Question 1126030
{{{f(x) = x^2 + 8x + 22}}}

Find the vertex. 

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

{{{f(x) = (x^2 + 8x+b^2)-b^2 + 22}}}...complete square

{{{f(x) = (x^2 + 8x+b^2)-b^2 + 22}}}

since coefficients {{{a=1}}} and {{{2ab=8}}}->{{{2*1*b=8}}}->{{{b=4}}}

{{{f(x) = (x + 4)^2-4^2 + 22}}}

{{{f(x) = (x + 4)^2-16 + 22}}}

{{{f(x) = (x + 4)^2 + 6}}}

=> {{{h=-4}}} and {{{k=6}}} and vertex is at ({{{-4}}},{{{6}}})


Answer: ___({{{-4}}},{{{6}}})_____



State the range of the function. 

Answer: ________ { {{{f}}} element {{{R}}} : {{{f>=6}}} }



On what interval is the function increasing? 

Answer: ____({{{-4}}},{{{infinity}}})____



{{{drawing( 600, 600, -10, 10, -5, 10,
circle(-4,6,.12), locate(-4,6,V(-4,6)),
 graph( 600, 600, -10, 10, -5, 10,(x + 4)^2 + 6)) }}}