[latex](3x-1)^2-(x+2)^2 =0\ 9x^2-6x+1-x^2-4x-4=0\ 8x^2-10x-3=0\ 8x^2+2x-12x-3=0\ 2x(4x+1)-3(4x+1)=0\ (2x-3)(4x+1)=0\ x=dfrac{3}{2} vee x=-dfrac{1}{4} [/latex] Zatem miejsca zerowe funkcji f, to [latex]x=-dfrac{3}{2} vee x=dfrac{1}{4}[/latex]
[latex]g(x)= (3x-1)^2-(x+2)^2 \ \ g(x)= 9x^2-6x+1-(x^2+4x+4) \ \ g(x)= 9x^2-6x+1-x^2-4x-4 \ \ g(x)= 8x^2-10x-3 \ \ x _{1 y 2}= frac{-bpm sqrt{b^2-4ac}}{2a}quad a=8quad b= -10quad c= -3 \ \ x _{1 y 2}= frac{-(-10)pm sqrt{100-4(8)(-3)}}{2(8)} o x _{1 y 2}= frac{10pm sqrt{100+96}}{16} o x _{1 y 2}= frac{10pm 14}{16} \ \ x_1= frac{10+14}{16} o x_1= frac{24}{16} o x_1= frac{3}{2} \ \ x_2= frac{10-14}{16} o x_2= frac{-4}{16} o x_2= - frac{1}{4} [/latex] [latex]g(x)= (3x-1)^2-(x+2)^2 \ \ x_1= frac{3}{2} qquad qquad x_2= - frac{1}{4} \ \ f(x) jest symetryczny do wykresu g(x) o OY \ \ f(x) : x_1= - frac{3}{2} qquad qquad x_2= frac{1}{4} [/latex] pozdrowienia!!!!
