Răspuns:
f(x)=(2m-5)x+5m++3
a) Pui conditia ca f(-2)=14
(2m-5)*(-2)+5m+3=14
-4m+10+5m+3=14
m+13=14=>
m=1
b) f(x)=(2*1-5)x+5*1+3=
(2-5)x+8= -3x+8
Intersectia cuOx
f(x)=0
-3x+8=0
x=3/8
A(3/8,0)
Intersectia cu Oy
f(0)=-3*0+8=0+8=8
B(0,8)
Trasezi dreapta AB
c)Pui conditia f(x)=x
-3x+8=x
-3x-x=-8
-4x= -8
x=2
f(2)=-3*2+8=-6+8=2
C(2,2)
Vezi atasament
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