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Určete Taylorův polynom druhého stupně funkce

$f(x)=\sin(3x-\frac{\pi}{3})-9.$ f(x)=\sin(3x-\frac{\pi}{3})-9. se středem v bodě $a=\frac{\pi}{3}$ a=\frac{\pi}{3}.

$-9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{3})^2$ -9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})+\frac{9\sqrt 3}{4}(x-\frac{\pi}{3})^2

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$-9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{3})^2$ -9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 3}{2}(x-\frac{\pi}{3})^2

$-9+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{3})-\frac{\sqrt 3}{4}(x-\frac{\pi}{3})^2$ -9+\frac{\sqrt 3}{2}-\frac{1}{2}(x-\frac{\pi}{3})-\frac{\sqrt 3}{4}(x-\frac{\pi}{3})^2

$\frac{-19}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{9}{4}(x-\frac{\pi}{3})^2$ \frac{-19}{2}-\frac{3\sqrt 3}{2}(x-\frac{\pi}{3})-\frac{9}{4}(x-\frac{\pi}{3})^2

$-9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{3})^2$ -9+\frac{\sqrt 3}{2}-\frac{3}{2}(x-\frac{\pi}{3})-\frac{9\sqrt 3}{4}(x-\frac{\pi}{3})^2

$-9+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{3})-\frac{9\sqrt 3}{2}(x+\frac{\pi}{3})^2$ -9+\frac{\sqrt 3}{2}-\frac{3}{2}(x+\frac{\pi}{3})-\frac{9\sqrt 3}{2}(x+\frac{\pi}{3})^2

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Matematika - LS 24/25

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