First, factor the leading coefficient 3 out of the first to terms to write the function as #f(x)=3(x^2-18x)+241#.
Now complete the square with #x^2-18x# to write it as #x^2-18x+81-81# (take the #-18#, divide it by 2 to get #-9#, and square that to get #81#, which gets both added and subtracted).
Since #3*(-81)=-243#, we can say that #f(x)=3(x^2-18x+81)+241-243=3(x-9)^2-2#.
This means the vertex is at #(x,y)=(9,-2)# and is the low point of the parabola (the parabola opens upward since the leading coefficient is positive) and the axis of symmetry is the vertical line #x=9#.
The #x#-intercepts can be found either with the quadratic formula used on the equation or by solving #3(x-9)^2-2=0# to get #x-9=\pm sqrt{2/3}# so that #x=9\pm sqrt{2/3}=\frac{27\pm \sqrt{6}}{3}\approx 8.18\mbox{ or } 9.82#.
To help graph the function, you can plot some other points. For example, #f(8)=f(10)=3-2=1# and #f(7)=f(11)=3*4-2=10#.
