Chpters Five Notes SN AA U1C5 Nme Period Section 5-: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles tht ll terms hve in common. This mens nothing cn divide into ech prt of term ecept 1. Tht fctor tht goes into every term is clled the gretest common fctor (GCF). Emple 1: Fctor the epression 1y 3 1y. First, look to see if ny numers or vriles re in common to ech term. In this emple, 7 nd y re common to ech term. Tht mens the gretest common fctor (GCF) is 7y. If you pull tht out in front nd divide out these from ech term, 1y 3 1y = 7y (3y ) Bck in lger, you lso lerned out fctoring trinomils ( polynomil with three terms) into inomils ( polynomil with two terms). As you should lredy know, ( + m)( + n) = + n + m + mn. This is form of the distriutive property known s the etended distriutive property. ( + m)( + n) + n + m + mn + (n + m) + mn + (m + n) + mn This works in reverse s well. In other words, + (m + n) + mn = ( + m)( + n). This type of reverse is clled fctoring. To fctor some trinomil + + c into ( + m)( + n), = m + n nd c = mn. Emple 1: Fctor the trinomil + 5 +. Emple : Fctor the trinomil c + c 80. Emple 3: Fctor the trinomil 1 + 5. pge 1 SN AA U1C5
Sometimes fctoring is it more complicted. For emple, if you hve (k + m)(l + n), (k + m)(l + n) = kl + kn + ml + mn. This is form of the distriutive property known s the etended distriutive property s well. (k + m)(l + n) kl + kn + ml + mn (kl) + (kn + ml) + mn This works in reverse s well. In other words, (kl) + (kn + ml) + mn = (k + m)(l + n). To fctor some trinomil + + c into (k + m)(l + n), = kl, = kn + ml, nd c = mn. Emple : Fctor the trinomil 1v 5v 7. Emple 5: Fctor the trinomil 6 9 + 5. There re esier wys to fctor for certin polynomils nd inomils if there re specil ptterns. Pttern Nme Pttern Emples Difference of two 9 = ( + 3)( 3) squres 9 16 = (3 + )(3 ) Perfect squre trinomil z 1z + 36 = (z 6)(z 6) = (z 6) 16g 0g + 5 = (g 5)(g 5) = (g 5) y + 1y + 9 = (y + 7)(y + 7) = (y + 7) t + 0t + 5 = (t + 5)(t + 5) = (t + 5) pge SN AA U1C5
Emple 6: Fctor the epression - 5. Emple 7: Fctor the epression 81c 198c + 11. Emple 8: Fctor the epression 8y 5 8y 60y 3. Emple 9: Fctor the epression v 3 + v v. In review, the est wy to fctor is s follows: Fctor out the gretest common fctor (GCF) y pulling it out in front nd dividing it from ech term, ll of which terms will go in prentheses. Fctor wht is left into seprte sets of prentheses. Section 5-5: Qudrtic Equtions As you proly know, nything times zero is equl to zero. 5 0 = 0 0 = 0 If you hve y = 0, or y could e equl to zero. This is the zero product property. Zero Product Property: Let A nd B e rel numers or lgeric epressions. If AB = 0, then A = 0 or B = 0. This could e epnded so tht if ABC = 0, then A = 0, B = 0, or C = 0. This cn e pplied to fctored terms in qudrtic equtions. Wht relly helps to mke this work is when the qudrtic eqution for which you re solving is in the stndrd form of qudrtic eqution. Stndrd form of qudrtic eqution: + + c = 0 where 0 pge 3 SN AA U1C5
Emple 1: Solve the eqution 3 =. Emple : Solve the eqution 5 13 + 6 = 0. Emple 3: Solve the eqution 3 3 + 3 + 90 = 0. Emple : Solve the eqution y y 8 = y + y Emple 5: Solve the eqution (w + 6) = 3(w + 1) w pge SN AA U1C5
The process tht hs een used in the pst few emples hs een where the stndrd form of qudrtic eqution is equl to zero. For y = + + c, we cn use similr process of finding zeros y letting y = 0. Emple 6: Solve the qudrtic function + 7 + 1 = 0. Turn on your clcultor. Press the Y= utton. Under \Y1, type in X + 7X + 1. Press ENTER t the ottom. Following this, you cn grph it. You will see ig U, known s prol, tht is opening upwrd. Now press the nd" utton nd then press the GRAPH utton to open the tle of points for the prol. Since you re solving for when y = 0, look up nd down on the tle for where y = 0. y = 0 eist for when = 3 nd =. Therefore,the zeros for this function re 3 nd. In some cses, you my end up with squre root. As review for squre roots: r is squre root of numer if r = s. Positive versions of s hve two squre roots: s nd s. For s, the rdicl sign is the. The rdicnd is the s eneth the rdicl sign. The rdicl is the epression s. Let s rememer some properties of squre roots. 6 8, ut you cn lso relize tht 6 16 8 This is wht is referred to s the product property, mening, or. When simplifying squre roots, look for the iggest squre fctor of the numer nd tke the squre root of it. If it is the iggest squre root, the other numer will e left s rdicl. Emple 7: Simplify 18. Another useful property is when you hve frction under rdicl. 100 100 100 10 5 5, ut you cn lso relize tht 5. This is wht is referred to s the quotient property, mening, or. Also, rdicls should NEVER e left in the denomintor of frction. A squre root epression is sid to e simplified if the following conditions re true: No rdicnd (numer under the squre root sign) hs perfect-squre fctor other thn 1. No rdicl is in the denomintor. pge 5 SN AA U1C5
If one hs the following epression where s is numer with no perfect-squre fctors other thn one, one cn rtionlize the epression (eliminte the rdicl in the denomintor) y multiplying the frction y the denomintor s vlue over the denomintor s vlue (the sme s multiplying y 1). In other words,. Emple 8: Simplify 36. 30 You cn use squre roots to solve qudrtic equtions, ut you need to get the eqution in the form of = s where s > 0. If you solve this eqution, you will ctully get two nswers: s nd s. To sve spce, the two nswers re often written together s s. s is red s plus or minus the squre root of s. Emple 9: Solve 7 8 = 0. When deling with motion nd projectiles, qudrtic equtions ecome etremely useful. When n oject is dropped, ssuming ir resistnce is negligile, the oject flls t rte of 16 feet per squre second (16 ft/s ). If h is the finl height of the oject in feet, h 0 is the originl height of the oject in feet, nd t is time in seconds, the finl height ecomes function of time s shown in the eqution h = 16t + h 0. pge 6 SN AA U1C5
Emple 10: According to legend, in 1589 the Itlin scientist Glileo Glilei dropped two rocks of different weights from the top of the Lening Tower of Pis. He wnted to show tht the rocks would hit the ground t the sme time. Given tht the tower s height is out 177 feet, how long would it hve tken the rocks to hit the ground? Section 5-6: Comple Numers Mny moons go, you lerned tht there is no rel solution to the squre root of negtive numer ecuse no rel numer, when multiplied y itself, will give negtive numer. However, rel numers re not the only kinds of numers. Something clled the imginry unit ( i ) llows us to tke the squre roots of negtive numers. By definition, i = 1, mening tht i 1. Properties of Imginry Numers Property Emple If r is positive rel numer, then 7 1 7 i 7 r 1 r i r If r is positive rel numer, then i 7 i 7 i 9 1 7 7 r 1 r i r When writing numers with imginry units in them, they re written in the stndrd form + i where nd re oth rel numers. + i is form of comple numer in which is the rel prt of the comple numer nd i is the imginry prt of the comple numer. Type of comple numer Form of numer Emple Rel numer + 0i, 3, 6,, 5 Imginry numer + i, 0 3 i 5, 3i Pure imginry numer 0 + i, 0 3i, 5i, i 3 When one grphs rel numers on rel numer line, every rel numer corresponds to specific point on the numer line. In the sme wy, ech pure imginry numer corresponds to specific point on n imginry numer line. Since comple numers re comintion of oth rel nd imginry numers, one cn grph numer on comple plne. pge 7 SN AA U1C5
i (0 + i) 3 ( 3 + 0i) 3i Comple numers re equl if ech prt of the comple numer is equl to the corresponding prt of the other comple numer. For emple, if + i = 3 7i, then = 3 nd = 7. To dd nd sutrct comple numers, you must comine the rel prts s eing common terms nd seprtely comine the imginry prts s eing common terms. Emple 1: Write the epression ( + 3i) + (7 + i) s comple numer in stndrd form. Emple : Write the epression i + (8 i) (5 9i) s comple numer in stndrd form. When multiplying comple numers, the sme rules pply s when multiplying rel numers; one still uses the distriutive nd etended distriutive properties. However, s stted efore, i = - 1 nd i = 1. Emple 3: Write the epression 10i( + 7i) s comple numer in stndrd form. pge 8 SN AA U1C5
Emple : Write the epression (15-8i) s comple numer in stndrd form. Emple 5: Write the epression (7 + 5i)(7 5i) s comple numer in stndrd form. Notice tht 7 + 5i nd 7 5i hve the form + i nd i nd, when multiplied, give rel numer. These two forms of numers re clled comple conjugtes nd cn e used to get rid of imginry numers in denomintors of frctions (s they should not e there if frctions re to e simplified). Emple 6: Write the epression i 1 i s comple numer in stndrd form. Emple 7: Solve the eqution 3 = 81. pge 9 SN AA U1C5
1 v 8. Emple 8: Solve the eqution 3 7 Section 5-7: Completing the Squre When you took lger, s well s during the course of this semester, you lerned tht perfect squre is of the form + + = ( + )( + ). However, when you hve qudrtic eqution of the form +, it is often helpful to hve tht lst term to help fctor nd solve. We cn use the digrm shown to complete the squre so tht we hve perfect squre trinomil. pge 10 SN AA U1C5
If we split up the re of nd move hlf of it over, the only other re left if squre of the quntity of hlf of. In other words, the lst term (c) comes from c.. Therefore, to complete the squre of some +, Emple 1: Find the vlue of c tht mkes the epression + 18 + c perfect squre trinomil. Then write the epression s the squre of inomil. Emple : Solve the eqution + = 9 y completing the squre. Emple 3: Solve the eqution u u = u + 35 y completing the squre. pge 11 SN AA U1C5
If there is coefficient in front of the term of qudrtic eqution, this cn complicte things. The esiest thing to do is divide it from oth sides to mke the process of completing the squre esier. Emple : Solve the eqution 6 + 8 + 300 = 0 y completing the squre. You lerned, ck when you did solute vlue functions, tht you could esily find the verte of n solute vlue function if it ws in the form y = h + k where (h, k) is the verte. You cn lso do the sme for qudrtic functions, ecept the form is y = ( h) + k, nd you cn do it y completing the squre s well. Emple 5: Write the qudrtic function y = + 16 + 1 in verte form nd identify the verte. pge 1 SN AA U1C5
pge 13 SN AA U1C5 Section 5-8: The Qudrtic Formul Almost entirely throughout this chpter, the qudrtic eqution hs een discussed: + + c = 0. Solutions for of this eqution hve een found y fctoring, finding squre roots, nd completing the squre. However, if ll these methods re comined, cn e solved for ANY form of the qudrtic eqution + + c = 0. + + c = 0 Stndrd form of qudrtic eqution 0 c Dividing ech side y c Sutrcting ech side y c c Completing the squre c Fctoring the left side c Simplifying the eponent on the right side c Getting common denomintor c Comining frctions c Tking the squre root of ech side c Quotient property c Simplifying the rdicl of ech side c Sutrcting from ech side c Comining frctions This is the qudrtic formul, formul tht cn e used to solve for ny vlue of (rel or imginry) s long s the qudrtic eqution eing solved is in the stndrd form of + + c = 0. Emple 1: Use the qudrtic formul to solve the eqution 7 + + 9 = 0.
Emple : Use the qudrtic formul to solve the eqution + 8 = 9. Emple 3: Use the qudrtic formul to solve the eqution = 8 35. As you my hve noticed, the rdicl in the qudrtic formul determines the numer nd kinds of solutions tht will result for. The epression in the rdicl ( c) is clled the discriminnt. Below is summry how you cn know the possile nswers resulting from solving the discriminnt. The Discriminnt nd Its Properties Discriminnt Vlue Numer of nswers Type of nswer(s) c > 0 c < 0 c = 0 pge 1 SN AA U1C5
Emple : Find the discriminnt of the qudrtic eqution + 3 6 = 0 nd give the numer nd type of solutions of the eqution. For n oject tht is thrown, lunched, or thrown, there is physics eqution tht models reltionship etween finl height of n oject (h) in feet, initil height of n oject (h 0 ) in feet, initil verticl velocity (v 0 ) in feet per second, nd time in motion (t) in seconds: h = 16t + v 0 t + h 0. To solve for time, one must use the qudrtic formul to solve for t. Emple 5: In July of 1997, the first Cliff Diving World Chmpionships were held in Brontllo, Switzerlnd. Prticipnts performed crotic dives from height of up to 9 feet. Suppose cliff diver jumps from this height with n initil upwrd velocity of 5 feet per second. How much time does the diver hve to perform crotic mneuvers efore hitting the wter? Do you necessrily lwys wnt to use the qudrtic formul? Sometimes it is esier to use fctoring, finding squre roots, or completing the squre. Emple 6: Solve the eqution = 1 using whtever method you choose. pge 15 SN AA U1C5
Emple 7: Solve the eqution 7 + 15 = + 5 using whtever method you choose. Emple 8: Solve the eqution 6 + 17 = 5 using whtever method you choose. pge 16 SN AA U1C5
Section 5-1: Modeling Dt with Qudrtic Functions Section 5-: Properties of Prols Section 5-3: Trnsforming Prols Erlier in this chpter, you lerned out qudrtic functions. A qudrtic function hs the form y = + + c where 0. When you grph qudrtic function, you get U-shped figure tht is clled prol. The lowest point on prol tht opens up (y = ) or the highest point on prol tht opens down (y = ) is clled the verte. For these two grphs, they re symmetric out the y-is. The y-is goes through the verte nd is n is of symmetry. Since the grph of prol comes from y = + + c, we cn use the qudrtic formul to find few things out prol. As you should recll from the lst section we discussed, the qudrtic c formul is when + + c = 0 nd 0. Since there is only one verte, to find the -coordinte of the verte, the rdicl prt of the qudrtic formul is 0. This mens tht the formul for the -coordinte is. Once the -coordinte hs een found, one cn sustitute tht vlue in for in y = + + c to find the y-coordinte of the verte. Since we know the verte of prol goes through the verte verticlly, the formul for the is of symmetry is. There re few other spects of prols to study. As you sw in the originl grphs, is positive in grph of prol of the form y = + + c, the grph opens up. If is negtive in prol of the form y = + + c, the grph opens down. Also, the grph of y = + + c cn widen or nrrow. If < 1, then the grph is wider thn the grph of y =. If is > 1, then the grph is nrrower thn the grph of y =. The grph of y = + + c is prol with these chrcteristics: The prol opens up if > 0 nd opens up if < 0. The prol is wider thn the grph of y = if < 1 nd nrrower thn the grph of y = if > 1. The -coordinte of the verte is The is of symmetry is the verticl line. pge 17 SN AA U1C5
Emple 1: Grph the qudrtic function y = 1 + 19. Lel the verte nd is of symmetry. Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte Emple : Grph the qudrtic function 1 y 3. Lel the verte nd is of symmetry. 6 Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte pge 18 SN AA U1C5
There re three different wys of epressing qudrtic functions: Chrcteristic Stndrd form Verte form Intercept form Form of qudrtic function y = + + c y = ( - h) + k y = ( - p)( - q) Verte or -intercepts Ais of symmetry Emple 3: Grph the qudrtic function y = ( ) 1. Lel the verte nd is of symmetry. Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte pge 19 SN AA U1C5
Emple : Grph the qudrtic function y = 3( + ) + 5. Lel the verte nd is of symmetry. Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte Emple 5: Grph the qudrtic function y = ( + 1)( 1). Lel the verte nd is of symmetry. Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte pge 0 SN AA U1C5
Emple 6: Grph the qudrtic function y = 3( ). Lel the verte nd is of symmetry. Also indicte the y-intercept. Verte: Ais of symmetry: y-intercept: y Verte pge 1 SN AA U1C5