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{"id":23920,"date":"2018-03-27T11:45:28","date_gmt":"2018-03-27T04:45:28","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=23920"},"modified":"2018-03-27T11:45:28","modified_gmt":"2018-03-27T04:45:28","slug":"dai-so-chuong-1-bai-4-do-thi-cua-ham-so-va-phep-tinh-tien-he-toa-do","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/dai-so-chuong-1-bai-4-do-thi-cua-ham-so-va-phep-tinh-tien-he-toa-do\/","title":{"rendered":"\u0110\u1ea1i s\u1ed1 – Ch\u01b0\u01a1ng 1 – B\u00e0i 4: \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 v\u00e0 ph\u00e9p t\u1ecbnh ti\u1ebfn h\u1ec7 t\u1ecda \u0111\u1ed9"},"content":{"rendered":"

B\u00e0i 29 (trang 27 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

X\u00e1c \u0111\u1ecbnh \u0111\u1ec9nh I c\u1ee7a m\u1ed7i parabol (P) sau. Vi\u1ebft c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u00ednh ti\u1ebfn theo vect\u01a1 OI v\u00e0 vi\u1ebft Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol (P) \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY.<\/p>\n

a)y=2x2<\/sup>-3x+1<\/p>\n

b) y=1\/2 x2<\/sup>-x-3<\/p>\n

c) y=x-4x2<\/sup><\/p>\n

d) y=2x2<\/sup>-5<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

\u0110\u1ec9nh I c\u00f3 t\u1ecda \u0111\u1ed9 I(3\/4; -1\/8)<\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u00ednh ti\u1ebfn theo vect\u01a1 OI l\u00e0:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY l\u00e0:<\/p>\n

\u0110\u1ec9nh I(1; -7\/2)<\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ecd trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI l\u00e0:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY l\u00e0:<\/p>\n

\u0110\u1ec9nh I(1\/8;1\/16)<\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI l\u00e0:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY l\u00e0:<\/p>\n

\u0110\u1ec9nh I(0; -5)<\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI l\u00e0:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY l\u00e0:<\/p>\n

Y-5=2X2<\/sup>-5 hay Y=2X2<\/sup><\/p>\n

\n
\n

B\u00e0i 30 (trang 27 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b><\/span><\/p>\n

Cho h\u00e0m s\u1ed1 y=f(x)=x3<\/sup>-3x2<\/sup>+1<\/p>\n

a)X\u00e1c \u0111\u1ecbnh \u0111i\u1ec3m I thu\u1ed9c \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1 \u0111\u00e3 cho bi\u1ebft r\u1eb1ng ho\u00e0nh \u0111\u1ed9 c\u1ee7a \u0111i\u1ec3m I l\u00e0 nghi\u1ec7m c\u1ee7a Ph\u01b0\u01a1ng tr\u00ecnh f\u2019\u2019(x)= 0.<\/p>\n

b)Vi\u1ebft c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn vect\u01a1 OI v\u00e0 vi\u1ebft Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a \u0111\u01b0\u1eddng cong v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY. T\u1eeb \u0111\u00f3 suy ra b\u1eb1ng I l\u00e0 t\u00e2m \u0111\u1ed1i x\u1ee9ng \u0111\u01b0\u1eddng cong (C).<\/p>\n

c)Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn c\u1ee7a \u0111\u01b0\u1eddng cong (C) t\u1ea1i \u0111i\u1ec3m I \u0111\u1ed1i v\u1edbi h\u1ec7n t\u1ecda \u0111\u1ed9 Oxy. Ch\u1ee9ng minh r\u1eb1ng tr\u00ean kho\u1ea3ng (-\u221e;1) \u0111\u01b0\u1eddng cong (C) n\u1eb1m ph\u00eda d\u01b0\u1edbi ti\u1ebfp tuy\u1ebfn t\u1ea1i I c\u1ee7a (C) v\u00e0 tr\u00ean kho\u1ea3ng (1; +\u221e) \u0111\u01b0\u1eddng cong (C) n\u1eb1m ph\u00eda tr\u00ean ti\u1ebfp tuy\u1ebfn \u0111\u00f3.<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

f’ (x)=3x2<\/sup>-6x<\/p>\n

f” (x)=6x-6;f” (x)=0 <=> x=1 =>f (1) = -1<\/p>\n

V\u1eady I(1; -1)<\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a (C) \u0111\u1ed1i v\u1edbi h\u1ec7 tr\u1ee5c IXY l\u00e0:<\/p>\n

Y-1=(X+1)3<\/sup>-3(X+1)2<\/sup>+1 hay Y=X3<\/sup>-3X<\/p>\n

V\u00ec h\u00e0m s\u1ed1 Y=X3<\/sup>-3X l\u00e0 h\u00e0m s\u1ed1 l\u1ebb n\u00ean \u0111\u1ed3 th\u1ecb c\u1ee7a n\u00f3 nh\u1eadn g\u1ed1c t\u1ecda \u0111\u1ed9 I l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n

* Ti\u1ebfp tuy\u1ebfn v\u1edbi (C) t\u1ea1i I(1; -1) \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxy l\u00e0:<\/p>\n

y=f’ (1)(x-1)+f(1) v\u1edbi f\u2019(1) = -3; f(1) = -1<\/p>\n

N\u00ean Ph\u01b0\u01a1ng tr\u00ecnh ti\u1ebfp tuy\u1ebfn: y-3(x-1)+(-1) hay y = -3x + 2<\/p>\n

X\u00e9t hi\u1ec7u (x3<\/sup>-3x2<\/sup>+1)-(-3x+2)=(x-1)3<\/sup><\/p>\n

V\u1edbi x \u2208(-\u221e;1)=>(x-1)3<\/sup><0 n\u00ean \u0111\u01b0\u1eddng cong (C): y=x3<\/sup>-33<\/sup>+1 n\u1eb1m ph\u00eda d\u01b0\u1edbi ti\u1ebfp tuy\u1ebfn y = -3x + 2<\/p>\n

V\u1edbi x \u2208(1; +\u221e)=>(x-1)3<\/sup>>0 n\u00ean \u0111\u01b0\u1eddng cong (C): n\u1eb1m ph\u00eda tr\u00ean ti\u1ebfp tuy\u1ebfn t\u1ea1i I.<\/p>\n

B\u00e0i 31 (trang 27 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

Cho \u0111\u01b0\u1eddng cong (E):<\/p>\n

v\u00e0 \u0111i\u1ec3m I(-2; 2). Vi\u1ebft c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn vect\u01a1 OI v\u00e0 vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a \u0111\u01b0\u1eddng cong (E ) \u0111\u1ed1i v\u1edbi h\u1ec7 IXY. T\u1eeb \u0111\u00f3 suy a I t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a (E ).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo OI:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh (E ) trong h\u1ec7 t\u1ecda \u0111\u1ed9 IXY:<\/p>\n

V\u00ec Y=-1\/X l\u00e0 h\u00e0m s\u1ed1 l\u1ebb n\u00ean (C) nh\u1eadn g\u1ed1c t\u1ecda \u0111\u1ed9 I l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n

B\u00e0i 32 (trang 28 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

X\u00e1c \u0111\u1ecbnh t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb m\u1ed7i h\u00e0m s\u1ed1 sau \u0111\u00e2y:<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

(*) l\u00e0 c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI v\u1edbi I(1; 1) (\u0111\u1ed1i v\u1edbi h\u1ec7 tr\u1ee5c Oxy)<\/p>\n

\u0110\u1ed1i v\u1edbi h\u1ec7 tr\u1ee5c IXY, h\u00e0m s\u1ed1 y=2\/Y l\u00e0 h\u00e0m s\u1ed1 l\u1ebb n\u00ean \u0111\u1ed3 th\u1ecb nh\u1eadn g\u1ed1c t\u1ecda \u0111\u1ed9 l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n

V\u1eady t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n

\u0110\u00e2y l\u00e0 c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 tr\u1ee5c t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo vect\u01a1 OI v\u1edbi I(-1; 3)<\/p>\n

V\u00ec Y=-5\/X l\u00e0 h\u00e0m l\u1ebb n\u00ean \u0111\u1ed3 th\u1ecb nh\u1eadn g\u1ed1c t\u1ecda \u0111\u1ed9 I l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n

V\u1eady t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n

B\u00e0i 33 (trang 28 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):<\/b>\u00a0<\/span><\/p>\n

Cho \u0111\u01b0\u1eddng cong (C) c\u00f3 Ph\u01b0\u01a1ng tr\u00ecnh<\/p>\n

trong \u0111\u00f3 a \u2260 0,c \u2260 0 v\u00e0 I(xo<\/sub>,yo<\/sub>) th\u00f5a m\u00e3n yo<\/sub>=axo<\/sub>+b. Vi\u1ebft c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u00ednh ti\u1ebfn theo vect\u01a1 OI v\u00e0 Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a (C) \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY. T\u1eeb \u0111\u00f3 suy ra r\u1eb1ng t\u00e2m \u0111\u1ed1i x\u1ee9ng c\u1ee7a \u0111\u01b0\u1eddng cong (C).<\/p>\n

L\u1eddi gi\u1ea3i:<\/b><\/p>\n

C\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u1ecbnh ti\u1ebfn theo OI v\u1edbi I(xo<\/sub>,yo<\/sub>) l\u00e0:<\/p>\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a (C) \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY.<\/p>\n

D\u00f3 h\u00e0m s\u1ed1 Y=aX+c\/X l\u00e0 h\u00e0m s\u1ed1 l\u1ebb n\u00ean \u0111\u1ed3 th\u1ecb (C) c\u1ee7a h\u00e0m s\u1ed1 nh\u1eadn g\u1ed1c t\u1ecda \u0111\u1ed9 t\u00e2m I l\u00e0m t\u00e2m \u0111\u1ed1i x\u1ee9ng.<\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 29 (trang 27 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 X\u00e1c \u0111\u1ecbnh \u0111\u1ec9nh I c\u1ee7a m\u1ed7i parabol (P) sau. Vi\u1ebft c\u00f4ng th\u1ee9c chuy\u1ec3n h\u1ec7 t\u1ecda \u0111\u1ed9 trong ph\u00e9p t\u00ednh ti\u1ebfn theo vect\u01a1 OI v\u00e0 vi\u1ebft Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a Parabol (P) \u0111\u1ed1i v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 IXY. a)y=2×2-3x+1 b) y=1\/2 x2-x-3 c) y=x-4×2 d) y=2×2-5 L\u1eddi […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1302],"tags":[1392,1393],"yoast_head":"\n\u0110\u1ea1i s\u1ed1 - Ch\u01b0\u01a1ng 1 - B\u00e0i 4: \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 v\u00e0 ph\u00e9p t\u1ecbnh ti\u1ebfn h\u1ec7 t\u1ecda \u0111\u1ed9<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/dai-so-chuong-1-bai-4-do-thi-cua-ham-so-va-phep-tinh-tien-he-toa-do\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"\u0110\u1ea1i s\u1ed1 - Ch\u01b0\u01a1ng 1 - B\u00e0i 4: \u0110\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 v\u00e0 ph\u00e9p t\u1ecbnh ti\u1ebfn h\u1ec7 t\u1ecda \u0111\u1ed9\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 29 (trang 27 sgk Gi\u1ea3i T\u00edch 12 n\u00e2ng cao):\u00a0 X\u00e1c \u0111\u1ecbnh \u0111\u1ec9nh I c\u1ee7a m\u1ed7i parabol (P) sau. 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