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p,q - współrzędne wierzchołka paraboli
f(x)=a(x+2)²+3
f(x)=a(x²+4x+4)+3
f(x)=ax²+4ax+4a+3
f(3)=0=9a+12a+4a+3
-3=25a
a=-3/25
p=-b/2a
-2=-b/2a
2=b/2a
b=4a
b=-12/25
f(x)=ax²+bx+c
f(3)=-3/25×9+3×(-12/25)+c
0=-27/25-36/25+c
c=63/25
Odp. f(x)=-3/25x²-12/25x+63/25
DANE
W=(-2,3)
x1=3
szukam y=ax²+bx+c
W(p,q)
p=-b/2a
q=f(p) lub q=-Δ/(4a)
-------------------------------
x1=3 → y(3)=0→9a+3b+c=0
-b/2a=-2→b=4a
a*b²/(4a²)-b²/2a+c=3
------------------------------
9a+12a+c=0
1/4*b²/a-1/2*b²/a+c=3
------------------------------
21a+c=0
-1/4*b²/a+c=3
---------------------------
c=-21a
-4a+c=3
---------------------------
-4a-21a=3
-25a=3
-------------------
a=-3/25=-0,12
b=4a=-12/25=-0,48
c=-21a=63/25=2,52
y=-0,12x²-0,48x+2,52
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