Question 946390
Your function is f(x)=300x-11x^2+2, restatable as f(x)=-11x^2+300x+2.
Adding the proper rendering tags, {{{f(x)=-11x^2+300x+2}}}.


The function f is a parabola opening downward, therefore having a maximum.  If f has real
roots, they can be used for finding the symmetry axis, and from it, the maximum of f.


Not clear which method you are allowed here.  One way is, complete the square to put into
standard form.  The other way is, find the roots of f, find the midpoint of those roots, and
use that value to find corresponding f.


I will start this using Completing The Square, which might possibly not be the easiest way.


{{{-11x^2+300x+2}}}
{{{-11(x^2-(300/11)x-2/11)}}}
{{{-11(x^2-(300/11)x+(150/11)^2-2/11-(150/11)^2)}}}
{{{-11((x-150/11)^2-2/11-(150/11)^2)}}}
{{{-11((x-150/11)^2-22/11^2-150^2/11^2)}}}
{{{-11((x-150)^2-(22+150^2)/11^2)}}}
Now distrubute the previously factored Negative Eleven.
{{{highlight(-11(x-150)^2+(22+150^2)/11)}}}


Now the equivalent form of your function is {{{highlight(f(x)=-11(x-150)^2+(22+150^2)/11)}}}
and I have simply not finished the computation for the constant term.  The vertex for the
maximum point is  at {{{highlight(x=150)}}},  {{{highlight(highlight(y=(22+150^2)/11))}}},  this being the maximum value.



You may need to study or review how to Complete The Square, and also about Standard Form for a parabola.