EIT ACADEMY MATH BY JUSTIN DICKMEYER, PE OF ENGINEERINTRAININGEXAM.COM

EIT ACADEMY MATH BY JUSTIN DICKMEYER, PE OF ENGINEERINTRAININGEXAM.COM The informtion contined in EIT ACADEMY, is ment to serve s comprehensive collection of subjects nd problems tht the uthor of this course hs pplied to substntilly increse success on the Engineer in Trining Exm. Summries, strtegies, tips nd tricks re only recommendtions by the uthor, nd completing this course does not gurntee tht one s results will exctly mirror our own results. The uthor of this exm hs mde ll resonble efforts to provide current nd ccurte informtion for the reders of this course. The uthor will not be held lible for ny unintentionl errors or omissions tht my be found. The mteril in EIT ACADEMY my include links to other informtion, products, or services by third prties. Third Prty mterils comprise of the products nd opinions expressed by their owners. As such, the uthor of this guide does not ssume responsibility or libility for ny Third Prty Mteril or opinions. The publiction of such Third Prty mterils does not constitute the uthors gurntee of ny informtion, instruction, opinion, products or service contined within the Third Prty Mteril. Use of recommended Third Prty Mteril does not gurntee tht your results, with EngineerInTriningExm.com will mirror our own. Publiction of such Third Prty Mteril is simply recommendtion nd expression of the uthors own opinion of tht mteril. Whether becuse of the generl evolution of the Internet, or the unforeseen chnges in compny policy nd editoril submission guidelines, wht is stted s fct t the time of this writing, my become outdted or simply inpplicble t lter dte. This my pply to the EngineerInTriningExm.com website pltform, s well s, the vrious similr compnies tht hve been referenced in this course, nd ny other complementry resource. Gret effort hs been exerted to sfegurd the ccurcy of this writing. Opinions regrding similr website pltforms hve been formulted s result of both personl experience, s well s the well documented experiences of others. 01 Dickmeyer Enterprises All Rights Reserve

A Note from Me to You: I wnt to begin by thnking you for your support of EngineerInTriningExm.com nd give you commendtions for tking inititive on crucil step in your preprtion for the EIT Exm, tking the EIT ACADEMY MATH course. I put together this course simply to help you focus on tking nd dominting the Engineer in Trining Exm in cost effective nd resourceful mnner. You my be fmilir with my story, but it wsn t too long go tht I myself begn the journey of becoming Registered Professionl Engineer. I ws motivted from the strt nd hd the desire to succeed, but found myself locked behind mind overwhelmed with the mssive tsk of studying for the Engineer in Trining exm. Looking for simple guidnce in prepring for the exm, I turned to the internet, only to find myself filing to uncover ny source of rel vlue dedicted to prepring for the EIT exm. With no time to spre, I delved forwrd on my journey creting my own pth to success. This course ws formulted s result of both personl experiences s well s well documented experiences of others, nd the purpose of it is simply to guide you. It is not system, nd definitely is not the only resource needed to tke nd pss the Engineer in Trining exm. However, it will llow you to complete the criticl tsk of reviewing nd working through criticl portion of the exm, Mth, reviewing concepts, reinforcing them through video, nd solving structured problems, further formulting your expecttions s you pproch test time. If I m ble to help just one person dominte the Engineer in Trining Exm, then I feel the time I hve spent to put together this course will be well worth it. Once you re complete with the course, you cn emil ny feedbck you my hve to Justin@EngineerInTriningExm.com, s I m lwys looking for better wys to serve my fellow engineers out there. Once gin, thnk you, I m honored nd grteful for your support. Good Luck! Justin Dickmeyer, PE

EIT ACADEMY MATH Anlyticl Geometry... 1 Lines... Qudrtic Formul... 9 Prbols... 19 Hyperbols... 7 Ellipses... 36 Circles... 4 Distnce Formul... 49 Lw of Sines... 56 Lw of Cosines... 63 Right Tringles... 69 Trigonometric Rtios... 74 Algebr... 8 Functions... 83 Power Functions... 90 Logrithms... 99 Complex Numbers... 106 Mtrix Addition nd Subtrction... 11 Mtrix Multipliction... 118 Mtrix Trnspose nd Determinte... 16 Elementry Row Opertions... 136 Mtrix Echelon Form... 144

Dot Product... 156 Cross Product... 163 Simultneous Liner Equtions... 169 Arithmetic Sequences nd Series... 183 Geometric Sequences nd Series... 19 Power Series... 198 Tylor nd Mclurin Series... 05 Probbility nd Sttistics... 1 Combintions... 13 Permuttions... 0 Lws of Probbility... 6 Counting Principle... 36 Mesures of Centrl Tendency... 43 Mesures of Dispersion... 51 Binomil Distribution nd Probbility... 60 Norml Distributions... 66 T Distributions... 75 Confidence Intervls... 84 Hypothesis Testing... 94 Liner Regression... 307 Differentil Clculus... 316 Derivtives... 317 Criticl Points... 37 Prtil Derivtives... 335 Curvture... 344

L Hopitl s Rule... 351 Integrtion by Prts... 359 Integrtion by Substitution... 368 Prtil Frctions... 376 Differentil Equtions... 385 Bernoulli Differentil Equtions... 386 Higher Order Homogenous Differentil Equtions... 395 Higher Order Non-Homogenous Differentil Equtions... 401 First Order Liner Differentil Equtions... 41 Seprble Differentil Equtions... 419

ANALYTICAL GEOMETRY Lines... Qudrtic Formul... 9 Prbols... 19 Hyperbols... 7 Ellipses... 36 Circles... 4 Distnce Formul... 49 Lw of Sines... 56 Lw of Cosines... 63 Right Tringles... 69 Trigonometric Rtios... 74

CONCEPT INTRODUCTION LINES CONCEPT INTRODUCTION One exmple of generl clss of equtions re liner equtions in two vribles. The two vribles re usully x nd y, but obviously don t need to be. Liner equtions dhere to the following rules: 1. The vribles (usully x nd y) pper only to the first power. The vribles my be multiplied only by rel number constnts 3. Any rel number term my be dded (or subtrcted, of course) 4. Nothing else is permitted! So if ny eqution contins things like x, y, 1, xy, squre roots, or ny other function x of x or y, then it is not stright line. To define line we need to specify two distinct pieces of informtion concerning tht line. A line cn be defined by specifying two distinct points tht the line psses through, or it cn be defined by giving one point tht it psses through nd somehow describing how the sloped of the line is. The slope of line is mesure of how tilted the line is nd cn be equted using the stndrd formul: Where (x 1, x ) nd (x, y ) re ny two distinct points on the line. It mkes no difference which two points re used for point 1 nd point. If they were switched, both the numertor nd the denomintor of the frction would be chnged to the opposite sign, giving exctly the sme result. A horizontl hs no slope, or: P ge psst don t forget to tke notes

CONCEPT INTRODUCTION Specify two distinct pieces of informtion concerning line, there re number of stndrd equtions tht cn be used to define the line in whole. SLOPE-INTERCEPT FORM: If liner eqution in two unknowns is written in the stndrd form: Where m nd b re ny two rel numbers, nd defines the stright line with slope of m nd y intercept equl to b. POINT-SLOPE FORM: Using the point-slope formul: y y 1 = m(x x 1 ) A line cn be defined if one distinct point nd the slope of the line is given. TWO-POINT FORM: Another wy to completely specify line is to give two different points tht the line psses through. If it is given tht line psses through the points (x 1, x ) nd (x, y ), then the two-point formul sttes tht: 3 P ge psst don t forget to tke notes

Concept Exmple: ENGINEERINTRAININGEXAM.COM CONCEPT INTRODUCTION The following problem introduces the concept reviewed within this module. Use this content s primer for the subsequent mteril. Determine the eqution in slope-intercept form for the line tht psses through the points (-1,) nd (,1) Solution: Recll tht to define line we need to specify two distinct pieces of informtion concerning tht line. In this cse, we hve two points, which is sufficient in defining the line tht runs through them. So the first step is to clculte the slope: Now determine the intercept by tking one of the points nd plugging it in to the generl slope-intercept eqution: Now tht we hve defined m nd b, we cn plug these vlues in to define the line in slope-intercept form: 4 P ge psst don t forget to tke notes

LINES VIDEO REVIEW ENGINEERINTRAININGEXAM.COM VIDEO REVIEW In this section of the EIT Acdemy Mth course, we will reinforce your understnding of the key concept covered in this workshop. In this video, we will discuss the topic t hnd by first grsping the definition nd then working through some exmples. Video Link: http://www.engineerintriningexm.com/fundmentls-of-engineering-exm-reviewlines 5 P ge psst don t forget to tke notes

PRACTICE PROBLEMS LINES PRACTICE PROBLEMS Complete the following problems to reinforce your understnding of the concept covered in this module. Problem 1: The y-intercept of line is 3 nd the x-intercept is, determine the eqution of the line in slope-intercept form. Problem : Given the following eqution of line in point-slope form: y+ 1 = 3( x ) Determine the y-intercept. Problem 3: Find the line prllel to the line y= x 1 which psses through the point (-5, 1). 6 P ge psst don t forget to tke notes

Solution 1: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Recll tht to define line we need to specify two distinct pieces of informtion concerning tht line. Although mybe not initilly obvious, in this cse, we hve two points, which is sufficient in defining the line. The two points re (0,3) nd (,0). Using these points, define the slope, which is: 3 0 3 m = = 0 At this point, we typiclly would plug in point, long with the slope, nd solve for the y intercept. The problem, however, lredy gives the y intercept s 3. Plugging this informtion in to the generl slope-intercept eqution, we get: 3 y= x= 3 Solution : To determine the y-intercept, simply multiply out the right hnd side of the eqution nd subtrct 1 from ech side to get it in to the more fmilir slope-intercept form: y+ 1 = 3( x ) y+ 1= 3x 6 y= 3x 7 Therefore, the y-intercept is -7. 7 P ge psst don t forget to tke notes

Solution 3: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Prllel lines hve the sme slope, so if the slope of the line tht is given is equl to, then: m = The line with slope of lso psses through the point (-5,1), plugging the vlues in to get the intercept gives us: b = 1 ( 5) = 11 So the eqution of the line prllel to the line y= x 1 which psses through the point (-5, 1) is: y= x+ 11 8 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION QUADRATIC FORMULA CONCEPT INTRODUCTION The stndrd form of qudrtic eqution is: x + bx + c = 0 A qudrtic eqution cn be solved by fctoring if, fter writing it in stndrd form, the qudrtic expression fctors. Completing this process cn be broken in to 5 steps. 1. Simplify ech side: This involves things like removing prentheses, frctions, dding like terms, etc.. Write in stndrd form x + bx + c = 0: If it is not in stndrd form, move ny terms to the pproprite side by using the ddition/subtrction property of equlity. Also, mke sure tht the squred term is written first left to right, the x term is second nd the constnt is third nd it is set equl to 0. 3. Fctor: 4. Use the Zero-Product Principle: If b = 0, then = 0 or b = 0. The only wy product cn become 0 is if t lest one of its fctors is 0. 5. Solve for the liner eqution set up in the previous step The qudrtic formul cn be used to solve ny qudrtic eqution. This is convenient nd consistent prctice due to qudrtic equtions being difficult to 9 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION fctor most of the time. The 5 steps to complete when using the qudrtic formul re: 1. Simplify ech side if needed. This involves things like removing prentheses, frctions, dding like terms, etc.. Write in stndrd form x + bx + c = 0: If it is not in stndrd form, move ny terms to the pproprite side by using the ddition/subtrction property of equlity. Also, mke sure tht the squred term is written first left to right, the x term is second nd the constnt is third nd it is set equl to 0. 3. Identify, b, nd c: When the qudrtic eqution is in stndrd form, x + bx + c = 0, then is the coefficient in front of the x term, b is the coefficient in front of the x term, nd c is the constnt term. 4. Plug the vlues, b, nd c into the qudrtic formul: x = b ± b 4c 5. Simplify the results if possible When qudrtic eqution is in stndrd form, x + bx + c = 0, the expression, b 4c, tht is found under the squre root prt of the qudrtic formul is clled the discriminnt. The discriminnt tells us how mny solutions there will be nd if the solutions re rel numbers or complex imginry numbers. The following tble defines the possibilities: 10 P ge psst don t forget to tke notes

Discriminnt, b 4c b 4c > 0 b 4c = 0 b 4c < 0 Solutions for qudrtic eqution ENGINEERINTRAININGEXAM.COM CONCEPT INTRODUCTION Two distinct rel solutions One rel solution Two complex distinct imginry solutions To find the discriminnt, the sme process is used for populting the qudrtic formul. Concept Exmple: The following problem introduces the concept reviewed within this module. Use this content s primer for the subsequent mteril. Determine the roots of the following qudrtic eqution: Solution: x + x = 7 The qudrtic formul cn be used to solve ny qudrtic eqution. Let s follow the 5 step process: 1. Simplify ech side if needed. This involves things like removing prentheses, frctions, dding like terms, etc. The eqution is simplified s given.. Write in stndrd form x + bx + c = 0: If it is not in stndrd form, move ny terms to the pproprite side by using the ddition/subtrction property of equlity. Also, mke sure tht the squred term is written first left to right, the x term is second nd the constnt is third nd it is set equl to 0. 11 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION All tht needs to be done to get the eqution in to stndrd form is to subtrct 7 from ech side so tht: x + x 7 = 0 3. Identify, b, nd c: When the qudrtic eqution is in stndrd form, x + bx + c = 0, then is the coefficient in front of the x term, b is the coefficient in front of the x term, nd c is the constnt term. For the given eqution, =1, b=, nd c=-7 4. Plug the vlues, b, nd c into the qudrtic formul: x = ± () 4(1)( 7) (1) 5. Simplify the results if possible = ± 3 To simplify, first reduce the squre root nd then do some cnceling so tht the roots re: x = ± (16) = ± 4 = ( 1 ± = 1 ± 1 P ge psst don t forget to tke notes

VIDEO REVIEW QUADRATIC FORMULA VIDEO REVIEW In this section of the EIT Acdemy Mth course, we will reinforce your understnding of the key concept covered in this workshop. In this video, we will discuss the topic t hnd by first grsping the definition nd then working through some exmples. Video Link: http://www.engineerintriningexm.com/fundmentls-of-engineering-review- %E%80%93-qudrtic-equtions/ 13 P ge psst don t forget to tke notes

PRACTICE PROBLEMS QUADRATIC FORMULA PRACTICE PROBLEMS Complete the following problems to reinforce your understnding of the concept covered in this module. Problem 1: Determine the roots of the following qudrtic eqution: 3q + 11 = 5 q Problem : Determine the roots of the following qudrtic eqution: Problem 3: 7t = 6 19 t Determine the solutions to the eqution: 3 1 = + 1 y y 14 P ge psst don t forget to tke notes

Solution 1: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Use the 5 step process to finding the roots using the qudrtic formul. 6. Simplify ech side if needed. This involves things like removing prentheses, frctions, dding like terms, etc. The eqution is simplified s given. 7. Write in stndrd form x bx c + + = 0 : If it is not in stndrd form, move ny terms to the pproprite side by using the ddition/subtrction property of equlity. Also, mke sure tht the squred term is written first left to right, the x term is second nd the constnt is third nd it is set equl to 0. All tht needs to be done to get the eqution in to stndrd form is to subtrct 5q from ech side so tht: 3q 5q+ 11 = 0 8. Identify, b, nd c: When the qudrtic eqution is in stndrd form, x + bx + c = 0, then is the coefficient in front of the x term, b is the coefficient in front of the x term, nd c is the constnt term. For the given eqution, =3, b=-5, nd c=11 9. Plug the vlues, b, nd c into the qudrtic formul: ( 5) ( 5) 4(3)(11) 5 107 5 107 q = ± = ± = ± (3) 6 6 i 15 P ge psst don t forget to tke notes

10. Simplify the results if possible ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Solution : To simplify, first reduce the squre root s much s you cn nd then do some cnceling if possible, the roots re: 5 ± 107 i q = 6 Use the 5 step process to finding the roots using the qudrtic formul. 1. Simplify ech side if needed. The eqution is simplified s given.. Write in stndrd form x bx c + + = 0 : All tht needs to be done to get the eqution in to stndrd form is to subtrct 6 19t from ech side so tht: 7t + 19t 6 = 0 3. Identify, b, nd c: For the given eqution, =7, b=19, nd c=-6 4. Plug the vlues, b, nd c into the qudrtic formul: ± ± t = = (7) 14 19 (19) 4(7)( 6) 19 59 5. Simplify the results if possible 16 P ge psst don t forget to tke notes

PRACTICE PROBLEMS The roots re t = nd 3 7 Solution 3: Use the 5 step process to finding the roots using the qudrtic formul. 1. Simplify ech side if needed. This eqution hs frctions in it, so the first thing we must do is to get rid of these. To do tht, identify the lest common denomintor, which is: LCD: yy ( ) Multiply both sides by the LCD: 3 1 yy ( )( ) ( 1) yy ( ) y = y + ( ) 3y= y + y y 3 = + y y y y. Write in stndrd form x bx c + + = 0 : All tht needs to be done now to get the eqution in to stndrd form is to combine terms on the left side of the eqution so tht: y 4y = 0 3. Identify, b, nd c: For the given eqution, =1, b=-4, nd c=- 17 P ge psst don t forget to tke notes

PRACTICE PROBLEMS 4. Plug the vlues, b, nd c into the qudrtic formul: ( ) ( ) ( )( ) ( ) 4± 4 41 4± 6 y = = 1 5. Simplify the results if possible The solutions re then: y = ± 6 18 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION PARABOLAS CONCEPT INTRODUCTION The most generl form of qudrtic function is, f ( x) = x + bx + c The grph of qudrtic eqution is clled prbol nd generlly tkes the shpe of U. Every prbol hs n imginry line tht runs down the center of it clled the xis of symmetry; where one ech side is mirror imge of the other. If one point on one side of the prbol is known, then the point directly on the other side is known bsed on the xis of symmetry. The highest or lowest point of prbol is clled the vertex nd the prbol my open up or down nd my or my not hve x-intercepts but they will lwys hve single y-intercept. There re two forms of the prbol to be fmilir with. This first form will mke defining the chrcteristics of the prbol firly esy, however, most prbols re not given in this form. The second form is the more common form nd will require little more work to define its chrcteristics. The first form of the prbol: ( ) ( ) f x = x h + k There re two pieces of informtion bout the prbol tht we cn instntly get from function in this form. First, if is positive then the prbol will open up nd if is negtive then the prbol will open down. Secondly, the vertex of the prbol is the point (, hk). Most prbols ren t given in this form. The more common form we see is the generl form: 19 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION f ( x) = x + bx + c The sign of will still determine whether or not the prbol opens upwrds or downwrds. To determine the vertex, the following equtions re used: b b, f ( ) To get the vertex, compute the x coordinte from nd b nd then plug the result into the function to get the y coordinte. The y-intercept is: ( ) f 0 = (0) + b(0) + c (0, c ) It s importnt to note tht n eqution given in the generl form cn be converted in to the first form by using the process of completing the squres. Concept Exmple: The following problem introduces the concept reviewed within this module. Use this content s primer for the subsequent mteril. Determine the vertex of the following prbol Solution: f ( x) = x + 4 Recll tht to determine the vertex, the following equtions re used: b b, f (,) 0 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION We first compute the x coordinte from nd b nd then plug the result into the function to get the y coordinte. In this problem, =1 nd b=0, therefore: b 0 0 = 1 = ( ) Now, f b = f ( 0) = ( 0) + 4= 4 So the vertex is locted t (0,4) 1 P ge psst don t forget to tke notes

VIDEO REVIEW PARABOLAS VIDEO REVIEW In this section of the EIT Acdemy Mth course, we will reinforce your understnding of the key concept covered in this workshop. In this video, we will discuss the topic t hnd by first grsping the definition nd then working through some exmples. Video Link: http://www.engineerintriningexm.com/fundmentls-of-engineering-exm-reviewprbols-prt-1/ http://www.engineerintriningexm.com/fundmentls-of-engineering-exm-reviewprbols-prt-/ P ge psst don t forget to tke notes

PRACTICE PROBLEMS PARABOLAS PRACTICE PROBLEMS Complete the following problems to reinforce your understnding of the concept covered in this module. Problem 1: Determine the vertex of the following prbol: Problem : ( ) f x = x + 4x + 4 Convert ech of the following into the form f ( x) = ( x h) + k ( ) f x = x 1x + 3 Problem 3: Convert ech of the following into the form f ( x) = ( x h) + k ( ) f x = x + 10x 1 3 P ge psst don t forget to tke notes

Solution 1: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Recll tht to determine the vertex, the following equtions re used: b b, f (,) We first compute the x coordinte from nd b nd then plug the result into the function to get the y coordinte. In this problem, =1 nd b=4, therefore: b 4 = 1 = ( ) Now, f b = f ( ) = ( ) + 4( ) + 4= 0 So the vertex is locted t (-,0) Solution : The first thing to do is to complete the squre. To do this, we need to ensure tht there is coefficient of 1 on the x term, such tht: f x x 6 x 3 ( ) = + Now tke one hlf of the coefficient of x term nd squre it: 6 = ( 3) = 9 4 P ge psst don t forget to tke notes

PRACTICE PROBLEMS To mintin the integrity of the originl eqution, mke sure to dd nd subtrct this quntity within the prenthesis so tht: 3 ( ) = ( 6 + 9 9 + f x x x The next step is to fctor the first three terms: ( ) = ( 3) 15 f x x Finlly distribute the bck through the eqution: ( ) f x = ( x 3) 15 The originl eqution is now in the form f ( x) = ( x h) + k Solution 3: The first thing to do is to complete the squre. To do this, we need to ensure tht there is coefficient of 1 on the x term, such tht: ( ) = ( 10 + 1) f x x x Now tke one hlf of the coefficient of x term nd squre it: 10 ( ) = = 5 5 To mintin the integrity of the originl eqution, mke sure to dd nd subtrct this quntity within the prenthesis so tht: ( ) = ( 10 + 5 5 + 1) f x x x The next step is to fctor the first three terms: 5 P ge psst don t forget to tke notes

( 5 4) ( ) ( ) f x = x ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS Finlly distribute the bck through the eqution: ( ) ( ) f x = x 5 + 4 The originl eqution is now in the form f ( x) = ( x h) + k 6 P ge psst don t forget to tke notes

CONCEPT INTRODUCTION HYPERBOLAS CONCEPT INTRODUCTION There re two stndrd forms for the hyperbol defined by the following chrcteristics: ( x h) ( y k) ( y k) ( x h) Stndrd Form = 1 = 1 b b Center ( hk, ) ( hk, ) Orienttion Opens left nd right Opens up nd down Vertices ( h+ k, ) nd ( h k, ) ( hk, + b) nd ( hk, b) Slope of Asymptotes Equtions of Asymptotes b ± y k b ( x h) b ± y= k± b x h = ± ( ) Hyperbols re two prbol-like shped pieces tht open either up nd down or left nd right. Just like prbols, ech of the pieces hs vertex. There re lso two lines tht my be illustrted long with the Hyperbol, these re clled symptotes. The symptotes re not officilly prt of the the hyperbol. The point where the two symptotes cross is clled the center of the hyperbol. Concept Exmple: The following problem introduces the concept reviewed within this module. Use this content s primer for the subsequent mteril. Determine the equtions for the symptotes of the ellipse defined by the eqution: ( x 3) ( y+ 1) = 1 5 49 7 P ge psst don t forget to tke notes

Solution: ENGINEERINTRAININGEXAM.COM CONCEPT INTRODUCTION When working towrds the eqution of the symptotes, it s importnt to recll the following chrcteristics of hyperbol: Stndrd Form ( x h) ( y k) ( y k) ( x h) = 1 = 1 b b Center ( hk, ) ( hk, ) Orienttion Opens left nd right Opens up nd down Vertices ( h+ k, ) nd ( h k, ) ( hk, + b) nd ( hk, b) Slope of Asymptotes b b ± ± Equtions of b y= k± ( x h) y= k± b ( x h) Asymptotes The y term hs the minus sign nd so it s known tht the hyperbol will be opening left nd right. The eqution is lso lredy in its stndrd form, so the first thing tht cn be defined is the center, which is: (h,k)=(3,-1) The vertices re then: (8,-1) nd (-,-1) Next determine the slopes of the symptotes. These re given s b ±, where b = 7 nd = 5, therefore the slopes re: 8 P ge psst don t forget to tke notes

7 ± 5 ENGINEERINTRAININGEXAM.COM CONCEPT INTRODUCTION With the center nd the slopes defined, define the equtions of the symptotes s: 7 y= 1 + ( x 3) 5 nd 7 y= 1 ( x 3) 5 9 P ge psst don t forget to tke notes

VIDEO REVIEW HYPERBOLAS VIDEO REVIEW In this section of the EIT Acdemy Mth course, we will reinforce your understnding of the key concept covered in this workshop. In this video, we will discuss the topic t hnd by first grsping the definition nd then working through some exmples. Video Link: http://www.engineerintriningexm.com/fundmentls-of-engineering-exm-reviewhyperbol/ 30 P ge psst don t forget to tke notes

PRACTICE PROBLEMS HYPERBOLAS PRACTICE PROBLEMS Complete the following problems to reinforce your understnding of the concept covered in this module. Problem 1: Determine the equtions for the symptotes of the ellipse defined by the eqution: y 9 + = ( x ) 1 Problem : Define the chrcteristics of the hyperbol defined by: ( x ) ( y+ 3) = 1 36 64 Problem 3: Determine the equtions for the symptotes of the ellipse defined by the eqution: x y = 1 9 16 31 P ge psst don t forget to tke notes

Solution 1: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS When working towrds the eqution of the symptotes, it s importnt to recll the following chrcteristics of hyperbol: Stndrd Form ( x h) ( y k) ( y k) ( x h) = 1 = 1 b b Center ( hk, ) ( hk, ) Orienttion Opens left nd right Opens up nd down Vertices ( h+ k, ) nd ( h k, ) ( hk, + b) nd ( hk, b) Slope of Asymptotes b b ± ± Equtions of b y= k± ( x h) y= k± b ( x h) Asymptotes The x term hs the minus sign nd so it s known tht the hyperbol will be opening up nd down. The eqution cn be revised little to represent more the stndrd form tht we re used to working with, this gives: ( y 0) ( x+ ) = 1 9 1 In its stndrd form, the center cn be determined, which is: (h,k)=(-,0) The vertices re then: (-,0) nd (-,-3) 3 P ge psst don t forget to tke notes

PRACTICE PROBLEMS b Next determine the slopes of the symptotes. These re given s ±, where b = 3 nd = 1, therefore the slopes re: ±3 With the center nd the slopes defined, define the equtions of the symptotes s: And y= 3( x+ ) y= 3( x+ ) Solution : From quick observtion, the y term hs the minus sign nd so it s known tht the hyperbol will be opening up nd down. The chrcteristics tht we will wnt to define re then: Stndrd Form ( x h) ( y k) = 1 b Center ( hk, ) Orienttion Opens left nd right Vertices ( h+ k, ) nd ( h k, ) Slope of Asymptotes b ± Equtions of b y= k± ( x h) Asymptotes The eqution is lredy in its stndrd form: ( x ) ( y+ 3) = 1 36 64 33 P ge psst don t forget to tke notes

The center is locted t: ENGINEERINTRAININGEXAM.COM PRACTICE PROBLEMS (,-3) Note tht = 6 nd b = 8 so the vertices re then: (8,-3) nd (-4,-3) The slopes of the symptotes re given by 8 4 ± =± 6 3 Solution 3: b ± : When working towrds the eqution of the symptotes, it s importnt to recll the following chrcteristics of hyperbol: Stndrd Form ( x h) ( y k) ( y k) ( x h) = 1 = 1 b b Center ( hk, ) ( hk, ) Orienttion Opens left nd right Opens up nd down Vertices ( h+ k, ) nd ( h k, ) ( hk, + b) nd ( hk, b) Slope of Asymptotes b b ± ± Equtions of b y= k± ( x h) y= k± b ( x h) Asymptotes The y term hs the minus sign nd so it s known tht the hyperbol will be opening left nd right. The eqution cn be revised little to represent more the stndrd form tht we re used to working with, this gives: 34 P ge psst don t forget to tke notes

PRACTICE PROBLEMS ( x 0) ( y 0) = 1 9 16 The center cn now be defined s: (h,k)=(0,0) The vertices re then: (0,0) Next determine the slopes of the symptotes. These re given s = 3, therefore the slopes re: b ±, where b = 4 nd 4 ± 3 With the center nd the slopes defined, define the equtions of the symptotes s: And y= 4 3 x 4 y= x 3 35 P ge psst don t forget to tke notes