Odpowiedź:
[tex]a)\ \ f(x)=2x^2-4x+3\\\\a=2\ \ ,\ \ b=-4\\\\x_{w}=-\frac{b}{2a}=-\frac{-4}{2\cdot2}=-\frac{-4}{4}=1\\\\y_{w}=f(x_{w})=f(1)=2\cdot1^2-4\cdot1+3=2\cdot1-4+3=2-4+3=5-4=1\\\\Wierzcholkiem\ \ paraboli\ \ jest\ \ punkt\ \ (1,1)[/tex]
[tex]b)\ \ f(x)=-x^2-3x+4\\\\a=-1\ \ ,\ \ b=-3\\\\x_{w}=-\frac{b}{2a}=-\frac{-3}{2\cdot(-1)}=-\frac{-3}{-2}=-\frac{3}{2}=-1\frac{1}{2}\\\\y_{w}=f(x_{w})=f(-\frac{3}{2})=-(-\frac{3}{2})^2-3\cdot(-\frac{3}{2})+4=-\frac{9}{4}+\frac{9}{2}+4=-\frac{9}{4}+\frac{18}{4}+\frac{16}{4}=\\\\=\frac{25}{4}=6\frac{1}{4}\\\\Wierzcholkiem\ \ paraboli\ \ jest\ \ punkt\ \ (-1\frac{1}{2},6\frac{1}{4})[/tex]
[tex]c)\ \ f(x)=\frac{1}{4}x^2-x+10\\\\a=\frac{1}{4}\ \ ,\ \ b=-1\\\\x_{w}=-\frac{b}{2a}=-\frac{-1}{2\cdot\frac{1}{4}}=-\frac{-1}{\frac{1}{2}}=1:\frac{1}{2}=1\cdot2=2\\\\y_{w}=f(x_{w})=f(2)=\frac{1}{4}\cdot2^2-2+10=\frac{1}{\not4}\cdot\not4+8=1+8=9\\\\Wierzcholkiem\ \ paraboli\ \ jest\ \ punkt\ \ (2,9)[/tex]
