Question

determinați coordonatele vârfului parabolei asociate funcției f(x)=2x^2-8x+1​

Answer 1

 

$\displaystyle\bf\\F(x)=2x^2-8x+1\\\\Varful~parabolei~este:~~V(x,y)\\\\unde:\\\\x=\frac{-b}{2a}=\frac{-(-8)}{2\cdot2}=\frac{8}{4}=\boxed{\bf2}\\\\y=\frac{-\Delta}{4a}=\frac{-(b^2-4ac)}{4a}=\frac{-((-8)^2-4\cdot 2\cdot c)}{4a}=\\\\=\frac{-((-8)^2-4\cdot2\cdot 1)}{4a}=\frac{-((-8)^2-8)}{4\cdot2}=\\\\=\frac{-(64-8)}{8}=\frac{-56}{8}=\boxed{\bf-7}\\\\\implies~~\boxed{\bf V(2;~-7)}$