Question 131896
To find the x-coordinate of the vertex, we can use this formula:


{{{x=-b/(2a)}}}



From the equation {{{f(x)=-2x^2+2x+6}}} we can see that a=-2 and b=2


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



{{{x=(-2)/-4}}} Multiply 2 and -2 to get -4




{{{x=1/2}}} Reduce



So the  x-coordinate of the vertex is  {{{x=1/2}}}. 



This also means that the line of symmetry is {{{x=1/2}}}.



Now lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(1/2)}}}


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



{{{f(1/2)=-2(1/2)^2+2(1/2)+6}}} Plug in {{{x=1/2}}}



{{{f(1/2)=-2(1/4)+2(1/2)+6}}} Square {{{1/2}}} to get {{{1/4}}}



{{{f(1/2)=-1/2+2(1/2)+6}}} Multiply -2 by {{{1/4}}} to get {{{-1/2}}}



{{{f(1/2)=-1/2+1+6}}} Multiply 2 by {{{1/2}}} to get 1



{{{f(1/2)=13/2}}} Combine like terms



So the vertex is *[Tex \LARGE \left(\frac{1}{2},\frac{13}{2}\right)]




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Answer:


This means that the x-coordinate of the vertex is *[Tex \LARGE \frac{1}{2}] and the y-coordinate of the vertex is *[Tex \LARGE \frac{13}{2}]