Odpowiedź:
[tex]f(x) = x^2+bx+c\\A = (-4; 29)\\B = (1; 6)\\\text{Wierzcholek paraboli: } W = (p; q)\\\text{Rownanie osi symetrii: } p = \frac{-b}{2a}\\\\\left \{ {{29=(-4)^2-4b+c} \atop {6=(1)^2+b+c}} \right.\\\left \{ {{29=16-4b+c} \atop {6=1+b+c /*(-1)}} \right. \\+\left \{ {{29=16-4b+c} \atop {-6=-1-b-c}} \right. \\23=15-5b /-15\\8=-5b /:(-5)\\\frac{8}{-5}=b\\b=-1,6\\6=1-(-1,6)-c\\6=1+1,6-c\\6=2,6-c /-2,6\\3,4=-c\\c=-3,4\\f(x)=x^2-1,6x-3,4\\p=\frac{-(-1,6)}{2*1}\\p=\frac{1,6}{2}\\p=0.8\\[/tex]
Rownanie osi symetrii: x=0,8
