Question 396135
The x-coordinate of the vertex of a parabola in the form {{{f(x)=ax^2+bx+c}}} is given by {{{-b/2a}}}.


The y-coordinate is then {{{f(-b/2a)}}}.


The line of symmetry passes through the vertex, so the equation is {{{x=-b/2a}}}.


The maximum or minimum is the value of the function at {{{-b/2a}}}.  Whether it is a maximum or minimum depends on whether the parabola opens up or down.  If it is concave up (makes a valley rather than a hill), the point is a minimum, otherwise it is a maximum.  You can tell which way the parabola opens by the sign on the lead coefficient.  if {{{a<0}}}, it is concave down, if {{{a>0}}}, it is concave up, and, of course, if {{{a=0}}} you don't have a parabola at all.


Let's look at your specific problem:


{{{f(x) = -2x^2+2x+1 }}}


First thing to note is that {{{a<0}}}, so this is a concave down parabola and the vertex is a {{{maximum}}}.


{{{-b/2a=-2/(2(-2))=1/2}}}, so the x-coordinate of the vertex is {{{1/2}}} and the equation of the line of {{{symmetry}}} is {{{x=1/2}}}.


The value of the function at {{{-b/2a}}}, denoted {{{f(-b/2a)}}} for your problem is {{{f(1/2)=-2(1/2)^2+2(1/2)+1 = (-1/2)+1+1=3/2}}} 

So the y-coordinate of the vertex and the {{{maximum}}} value of {{{f}}} is {{{3/2}}}



{{{drawing(600,600,-5,5,-5,5,
grid(1),
graph(600,600,-5,5,-5,5,-2x^2+2x+1),
circle(1/2,3/2,.05),
locate(.7,5,V(1/2,3/2)),
green(line(1/2,-5,1/2,5)))}}}