4 a n d b = 2 . c e n t r e = ( 0 , 0 ) v e r t i c e s = ( 0 , ± a , ) = ( 0 , ± 4 , ) f o c i = ( 0 , ± a e ) = ( 0 , ± 4 × 20 4 ) w h e r e e = a 2 + b 2 a = 16 + 4 4 = 20 4 = ( 0 , ± 20 ) = ( 0 , ± 4.4 ) ?Transverse axis of the equation y 2 − 4 x 2 − 16 = 0 is vertical. Now for labelling graph, we have to find out asymptotes. For y-intercepts, we letting x = 0 in the equation, and we have y 2 16 = 1 f r o m ( 1 ) s o l v i n g g i v e s u s y = − 4 a n d x = 4 Now, we have an east-west opening hyperbola,so the slopes of the asymptotes will be given ± a b = ± 4 2 = ± 2 The equation for the asymptotes, since they pass through (0, 0) are given by , x = − 2 y a n d x = 2 y for this equation we find out points as- x = 0 ⇒ y = 0 x = 4 ⇒ y = 2 x = 8 ⇒ y = 4 Now we draw graph as- Where vertices are ( 0 , 4 ) a n d ( 0 , − 4 ) Asymptotes crosses through centre (0, 0) Conclusion: Using the standard formula of hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 The required equation y 2 − 4 x 2 − 16 = 0 shows transverse axis = vertical.

To draw the graph. Solution: The graphical points of the equation y 2 − 4 x 2 − 16 = 0 is ( 0 , 4 ) a n d ( 0 , − 4 ) Explanation: Given: The equation is y 2 − 4 x 2 − 16 = 0 Formula used: The equation is in its standard form of the hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 Calculation: Resolving equation in form of hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 We get , y 2 − 4 x 2 − 16 = 0 ⇒ y 2 − 4 x 2 = 16 ⇒ y 2 16 − 4 x 2 16 = 1 D i v i d i n g e a c h t e r m b y 16 ⇒ y 2 16 − x 2 4 = 1 − − − − − − − ( 1 ) ⇒ y 2 4 2 − x 2 2 2 = 1 W r i t t e n i n t h e s tan d a r d f o r m o f h y p e r b o l a . As compared with equation y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 , We get value of a = 4 a n d b = 2 . c e n t r e = ( 0 , 0 ) v e r t i c e s = ( 0 , ± a , ) = ( 0 , ± 4 , ) f o c i = ( 0 , ± a e ) = ( 0 , ± 4 × 20 4 ) w h e r e e = a 2 + b 2 a = 16 + 4 4 = 20 4 = ( 0 , ± 20 ) = ( 0 , ± 4.4 ) ?Transverse axis of the equation y 2 − 4 x 2 − 16 = 0 is vertical. Now for labelling graph, we have to find out asymptotes. For y-intercepts, we letting x = 0 in the equation, and we have y 2 16 = 1 f r o m ( 1 ) s o l v i n g g i v e s u s y = − 4 a n d x = 4 Now, we have an east-west opening hyperbola,so the slopes of the asymptotes will be given ± a b = ± 4 2 = ± 2 The equation for the asymptotes, since they pass through (0, 0) are given by , x = − 2 y a n d x = 2 y for this equation we find out points as- x = 0 ⇒ y = 0 x = 4 ⇒ y = 2 x = 8 ⇒ y = 4 Now we draw graph as- Where vertices are ( 0 , 4 ) a n d ( 0 , − 4 ) Asymptotes crosses through centre (0, 0) Conclusion: Using the standard formula of hyperbola y 2 a 2 − x 2 b 2 = 1 , w h e r e a , b > 0 The required equation y 2 − 4 x 2 − 16 = 0 shows transverse axis = vertical.

ISE Intermediate Algebra

ISE Intermediate Algebra

6th Edition

ISBN: 9781264418756

Author: Miller, Julie, O'Neill, Molly, Hyde, Nancy

Publisher: McGraw Hill

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Chapter 9 Solutions

ISE Intermediate Algebra

Ch. 9.1 - Prob. 1PE Ch. 9.1 - Prob. 2PE Ch. 9.1 - Prob. 3PE Ch. 9.1 - Prob. 4PE Ch. 9.1 - Prob. 5PE Ch. 9.1 - Prob. 6PE Ch. 9.1 - Prob. 7PE Ch. 9.1 - Prob. 8PE Ch. 9.1 - Prob. 9PE Ch. 9.1 - Prob. 10PE

Ch. 9.1 - Prob. 11PE Ch. 9.1 - Prob. 12PE Ch. 9.1 - Prob. 13PE Ch. 9.1 - Prob. 14PE Ch. 9.1 - Prob. 15PE Ch. 9.1 - Prob. 16PE Ch. 9.1 - Prob. 17PE Ch. 9.1 - Prob. 18PE Ch. 9.1 - Prob. 19PE Ch. 9.1 - Prob. 20PE Ch. 9.1 - Prob. 21PE Ch. 9.1 - Prob. 22PE Ch. 9.1 - Prob. 23PE Ch. 9.1 - Prob. 24PE Ch. 9.1 - Prob. 25PE Ch. 9.1 - Prob. 26PE Ch. 9.1 - Prob. 27PE Ch. 9.1 - Prob. 28PE Ch. 9.1 - Prob. 29PE Ch. 9.1 - Prob. 30PE Ch. 9.1 - Prob. 31PE Ch. 9.1 - Prob. 32PE Ch. 9.1 - Prob. 33PE Ch. 9.1 - Prob. 34PE Ch. 9.1 - Prob. 35PE Ch. 9.1 - Prob. 36PE Ch. 9.1 - Prob. 37PE Ch. 9.1 - Prob. 38PE Ch. 9.1 - Prob. 39PE Ch. 9.1 - Prob. 40PE Ch. 9.1 - Prob. 41PE Ch. 9.1 - Prob. 42PE Ch. 9.1 - Prob. 43PE Ch. 9.1 - Prob. 44PE Ch. 9.1 - Prob. 45PE Ch. 9.1 - Prob. 46PE Ch. 9.1 - Prob. 47PE Ch. 9.1 - Prob. 48PE Ch. 9.1 - Prob. 49PE Ch. 9.1 - Prob. 50PE Ch. 9.1 - Prob. 51PE Ch. 9.1 - Prob. 52PE Ch. 9.1 - Prob. 53PE Ch. 9.1 - Prob. 54PE Ch. 9.1 - Prob. 55PE Ch. 9.1 - Prob. 56PE Ch. 9.1 - Prob. 57PE Ch. 9.1 - Prob. 58PE Ch. 9.1 - 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