Odpowiedź:
[tex]\huge\boxed{\left(\dfrac{-5+2\sqrt5}{2},\ -\dfrac{\sqrt5-2}{4}\right)}[/tex]
Szczegółowe wyjaśnienie:
Postać ogólna funkcji kwadratowej:
f(x) = ax² + bx + c
(p, q) - współrzędne wierzchołka
p = -b/2a
q = f(p) = -Δ/4q
Δ = b² - 4ac
Mamy funkcję:
[tex]f(x)=\dfrac{1}{\sqrt5-2}x^2+\sqrt5x+\dfrac{1}{\sqrt5+2}\\\\a=\dfrac{1}{\sqrt5-2},\ b=\sqrt5,\ c=\dfrac{1}{\sqrt5+2}[/tex]
[tex]p=\dfrac{-\sqrt5}{2\cdot\dfrac{1}{\sqrt5-2}}\\\\p=\dfrac{-\sqrt5}{\dfrac{2}{\sqrt5-2}}\\\\p=-\sqrt5\cdot\dfrac{\sqrt5-2}{2}\\\\\huge\boxed{p=\dfrac{-5+2\sqrt5}{2}}[/tex]
[tex]\Delta=\left(\sqrt5\right)^2-4\cdot\dfrac{1}{\sqrt5-2}\cdot\dfrac{1}{\sqrt5+2}=5-\dfrac{4}{(\sqrt5-2)(\sqrt5+2)}\\\\=5-\dfrac{4}{(\sqrt5)^2-2^2}=5-\dfrac{4}{5-4}=5-\dfrac{4}{1}=5-4=1[/tex]
[tex]q=\dfrac{-1}{4\cdot\dfrac{1}{\sqrt5-2}}\\\\q=\dfrac{-1}{\dfrac{4}{\sqrt5-2}}\\\\\huge\boxed{q=-\dfrac{\sqrt5-2}{4}}[/tex]
