Odpowiedź:
[tex]ad.~a)~~W=(1,-4)\\\\ad.~b)~~x=1\\\\ad.~c)~~Z_{w} = < -4;+\infty)\\\\ad.~d)~~f(x) < 0~~dla~~x\in (-1;3)\\\\ad.~e)~~f(x)\searrow~~gdy~~x\in (-\infty;1 > ~~,~~f(x)\nearrow~~gdy~~x\in < 1;+\infty )[/tex]
Szczegółowe wyjaśnienie:
[tex]f(x)=x^{2} -2x-3\\\\a=1,~~b=-2,~~c=-3\\\\\Delta=b^{2} -4ac\\\\\Delta =(-2)^{2} -4\cdot 1\cdot (-3)=4+12=16\\\\\sqrt{\Delta} =\sqrt{16} =4\\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a} \\\\x_{1} =\dfrac{2-4}{2} =-1~~\lor~~x_{2} =\dfrac{2+4}{2} =3\\\\M_{o} =\{ -1,3 \}[/tex]
[tex]ad.~~a\\\\W=(p,q)~~-~~wspolrzedne~~wierzcholka\\\\p=-\dfrac{b}{2a} \\\\p=-\dfrac{(-2)}{2\cdot 1} =1\\\\q=-\dfrac{\Delta}{4a} \\\\q=-\dfrac{16}{4\cdot 1 } =-4\\\\\boxed{W=(1,-4)}[/tex]
[tex]ad.~~b\\\\x=p=-\dfrac{b}{2a} ~~-~~os~~symetrii~~funkcji~~kwadratowej\\\\\boxed{x=1}~~-~~rownanie~~osi~~symetrii[/tex]
[tex]ad.~~c\\\\a=1~~\Rightarrow~~a > 0~~\Rightarrow~~ramiona ~~padaboli~~skierowane ~~do~~gory\\\\q=-4\\\\a=-1~~\land~~q=-4~~\Rightarrow~~\boxed{Z_{w} = < -4, +\infty )}[/tex]
[tex]ad.~~d\\\\f(x) > 0~~dla~~x\in (-\infty;1-)\cup (3;+\infty)\\\\\boxed{f(x) < 0~~dla~~x\in (-1;3)}[/tex]
[tex]ad.~~e\\\\\boxed{f(x)\searrow~~gdy~~x\in (-\infty;1 > }\\\\\boxed{f(x)\nearrow~~gdy~~x\in < 1;+\infty )}[/tex]