HE SO GOC CUA ẹệễỉNG
THANG y = ax + b (
a 0)
!"#$%&'('%)*+,+
/01234/015234
6
2
0
2 7 54
0 4 7
2 7 4
0 4 7
8794:
85497:
8794:
8497:
54
0
1
2
3
4
2
0
6
0
1
5
2
3
4
;
;
<=>01-23?8-@7:
AB%(CDCE*%E+,FG''H-"IJDF$%KDF01-23?8-@7:
-BG'$LM?NC"IJDF$%KDF01-23?O)$PQ'62
54
0
1
2
3
4
4
;
A
;
4
0
1
7
R
2
3
4
5S
5A
0
1
4
2
3
4
T
;
2
6
0
?BE+,FG'
* Xeùt ño àthò caùc
haøm soá sau :
y = 0,5x+2
y = x+2
y = 2x+2
/%C%E+,'H-"#$%&%)*+,
8-@7:$%UFG'$LM?NC"#
$%&O)$PQ'62V)O)%E+,
D%IDFOWD
6
0
2
8794:
0
1
5
7
R
2
3
4
S
T
X
0
1
5
2
3
4
4
4
X
0
1
5
4
2
3
4
A
A
X
* Xeùt ño àthò caùc
haøm soá sau :
y = -0,5x+2
y = -x+2
y = -2x+2
/%C%E+,'H-"#$%&%)*+,
8-@7: !$%UFG'$LM?NC"#
$%&O)$PQ'62V)"O)%E+,
D%IDFOWD#$
=Y
/E+,-V)%'H-"IJDF
$%KDF01-23?8-@7:
/E+,?V)$.DF"ZFG''H-
"IJDF$%KDF01-23?8-@7:
TB)C$[\
4B]^Q8P-DF_3`B:
CaD8&':%Mb'8(:O)Mc$P,DF
O))*+,(
d.
AB,TV)%E+,FG''H-"#$%&%)*+,
01T2e
4BE+,FG''H-"IJDF$%KDF
0154Rf23AV)e4Rf2
TB,SV)%E+,FG''H-"#$%&%)*+,
= −
3
y 4 x
2
54Rf
3
2
−
T
SB)C$[\4_B`%M%)*+,?['D%g$01-23T
-Bh('"&D%%E+,FG'-R?Ci$PjDF"#$%&'H-%)*+,
"Ck "Cl*849m:
?B!"#$%&'H-%)*+,
CnC
2 7 54
0 T 7
3
b/. Ve ừ y = x 3
2
+
6
2
0
a/ .Theo ủe baứi ta coự:
6 = a.2 +3
6-3 = 2a
3 = 2a
3
a =
2
3
Vaọyheọ soỏ goựclaứ
2
85497:
879T:
3
y = x 3
2
+
SB)C$[\4`-B`
!"#$%&%)*+,015423T
CnC
2 7 AR A
0 T 7 A
8AR97:
879T:
0
1
5
4
2
3
T
A
A
6
2
0
/oD$%C%M%)*+,?['D%g$018*5A:23A
-Bh('"&D%(-./0 ?Ci$PjDF"#$%&'H-%)*+,
+MDF+MDFOpC"IJDF$%KDF0125
?Bh('"&D%%11234/'H-%)*+,$U*"Iq'
CnC
2 7 5A
0 A 7
6
2
0
85A97:
879A:
0
1
2
3
A
a/ .Theo đe àbài ta có:
m-1 0 m 1
m-1 = 1 m = 2
Vậykhi m = 2 thì đồthò
hàmsố y (m 1)x 1
songsongvớiđườngthẳng
y x 5.
≠ ≠
⇔
= − +
= −
?B/E+,FG'V)-1A
/!0123A
a
D%)
/r'$%.Z'?)C
/hs*O]^QA34B_3`B
/t)*?)C$[\4`?B`B
/%.uD?&?)C$[\\%vDV.0ED$[\
/%w'%CED?)C$[\4_34`-O)M$[\
cD$%C%r'xyA