[tex] f(x)=-x^2+6x+4\\a=-1,\ \ b=6, \ \ c=4 \\Before \ graphing \ we \ rearrange \ the \ equation, \ from \ this:\\\\f(x) = ax^2 + bx + c\\ To \ this:\\f(x) = a(x-h)2 + k\\Where:\\h =\frac{ -b}{2a} , \ \ \ k = f( h ) \\\\h=\frac{-6}{2\cdot (-1)}=\frac{6}{2}=3 \\\\k=f(3)
=-3^2+6 \cdot 3+4=-9+18+4=13 \\\\f(x)=-(x-3)^2+1[/tex]
[tex]\small vertex: (k,h) = (3,13) \\And\ the\ curve \ is \ symmetrical \ about \ the \ axis \ that \ passes \ through : \\ x=h \\x =3 \\ "a "\ is \ negative, \ so \ it \ is \ an \ "down" \ graph .[/tex]
