Răspuns :
Aflam ecuatia tangentei in punctul A(2,1)
A(x₀,y₀)
y-y₀=f'(x₀)(x-x₀)
y-1=f'(2)(x-2)
Calculam f'(x):
[tex]f'(x)=\frac{2(ax^2+bx)-(2x-1)(2ax+b)}{(ax^2+bx)^2} \\\\f'(2)=\frac{2(4a+2b)-3(4a+b)}{(4a+2b)^2} =\frac{-4a+b}{(4a+2b)^2} \\\\[/tex]
Ecuatia tangentei in A(2,1):
[tex]y-1=\frac{-4a+b}{(4a+2b)^2} (x-2)\\\\y=\frac{-4a+b}{(4a+2b)^2} (x-2)+1[/tex]
Panta unei drepte de ecuatie ax+by+c
[tex]m=-\frac{a}{b}[/tex]
In cazul nostru:
[tex]m=\frac{4a-b}{(4a+2b)^2}[/tex]
Ecuatia dreptei pentru a doua bisectoare a axelor de coordonate:
y=-x
m=-1
Pentru ca dreptele sa fie paralele, pantele trebuie sa fie egale
[tex]\frac{4a-b}{(4a+2b)^2}=-1[/tex]
Stim ca f(2)=1
[tex]f(2)=\frac{3}{4a+2b}[/tex]
[tex]\frac{3}{4a+2b} =1\\\\4a+2b=3[/tex]
[tex]\frac{4a-b}{9} =-1[/tex]
4a-b=-9
4a+2b=3
Le scadem si obtinem:
-3b=-12
b=4
4a-4=-9
4a=-5
[tex]a=-\frac{5}{4}[/tex]
