[tex]f(x)=ax^2+bx+c\\\\f(x)=-x^2+4x-1\\\\a=-1,\ b=4,\ c=-1\\\\a < 0\ \text{then the parabola ope}\text{n down,}\\\text{ terefore the maximum value is in vertex}.\\\\\text{The formula of a vertex}\\\\(h,\ k),\ \text{where}\ h=\dfrac{-b}{2a},\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[/tex]
[tex]h=\dfrac{-4}{2(-1)}=\dfrac{-4}{-2}=2\\\\k=f(2)=-2^2+4(2)-1=-4+8-1=3[/tex]
[tex]\text{Answer: the maximum value is euqal 3 for x = 2}[/tex]
