Question 90267
The maximum value occurs at the vertex of the graph

First determine the x-value of the vertex:



{{{x=-b/(2a)}}} Here is the general formula to find the x-value of the vertex


From the equation {{{y=-x^2+6x}}} we can see that a=-1 and b=6


{{{x=(-6)/(2*-1)}}} Plug in b=6 and a=-1



{{{x=(-6)/-2}}} Multiply 2 and -1 to get -2


{{{x=3}}} Reduce




So the x-coordinate of the vertex is {{{x=3}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(3)}}}


{{{f(x)=-x^2+6x}}} Start with the given polynomial



{{{f(3)=-(3)^2+6(3)}}} Plug in {{{x=3}}}



{{{f(3)=-(9)+6(3)}}} Raise 3 to the second power to get 9



{{{f(3)=-(9)+18}}} Multiply 6 by 3 to get 18



{{{f(3)=9}}} Now combine like terms



So the vertex is (3,9)



So the maximum value is 9



If we graph {{{y=-x^2+6x}}}, we can clearly see the maximum value


{{{drawing(900,900,-7,13,-1,19,
grid( 1 ),
graph(900,900,-7,13,-1,19, -x^2+6x),

circle(3,9,0.05),
circle(3,9,0.08)
)}}}