Grphig Review Prt : Polomils Prbols Recll, tht the grph of f ( ) is prbol. It is eve fuctio, hece it is smmetric bout the bout the -is. This mes tht f ( ) f ( ). Its grph is show below. The poit ( 0,0) is clled its verte. From prt we lered tht the grph of ( ) hs the ect sme shpe s, but it hs bee shifted uits to the right d uit up. Hece, the verte of this prbol will ow be (,). See the grph below o the left. From prt we lso lered tht the grph of ( ) 9 will hve shpe similr to, but the verte will be t (,9) d becuse 0 / d / is egtive, the grph will be wider d ope dowwrd. See the grph below o the right. The stdrd form of prbol is ( h) k.
Emple Fid the specific equtio of the prbol (, ) d tht psses through (, ). ( h) k with verte The grph of will be prbol tht opes to the right d the grph of will be prbol tht opes to the left. Note tht the re ot fuctios. I this cse, the stdrd form of prbol is ( k) h.
Emple Sketch the regio tht is bouded b the prbols d. Be sure to iclude the coordites of the poits of itersectio for these two grphs. Emple Sketch the regio tht is bouded betwee the grphs d 6. Be sure to iclude the coordites of the poits of itersectio for these two grphs.
Cosider the followig three ws for writig the sme qudrtic fuctio: ( ) 8 0 ( )( ). I the first epressio, ou c immeditel see tht the verte is (,8), the grph opes dowwrd, d is steeper th.. I the secod epressio, ou c immeditel see tht the -itercept is ( 0, 0), the grph opes dowwrd, d is steeper th. Also, becuse the tget lie to the grph is horizotl t the verte, d the derivtive (from clculus) will be 0 t the -coordite of the verte, the ou c fid this -coordite b solvig 0 for. Becuse of this, the -coordite of the verte for the equtio b c is lws b /.. I the third epressio, ou c immeditel see tht the -itercepts re (,0) d (,0), the grph opes dowwrd, d is steeper th. Also, due to smmetr, the -coordite of the verte will be hlfw betwee d, so it will be. Emple Three prbols re sketched below. Fid their equtios. You m ssume tht the coordites of the vertices d itercepts re itegers.
Polomils of Higher Degree Whe workig with polomils of higher degree, there re certi ttributes tht c help ou to quickl sketch their grphs. Ed Behvior Cosider the grphs of,,, d show below. The ll hve -itercept of 0. Whe 0, the 0 d whe 0, the 0 for the eve-degree fuctios d 0 for the odd-degree fuctios. Whe, the d whe 0 the (see the grph bove o the right). for the eve - degree fuctios lim d lim for the odd - degree fuctios Cosider the grphs of,,, d show below. lim d lim for the eve - degree fuctios for the odd - degree fuctios
6 The behvior of fuctio s is clled its ed behvior. I the cse of polomils, the ed behvior c be geerlized. Suppose tht 0... where 0, the whe is eve 0 whe 0 whe lim d whe is odd 0 whe 0 whe lim 0 whe 0 whe lim To see wh this is true, fctor out of 0... 0.... The limit of the epressio i pretheses is s, hece s,. This mes tht 0... hs the sme ed behvior s. Hece, polomil with ledig term of eve degree hs the sme ed behvior i ech directio d polomil with ledig term of odd degree hs the opposite ed behviors i the two directios. Emple 800 lim 800 lim 800 lim 800 lim
7 Emple Three polomils re sketched below. For ech polomil, determie whether the ledig term hs eve or odd degree d whether the ledig coefficiet is positive or egtive. The grphs show ll of the plces where directios d cocvit re switched. f (): g() : h() : Behvior Ner Roots The grph of is show below. It hs -itercepts of ( 0,0) d (,0). Becuse the polomil hs fctor of, the the root 0 is repeted zero of multiplicit d becuse the polomil hs fctor of, the the root hs multiplicit. Notice tht the grph bouces off the -is t ( 0,0) d psses through the - is t (,0). The resos for this re tht 0 is root of eve multiplicit d is root of odd multiplicit: The fuctio will ot chge sigs immeditel fter 0 becuse The fuctio will chge sigs immeditel fter becuse does ot. does.
8 Whe we looked t the grphs of,,, d, we oticed tht the grphs with higher degree were fltter (more horizotl) er 0. A similr tpe of thig hppes where the grphs of polomils touch the - is; s the multiplicit of root icreses, the grph flttes er the root. Also, the slope of the tget lie to the polomil will be zero t roots with multiplicit greter th. The grph of h( ) 0( ) ( ) is show below. Notice how the grph is flt t d becuse these re roots with multiplicit greter th d the grph is ot s flt t d becuse these re roots with multiplicit. Also, becuse the multiplicit of is eve the grph bouces off the -is t (,0) d becuse the multiplicit of the roots, d re odd, the grph crosses through the -is t (,0), (,0) d (,0). Creful: It is possible for the grph of polomil to fltte er root of multiplicit d eve look s though the slope of the tget lie is 0 t tht root. Cosider the polomil g ( ) ( ) ( ). It switches cocvit t 0 d multiplig g () b / 0 flttes the grph further; mkig it pper s though the multiplicit t 0 is greter th.
9 Emple The polomil sketched below hs degree of either 9 or 0 (ou will eed to determie which). The bsolute vlue of the ledig term is / 00. It hs ectl roots d the re ll itegers show i the grph. Fid the equtio of this polomil. The grph shows ll of the plces where directios d cocvit re switched. Cocvit A polomil of degree c switch its cocvit t most times. Oe w to see wh this is true is to cosider its secod derivtive. The secod derivtive will be polomil of degree. Hece, the secod derivtive hs t most roots. This mes tht the secod derivtive c chge sig (from positive to egtive or from egtive to positive) t most times. Hece, the origil polomil c switch cocvit (from cocve up to cocve dow or from cocve dow to cocve up) t most times. Emple Wht is the miimum degree of the polomil show below?
0 Strteg Whe grphig the polomil m m m r r r k...... 0 k, ou c strt b plottig the zeroes d remember tht If the multiplicit of r i is odd the grph will cross through the -is t r i d if the multiplicit of r i is eve the grph will bouce off the - is t ri. The grph flttes out (becomes horizotl) t ri s the multiplicit of this zero icreses. The polomil will hve the sme ed behvior s. The polomil will switch cocvit t most times.
Solutios Emple Fid the equtio of the prbol tht psses through (, ). ( h) k with verte (, ) d Becuse the verte is (, ) the prbol hs stdrd form ( ). Becuse it psses through (, ): ( ) ( ) Emple Sketch the regio tht is bouded b the prbols d. Be sure to iclude the coordites of the poits of itersectio for these two grphs. These re both prbols tht ope upwrd d hve vertices of ( 0,0) d ( 0,), respectivel. The grph of will be steeper th the grph of, so eve though is below whe 0, the grphs will itersect. Their itersectio c be foud b settig them equl to ech other: Hece, the itersect t the poits (, ) d (,). The regio is show below.
Emple Sketch the regio tht is bouded betwee the grphs d 6. Be sure to iclude the coordites of the poits of itersectio for these two grphs. The grph of is lie tht hs itercepts t ( 0, ) d (,0). The grph of 6 ( ) will be prbol tht opes to the right with verte of,0) 0, 6 To solve for where the itersect, we c replce i the secod equtio with (from the first equtio) d we get: ( d -itercepts of ( ) 6 6 0 ( )( ) 0 Hece, the itersect whe d whe : (, ) d (, ). The regio is show below. Emple Three prbols re sketched below. Fid their equtios. You m ssume tht the coordites of the vertices d itercepts re itegers. The grph of f () hs verte t (,0) d it opes upwrd. So, it hs the form ( ) where 0 ( 0,), the (0 ) 9 /. Hece,. Becuse it hs -itercept of f ( ). ) ( The grph of g() hs -itercepts of (,0) d (,0), d the grph opes dowwrd. So, it hs the form ( )( ) where 0. Becuse it hs -itercept of ( 0,0), the 0 (0 )(0 ) 0. Hece, g ( ) ( )( ). Notice, it ws ot ecessr to use its verte of (,).
Emple The grph of h() hs verte t (, ), -itercept of ( 0, 9), d it opes dowwrd. So, it hs the form ( ) where 0. Becuse it hs -itercept of ( 0, 9), the 9 (0 ) 8 6 /. Hece, h ( ) ( ). lim 800 lim 800 lim 800 lim 800 Emple Three polomils re sketched below. For ech polomil, determie whether the ledig term hs eve or odd degree d whether the ledig coefficiet is positive or egtive. f (): Odd, Positive g() : Eve, Negtive h() : Odd, Negtive Emple The polomil sketched below hs degree of either 9 or 0 (ou will eed to determie which). The bsolute vlue of the ledig term is / 00. It hs ectl roots d the re ll itegers show i the grph. Fid the equtio of this polomil.
Becuse the grph hs the sme ed behvior i both directios, it hs eve degree; hece 0. Becuse it opes dowwrd, the ledig coefficiet is egtive; hece it is / 00. The ol possible roots re,, 0 d. Becuse the grph flttes out d crosses through the -is t, the the multiplicit of this root is odd d greter th ; hece it is t lest. Becuse the grph does ot look s though it flttes out t but does cross through the -is t, the the multiplicit of this root is odd d probbl equl to. Becuse the grph flttes out t 0 d ppers to bouce off of the -is t 0, the the multiplicit of this root is eve d greter th ; hece it is t lest. Becuse the grph flttes out t d ppers to bouce off of the -is t, the the multiplicit of this root is eve d greter th ; hece it is t lest. If ou dd up the miimum multiplicities ou get 8. This cot be correct becuse we kow the degree is 0, hece oe of these eeds to icrese b. Becuse the grph looks flttest t 0, I would guess tht tht root hs the highest multiplicit. This gives me polomil of: f ( ) ( ) ( ) 00 You could plot some poits to see tht this is the ol choice tht is resoble uder the give costrits. Emple Wht is the miimum degree of the polomil show below? The grph looks like it switches cocvit t bout. 8,., 0.,, d. If it switches cocvit times, its miimum degree is 7.