Odpowiedź:
[tex]a)\\\\f(x)=(x-3)^2\\\\f(x)=x^2-6x+9\\\\\\b)\\\\f(x)=-\frac{3}{4}(x+6)^2+12\\\\f(x)=-\frac{3}{4}(x^2+12x+36)+12=-\frac{3}{4}x^2-9x-27+12=-\frac{3}{4}x^2-9x-15\\\\f(x)=-\frac{3}{4}x^2-9x-15\\\\\\c)\\\\f(x)=-(x-\sqrt{3})^2+3\\\\f(x)=-(x^2-2\sqrt{3}x+3)+3=-x^2+2\sqrt{3}x-3+3=-x^2+2\sqrt{3}x\\\\f(x)=-x^2+2\sqrt{3}x[/tex]