Trn S Tựng Hm s bc nht bc hai CHNG II II CHNG HM S S BC BC NHT NHT V V BC BC HAI HAI HM HM S S I.I HM nh ngha Cho D R, D Hm s f xỏc nh trờn D l mt qui tc t tng ng mi s x D vi mt v ch mt s y R x gl bin s (i s), y gl giỏ tr ca hm s f ti x Kớ hiu: y = f(x) D gl xỏc nh ca hm s T = { y = f ( x ) x D} gl giỏ tr ca hm s Cỏch cho hm s Cho bng bng Cho bng biu Cho bng cụng thc y = f(x) Tp xỏc nh ca hm s y = f(x) l hp tt c cỏc s thc x cho biu thc f(x) cú ngha th ca hm s th ca hm s y = f(x) xỏc nh trờn D l hp tt c cỏc im M ( x; f ( x ) ) trờn mt phng to vi mi x D Chỳ ý: Ta thng gp th ca hm s y = f(x) l mt ng Khi ú ta núi y = f(x) l phng trỡnh ca ng ú S bin thiờn ca hm s Cho hm s f xỏc nh trờn K Hm s y = f(x) ng bin (tng) trờn K nu x1 , x2 K : x1 < x2 f ( x1 ) < f ( x2 ) Hm s y = f(x) nghch bin (gim) trờn K nu x1 , x2 K : x1 < x2 f ( x1 ) > f ( x2 ) Tớnh chn l ca hm s Cho hm s y = f(x) cú xỏc nh D Hm s f gl hm s chn nu vi x D thỡ x D v f(x) = f(x) Hm s f gl hm s l nu vi x D thỡ x D v f(x) = f(x) Chỳ ý: + th ca hm s chn nhn trc tung lm trc i xng + th ca hm s l nhn gc to lm tõm i xng VN 1: Tỡm xỏc nh ca hm s Tỡm xỏc nh D ca hm s y = f(x) l tỡm tt c nhng giỏ tr ca bin s x cho biu thc f(x) cú ngha: D = { x R f ( x ) coự nghúa} iu kin xỏc nh ca mt s hm s thng gp: P( x ) 1) Hm s y = : iu kin xỏc nh: Q(x) Q( x ) 2) Hm s y = R( x ) : iu kin xỏc nh: R(x) Chỳ ý: + ụi ta s dng phi hp cỏc iu kin vi + iu kin hm s xỏc nh trờn A l A D A + A.B B Baứi Tỡnh giỏ tr ca cỏc hm s sau ti cỏc im ó ch ra: Trang www.MATHVN.com Hm s bc nht bc hai Trn S Tựng a) f ( x ) = x Tớnh f(0), f(2), f(2), f(3) b) f ( x ) = x Tớnh f(2), f(0), f(3), f(2) x 3x + c) f ( x ) = x + x Tớnh f(2), f(2), f(0), f(1) x x < d) Tớnh f(2), f(0), f(1), f(2) f(3) f ( x ) = x + x x x > e) Baứi a) d) g) Baứi a) x < f ( x ) = x = Tớnh f(2), f(1), f(0), f(2), f(5) x > Tỡm xỏc nh ca cỏc hm s sau: 2x + x b) y = y= 3x + 2x x x y= y= e) x 3x + 2 x 5x + 2x +1 x y= y= h) ( x 2)( x x + 3) x3 + Tỡm xỏc nh ca cỏc hm s sau: b) y = x y = 2x d) y = x + g) y = Baứi a) b) c) x e) y = 2x ( x 2) x Tỡm a hm s xỏc nh trờn K ó ch ra: 2x + y= ; K = R x2 6x + a 3x + y= ; K = R x 2ax + K = (0; +) y = x a + 2x a ; d) y = x 3a + + xa ; K = (0; +) x + a x + 2a ; x a +1 + x + 2a + ; f) y = xa e) y = x + a + + ; xa e) y = www.MATHVN.com www.MATHVN.com f) y = i) y = x+4 3x x2 + x + 1 x + 2x2 c) y = x + x + f) y = x + x + ( x + 2) x h) y = x + c) y = x i) y = x + + x2 S: a > 11 S: < a < S: a S: a K = (1; 0) S: a hoc a K = (1; 0) S: a K = (1; +) S: a Trang Trn S Tựng Hm s bc nht bc hai VN 2: Xột s bin thiờn ca hm s Cho hm s f xỏc nh trờn K y = f(x) ng bin trờn K x1 , x2 K : x1 < x2 f ( x1 ) < f ( x2 ) x1 , x2 K : x1 x2 f ( x2 ) f ( x1 ) >0 x2 x1 y = f(x) nghch bin trờn K x1 , x2 K : x1 < x2 f ( x1 ) > f ( x2 ) x1 , x2 K : x1 x2 f ( x2 ) f ( x1 ) 0, hm s ng bin trờn R + Khi a < 0, hm s nghch bin trờn R th l ng thng cú h s gúc bng a, ct trc tung ti im B(0; b) Chỳ ý: Cho hai ng thng (d): y = ax + b v (d): y = ax + b: + (d) song song vi (d) a = a v b b + (d) trựng vi (d) a = a v b = b + (d) ct (d) a a Hm s y = ax + b (a 0) b x ax + b a y = ax + b = b (ax + b) x < a Chỳ ý: v th ca hm s y = ax + b ta cú th v hai ng thng y = ax + b v y = ax b, ri xoỏ i hai phn ng thng nm phớa di trc honh Baứi V th ca cỏc hm s sau: a) y = x b) y = x + c) y = Baứi Tỡm to giao im ca cỏc cp ng thng sau: a) y = x 2; x d) y = x y = 2x + b) y = x + 2; y = 4( x 3) x x y = x c) y = x; d) y = ; y= Baứi Trong mi trng hp sau, tỡm giỏ tr k th ca hm s y = x + k ( x + 1) : a) i qua gc ta O b) i qua im M(2 ; 3) c) Song song vi ng thng y = 2.x Baứi Xỏc nh a v b th ca hm s y = ax + b : a) i qua hai im A(1; 20), B(3; 8) b) i qua im M(4; 3) v song song vi ng thng d: y = x + y = x + c) Ct ng thng d1: ti im cú honh bng v ct ng thng d2: y = x + ti im cú tung bng d) Song song vi ng thng y = x v i qua giao im ca hai ng thng y = x + v y = x + Baứi Trong mi trng hp sau, tỡm cỏc giỏ tr ca m cho ba ng thng sau phõn bit v ng qui: a) y = x; y = x 3; y = mx + y = 3x + m b) y = 5( x + 1); y = mx + 3; www.MATHVN.com www.MATHVN.com Trang 10 Trn S Tựng Hm s bc nht bc hai c) y = x 1; y = x; y = (3 2m) x + d) y = (5 3m) x + m 2; y = x + 11; y = x + e) y = x + 5; y = x 7; y = (m 2) x + m + Baứi Tỡm im cho ng thng sau luụn i qua dự m ly bt c giỏ tr no: a) y = 2mx + m b) y = mx x c) y = (2m + 5) x + m + d) y = m( x + 2) e) y = (2m 3) x + f) y = (m 1) x 2m Baứi Vi giỏ tr no ca m thỡ hm s sau ng bin? nghch bin? a) y = (2m + 3) x m + b) y = (2m + 5) x + m + c) y = mx x d) y = m( x + 2) Baứi Tỡm cỏc cp ng thng song song cỏc ng thng cho sau õy: x a) 3y x + = b) y = 0,5 x c) y = + d) y + x = e) x y = f) y = 0,5 x + Baứi Vi giỏ tr no ca m thỡ th ca cỏc cp hm s sau song song vi nhau: m 2(m + 2) 3m 5m + a) y = (3m 1) x + m + 3; y = x b) y = x+ ; y= x m m 3m + 3m + c) y = m( x + 2); y = (2m + 3) x m + Baứi 10 V th ca cỏc hm s sau: x x x x < < x < x a) y = b) y = x x x x c) y = x + d) y = x f) y = x + x g) y = x x 1 2x + + 2 h) y = x + x + x + e) y = Baứi 11 a) Trang 11 www.MATHVN.com Hm s bc nht bc hai Trn S Tựng III HM HM S S BC BC HAI HAI III y = ax + bx + c (a 0) Tp xỏc nh: D = R S bin thiờn: b b th l mt parabol cú nh I ; ữ, nhn ng thng x = lm trc i 2a 4a 2a xng, hng b lừm lờn trờn a > 0, xuụng di a < Chỳ ý: v ng parabol ta cú th thc hin cỏc bc nh sau: b Xỏc nh to nh I ; ữ 2a 4a b Xỏc nh trc i xng x = v hng b lừm ca parabol 2a Xỏc nh mt s im c th ca parabol (chng hn, giao im ca parabol vi cỏc trc to v cỏc im i xng vi chỳng qua trc trc i xng) Cn c vo tớnh i xng, b lừm v hỡnh dỏng parabol v parabol Baứi Xột s bin thiờn v v th ca cỏc hm s sau: a) y = x x b) y = x + x + c) y = x + x d) y = x + x e) y = x x + f) y = x x + Baứi Tỡm to giao im ca cỏc cp th ca cỏc hm s sau: a) y = x 1; b) y = x + 3; y = x2 2x y = x2 4x + c) y = x 5; y = x2 4x + e) y = x x + 1; y = 3x + x Baứi Xỏc nh parabol (P) bit: d) y = x x 1; y = x x + f) y = x + x + 1; y = x + x b) (P): y = ax + bx + i qua im A(1; 9) v cú trc i xng x = a) (P): y = ax + bx + i qua im A(1; 0) v cú trc i xng x = c) (P): y = ax + bx + c i qua im A(0; 5) v cú nh I(3; 4) d) (P): y = ax + bx + c i qua im A(2; 3) v cú nh I(1; 4) e) (P): y = ax + bx + c i qua cỏc im A(1; 1), B(1; 3), O(0; 0) f) (P): y = x + bx + c i qua im A(1; 0) v nh I cú tung bng Baứi Chng minh rng vi mi m, th ca mi hm s sau luụn ct trc honh ti hai im phõn bit v nh I ca th luụn chy trờn mt ng thng c nh: www.MATHVN.com www.MATHVN.com Trang 12 Trn S Tựng Hm s bc nht bc hai m2 b) y = x 2mx + m 1 Baứi V th ca hm s y = x + x + Hóy s dng th bin lun theo tham s m, s im chung ca parabol y = x + x + v ng thng y = m a) y = x mx + Baứi V th ca cỏc hm s sau: a) y = x x + b) y = x ( x ) c) y = x x x 2 x + x neỏu x x < neỏu x < d) y = e) y = f) y = 2 x x neỏu x x + x + neỏu x < x x x Baứi a) BI TP ễN CHNG II Bi Tỡm xỏc nh ca cỏc hm s sau: a) y = x y= a) d) Bi a) x+4 b) y = x 1+ x x e) y = x + + 2x x c) 3x x x2 x + x d) y = Bi x2 + 2x + x Xột s bin thiờn ca cỏc hm s sau: x +1 trờn (1; +) y = x + x trờn (; 2) b) y = x 1 e) y = y = 2x x Xột tớnh chn l ca cỏc hm s sau: x + x2 b) y = + x + x y= x2 f) y = c) y = f) y = 2x x x x x +3 trờn (2; +) x c) y = x ( x + x ) x +1 + x x x e) y = f) y = x 2 x +1 x x +1 Bi Gi s y = f(x) l hm s xỏc nh trờn i xng D Chng minh rng: a) Hm s F ( x ) = [ f ( x ) + f ( x )] l hm s chn xỏc nh trờn D b) Hm s G( x ) = [ f ( x ) f ( x )] l hm s l xỏc nh trờn D c) Hm s f(x) cú th phõn tớch thnh tng ca mt hm s chn v mt hm s l Bi Cho hm s y = ax + bx + c (P) Tỡm a, b, c d) y = Tỡm a, b, c tho iu kin c ch Kho sỏt s bin thiờn v v th (P) ca hm s va tỡm c Tỡm m ng thng d ct (P) ti hai im phõn bit A v B Xỏc nh to trung im I ca on AB Trang 13 www.MATHVN.com Hm s bc nht bc hai Trn S Tựng a) (P) cú nh S ; ữ v i qua im A(1; 1); d: y = mx b) (P) cú nh S(1; 1) v i qua im A(0; 2); d: y = x + m Bi a) www.MATHVN.com www.MATHVN.com Trang 14