Question 763483


{{{f(x)= -x^2+ 2x+ 6}}}....since coefficient in front of {{{x^2}}} is {{{-1}}}, parabola opens downward and the vertex is its maximum 

write function in vertex form: {{{y = a(x-h)^2+k}}} where {{{h}}} and {{{k}}} are coordinates of the vertex

{{{y= -x^2+ 2x+ 6}}}.......complete the square, subtract {{{6}}} from both sides

{{{y-6= -x^2+ 2x+ 6-6}}} 

{{{y-6= -x^2+ 2x}}} ...factor out the leading coefficient {{{-1}}}

{{{y-6=-1(x^2-2x)}}}....take half of the {{{x}}} coefficient {{{-2}}} to get {{{-1}}}, add and subtract

{{{y-6=-1(x^2-2x+1-1)}}}

{{{y-6=-1((x^2-2x+1)-1)}}}

{{{y-6=-1((x-1)^2-1)}}}

{{{y-6=-(x-1)^2-(-1)}}}

{{{y-6=-(x-1)^2+1}}}

{{{y=-(x-1)^2+1+6}}}

{{{y=-(x-1)^2+7}}}

Vertex: {{{h = 1}}}, {{{k =7}}}

so, the maximum {{{y}}} value on the graph is {{{7}}}


{{{ graph( 600, 600, -10, 10, -10, 10, -(x-1)^2+7) }}}