Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]y = -x^{2}+2x-5\\\\a = -1, \ b = 2, \ c = -5[/tex]
Współrzędne wierzchołka paraboli:
[tex]W = (p,q)\\\\p = \frac{-b}{2a} = \frac{-2}{2\cdot(-1)} = \frac{-2}{-2} = 1\\\\q = f(p) = f(1) = -1^{2}+2\cdot1 - 5 = -1+2-5 = -4\\\\W = (1,-4) \ \ \rightarrow \ \ x_{W} = 1, \ y_{W} = -4\\\\A = (2,-1) \ \ \rightarrow \ \ x_{A} = 2, \ y_{A} = -1\\\\|WA| = \sqrt{(x_{A}-x_{W})^{2}+(y_{A}-y_{W})^{2}}[/tex]
[tex]|WA|=\sqrt{(2-1)^{2}+(-1-(-4))^{2}} = \sqrt{1^{2}+3^{2}} = \sqrt{1+9} = \boxed{\sqrt{10}}[/tex]
