Question 635819

general form:

{{{f(x)=ax^2+bx+c}}}...where {{{a}}} cannot be zero

you are given:

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

 the {{{x-coordinate}}} of the vertex is {{{-b/2a}}}

 We need only to find that coordinate, and then find the {{{y-coordinate}}} that goes with it by using that value for {{{x}}} in our equation for {{{f(x)}}}.

so, {{{x=-b/2a}}} where {{{a=-1}}} and {{{b=-2}}}

{{{x=-(-2)/2(-1)}}}

{{{x=2/-2}}}

{{{x=-1}}}.....now find {{{y}}}

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

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


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


{{{f(-1)=0}}}


 the vertex is at ({{{-1}}}, {{{0}}})



{{{ graph( 600, 600, -5, 5, -5, 5, -x^2-2x-1) }}}